Introduction - READ THIS FIRST

These are the notes that I write for myself before class. They are meant as reminders,not as a written introduction to the material - that is what the textbook is for! Though Igenerally follow them during class, they may not contain everything that you are expectedto know and may contain some material that I don’t cover during lectures. As such, theyshould not be taken as an exhaustive list of what you are expected to know. They are justmy reminders to myself for class. I do think that they may be helpful as a study guideand reminder of what we’ve discussed. There will be review classes in which I give a morecomplete description of what you are expected to know for exams.

These notes borrow heavily from previous notes by Termeh Kousha and Steve Desjardins.However, I have not taught this class before, and so this document has not been read byanybody before this summer. In contrast, the textbook was worked on for many years bya large team of authors and has been read by many, many people. As a consequence:when my notes disagree with the textbook, trust the textbook! More importantly,Do not rely solely on these notes! Of course, I (and your classmates) also appreciateemails that point out any errors or typos you might notice, and I will work hard to ensurethat any mistakes are corrected quickly.

Finally, the material covered in a given lecture will not correspond perfectly to the lecturenotes for that lecture. In particular, I’ll take more time on a subject if it seems to beconfusing and shuffle material to the next class, but will generally not update these notes toreflect the shuffling. This means that I normally ‘fall behind’ the lecture notes throughoutthe term. You shouldn’t worry about this - it is built into the schedule. However, you shouldcontact your classmates for an overview of what was actually covered if you miss a class.

Guide to Funny Notation. I keep a short guide to things in my lecture notes that mightbe confusing at first. If something confused you, please let me know and I may add it.

(1) I put a question mark (?) into an equation if it is something that will be presentedas a “reasonable guess” or a common error in class. Normally, these equations arenot quite right. For example, if this were a pre-calculus class, I might write

(x+ y)2?? =???x2 + y2.

You know that this equation isn’t correct - but I’m sure we’ve all been tempted towrite it down at some point.

(2) I have some review quizzes in the middle of most classes. If you miss class, you canobtain solutions in office hours.

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Rough Course Schedule

This is the planned course schedule. The midterm and homework dates are fixed; thesubjects covered on any given day may drift a little depending on class interest

• July 4: Introduction; Rate of change; Precalculus Review.• July 7: Limits; Continuity (most of chapter 1 of textbook).• July 11: Derivatives: Definitions, Polynomials, Chain rule and Quotient Rule (chap-

ter 2 of textbook).• July 14: Homework 1 Due. Derivatives: Applications to Curve Sketching (part

of chapter 3 of textbook). Midterm 1 Review.• July 18: Midterm 1. More curve sketching. Applications to Optimization Prob-

lems (remainder of chapter 3 of textbook).• July 21: Derivatives: Inverse Functions; Exponential, Logarithmic and Other Spe-

cial Functions; Start of trigonometric functions (most of chapter 5 of textbook; someof chapter 4).• July 25: Homework 2 Due. Derivatives: Trigonometric Functions (remainder of

chapter 4 of textbook). Midterm 2 review.• July 28: Midterm 2. Vectors: Vectors: Introduction and Basic Properties (start

of chapter 6 of textbook).• August 1: Municipal Holiday.• August 4: Vectors: Dot and Cross Products (chapter 7 of textbook).• August 8: Homework 3 Due. Lines and Planes (half of chapter 8 of textbook).• August 11: Intersections of Lines and Planes (remainder of chapter 8 of textbook).• August 15: Final Exam Review.

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Lecture 1: July 4

Administrative Details and Introduction to the Course.

• If you don’t know anybody in this class, please sit in the front-right and say hello toyour neighbours! We’ll wait a few minutes for everybody to chat, and for latecomersto find the classroom.• Textbook: Calculus and Vectors 12, published by McGraw-Hill Ryerson.• Website: go to aix1.uottawa.ca/∼asmi28. The syllabus is there, and this is my

primary means of communicating with the class.• Office hours: 585 KED, office 201G. Monday and Friday from 9:00 AM to 10:30 AM.• Evaluation will be based on homework sets (15%), midterms on July 18 and July 28

(35%) and a final exam (50%).• The first homework set is posted on the website, and is due on July 14. All home-

work is due by the start of class on its due date.• Homework and exam solutions must contain your work and explanations, not just

your final answer. If you aren’t sure if your explanations ‘count,’ I would be happyto read a few that aren’t homework problems. I would also suggest sharing workedout problems with friends and seeing if they can understand what you’ve written.• You are encouraged to ask for worked solutions to relevant questions that are not on

the homework; I am happy to give them and add them to this document.

Course Overview. In this course, we will learn two important techniques: differentialcalculus and vectors and planes. Like the algebra that you learned in high school, both ofthese are pretty broadly useful ideas, and you will use them if you study any of physics,engineering, economics, statistics, etc. To mention a few applications, we’ll use calculus tohelp us plot complicated functions by hand and get a good understanding of how to go froman equation to a picture. We’ll use vectors and planes to solve equations that are similar to,but more complicated than, those you saw in high school

On Taking Math Courses at University. Math courses at university are very differentfrom those in high school, even when they are teaching the same material (as this one does).Most people can make the transition if they realize that they have to change their habits,but some people don’t realize this until later. Thus, I want to scare you a little bit rightnow.

The most obvious difference between university and high school is the pace of the course:in high school, you would learn this material over roughly 40 weeks of classes; here youwill learn the same material over 6. This means that you will have to pick up the materialroughly 7 times more quickly, and that you should be doing about 7 times as much studyingper week as you would in high school. 7 times is a lot!

Actually, for many people, this understates the difficulty of the transition. In high school,it is fine to learn by listening to the lectures, trying the homework, and only correctingmisunderstandings once you’ve received a marked homework set back. You have plenty oftime to notice your mistakes and correct your understanding. In this compressed course,that strategy will be a recipe for failure. Let’s game this out by assuming you’re unluckyand misunderstand something in the lecture next week on July 12. This might cause you tomake a mistake on the homework due on July 26. You’d then receive the marked homeworkon July 28, and could come talk to me in office hours on July 29. That is, if you follow this

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passive strategy, you would correct your mistake 17 days after making it. That’s fine in ahigh school course, but here it is roughly 40 percent of the term!

There is a document on the course website that gives much more advice about how tomake the transition to university courses. If you want to get a good mark in this course, Istrongly encourage you to read the first few pages as soon as you can. It won’t teach youmath, but it has good ideas about how to study.

With that scary stuff out of the way, some encouragement: with the glaring and unfortu-nate exception of the second lecture, most people find the material in this course to be prettystraightforward. Unlike almost all other math courses in university, there are only a smallnumber of ‘types’ of questions that will show up on exams, and each and every question onany exam will show up (with numbers changed a little) in both the lecture notes and in aDGD. Thus, there is a pretty limited body of questions that you have to be able to do, andyou will have excellent opportunities to test your own understanding.

Algebra Review. In this course, we expect you to be comfortable with the algebra prereq-uisites given in the course description. I’ll give a review this class. If it is at all confusing, Istrongly encourage you to go over the longer review document that is on my website.

Sets. We let R be the real numbers and let Z be the integers. We are often interested inintervals:

• [100, 300] means all the real numbers that are between 100 and 300 inclusively.• [100, 300) means all the real numbers that are between 100 and 300 including 100

but not including 300.• (100, 300) means all the real numbers that are strictly between 100 and 300.• [100,∞) means all the real numbers that are greater than or equal to 100.

Note: ∞ is not a number and so is never included in an interval.Pictorially:

Example 1. Draw a diagram of the intervals

(1) (−∞, 12](2) [−7,−3)

We have more general notation for sets. For finite sets, we just enumerate items:

S = {1, 3, 4, 12}.

We sometimes write infinite sets this way if the pattern is obvious:

S = {2, 4, 6, 8, . . .},but this can be misleading. For this reason, we have some more general notation that allowsus to avoid these confusing situations. This more general notation looks a lot like:

S = {x ∈ A : some condition on x}.For example, we might write

S1 = {x ∈ R : x > 5} = (5,∞)

S2 = {x ∈ R : x2 < 4} = (−2, 2)4

S3 = {x ∈ N : x is prime.} = {2, 3, 5, 7, 11, . . .}.

Using this sort of notation, we might write {2, 4, 6, 8, . . .} as

S = {2x : x ∈ N}.There are some very common ways to combine sets that get special notation:

S ∩ U = {x : x ∈ S and x ∈ U}S ∪ U = {x : x ∈ S or x ∈ U}

For example,

Example 2. Let S = {1, 2, 3, 5} and U = {1, 2, 12}. Calculate S ∪ U and S ∩ U :

S ∪ U = {1, 2, 3, 5, 12}S ∩ U = {1, 2}.

Polynomials. Tools:Quadratic Formula for solving ax2 + bx+ c = 0:

x =−b±

√b2 − 4ac

2a

Difference of Squares for factoring x2 − a2:x2 − a2 = (x− a)(x+ a)

Example 3. Solve for x.

−x2 + 15x− 20 = 3x− 44

Solving Inequalities.

Example 4. The price per unit, p, of a product is related to the demand, x, for the productby

p = 120− 0.03x.

Given that price and demand are both nonnegative, what values of p and x are possible?We would like to find all pairs (p, x) that satisfy both:

p = 120− 0.03x ≥ 0

x ≥ 0.

To solve this, we notice that p = p(x) is a decreasing function of x. Thus, the allowed valuesof x form a single interval, so it is enough to solve the equation p = 0. That equation issimple:

{0 = 120− 0.03x} =⇒ {x = 4000}.Thus, the solutions are

S = {(120− 0.03x, x) : x ∈ [0, 4000]}.5

Example 5. The profit, P , of producing a product is related to the price per unit, u, by

P = −15u2 + 105u− 150.

For what values of u is the profit positive?Notice that, unlike the previous example, P is not a decreasing or increasing function of

u! Nonetheless, we start by solving P (u) = 0:

u =−105±

√(105)2 − 4(−15)(−150)

30

=−105± 45

−30.

Thus, the solutions are {2, 5}. However, we aren’t don’t yet - how do these solutions relateto our inequality?

If we draw P (u), we notice that P (u) < 0 for u very large or very small. Thus, we have

{u : P (u) ≥ 0} = [2, 5].

Absolute Value. If a is a real number, then the absolute value of a is defined as

|a| ={

a if a ≥ 0−a if a ≤ 0

The absolute value of a number is like the “positive part” or “magnitude” of the number.If c ≥ 0 is a positive constant, then

|x− a| ≤ c ⇔ −c ≤ x− a ≤ c⇔ a− c ≤ x ≤ a+ c

The other inequality |x− a| ≥ c has two solutions:

x− a ≥ c or x− a ≤ −c

Examples

(1) Solve |25− x| ≥ 20.solution: By the second remark above, we have that either

25− x ≥ 20 or 25− x ≤ −20−x ≥ −5 −x ≤ −45

x ≤ 5 x ≥ 45

The set of solutions is represented below:

� -x x

5 45

6

(2) Solve∣∣1− 2x

3

∣∣ < 1.solution: ∣∣∣∣1− 2x

3

∣∣∣∣ < 1

−1 < 1− 2

3x < 1

−2 < −2

3x < 0

3 > x > 0

Thus the set of solutions is the interval (0, 3) and is represented as below:

h h

0 3Exponents. Rules and Notation:

(1) xn = x× x× . . .× x(2) x0 = 1(3) x−n = 1

xn

(4) x1/n = n√x

(5) xnxm = xn+m

(6) xn

xm= xn−m

(7) (xy)n = xnyn

(8)(xy

)n= xn

yn

(9) (xn)m = xnm

Example: Simplify (x5√x

3√x7

)−2Fractions and Rationalization. To rationalize an expression,

(1) If√a is in the denominator, multiply by

√a√a.

(2) If√a−√b is in the denominator, multiply by

√a+√b√

a+√b.

(3) If√a+√b is in the denominator, multiply by

√a−√b√

a−√b.

Note that in each case the term that we are multiplying by is equal to 1, therefore we arenot changing the expression, just altering its appearance.

Examples:

(1) Simplify 136+√10

.

solution:

13

6 +√

10=

13

6 +√

10·

(6−√

10

6−√

10

)=

13(6−√

10)

62 −√

102

7

=13(6−

√10)

26=

(6−√

10)

2.

(2) Simplify (x−1)√x+x

.

solution:(x− 1)√x+ x

=(x− 1)√x+ x

·(√

x− x√x− x

)=

(x− 1)(√x− x)

√x2 − x2

=(x− 1)(

√x− x)

x− x2= · · ·

· · · = (x− 1)(√x− x)

x(x− 1)=

√x− x−x

=x−√x

x= 1−

√x

x= 1− x−1/2.

Rational Functions. If P (x), Q(x) are polynomials, then f(x) = P (x)Q(x)

is called a rational

function. In this class, you will need to be able to simplify rational functions - that is, removecommon factors from P (x) and Q(x):

Example 6. Simplify the following functions:

f(x) =(x− 1)(x− 2)

(x− 1)(x− 3)

g(x) =x2 − 2x− 8

x2 − 16.

The first is easy:

f(x) =x− 2

x− 3.

To do the second, we must factor the numerator and denominator. By the quadratic formula,the solution to x2 − 2x− 8 = 0 are

x =2±√

4 + 32

2= 1± 3.

Thus,

x2 − 2x− 8 = (x− 4)(x+ 2).

Similarly,

x2 − 16 = (x− 4)(x+ 4).

Thus, we can write:

g(x) =(x− 4)(x+ 2)

(x− 4)(x+ 4)

=x+ 2

x+ 4.

Example 7 (Short Recap Quiz). (1) Solve x2 − 2x ≥ 1.(2) Solve |x− 5| ≥ |x|+ 1.

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(3) Simplify x2−6x+9x2−9 .

Functions. A function is a mathematical relationship between two variables (called the in-dependent variable and the dependent variable) such that each value of the independentvariable corresponds to a unique value of the dependent variable.

The domain of a function is the set of all the possible allowed values of the independentvariable.

The range of a function is the set of all the values of the dependent variable which corre-spond to some value of the independent variable in the domain.

Often mathematicians will use x to denote the independent variable, y to denote the de-pendent variable, and f to denote the function so y = f(x).

Question: Where do functions come from?

They come from our analysis of the perceived relationship between two variables (such asprofit and number of units sold). When a certain relationship is simple and well-studied,we may be able to find a function that exactly describes the relationship between the twovariables. If the relationship is complex, then statisticians can collect data on the relationshipand create a function that seems to be an acceptable fit to the data. There is often a tradeoffbetween the simplicity and the accuracy of the function.

Example 8. The profit, P , of producing x units of a product is given by

P (x) = 400x−√

40x− 1000− 60000.

(1) What is the domain of this function?(2) What is the profit when 275 units are produced?

To find the domain, we note:

• We can only produce a nonnegative number of units. Thus, x ≥ 0.• The term

√40x− 1000 only makes sense if x ≥ 25.

• The profit is negative for some values of x - but that is fine!

Thus, the best domain is [25,∞).When 100 units are produced,

P (100) = (400)(275)−√

11000− 1000− 60000

= 49900.

Example 9. Find the domain of the function

f(x) =x2

x− 1

This function makes sense when x 6= 1.

Composite Functions. Given two functions f and g, the composite, f ◦ g, is a new functionwhose values are f(g(x)).

Example 10. Let f(x) = 1 + x2 and g(x) = 2x− 1. Find f ◦ g and g ◦ f .We have

(f ◦ g)(x) = f(g(x)) = (1 + (2x− 1)2)9

and

(g ◦ f)(x) = g(f(x)) = 2(1 + x2)− 1.

Example 11. The profit, P (x), of selling x units of a product is given by

P (x) = 5x−√

4x− 100− 6000.

If the number of units sold, x, depends on the price per unit, q as

x(q) =50000

q,

find P as a function of q.We have

P (x(q)) = 550000

q−√

450000

q− 100− 6000.

Inverses. Two functions f and g are inverses if f(g(x)) = x and g(f(x)) = x for all x in thedomain of g and f respectively. Often the inverse of a function f is denoted f−1.

(Roughly speaking, f−1 undoes what f did to x.)Ex: f(x) = x3 and g(x) = x1/3 are inverses.Ex: f(x) = x2 has no inverse but f(x) = x2, x ≥ 0 has inverse f−1(x) =

√x.

We can find the inverse by solving for the independent variable and then appropriatelyrenaming the variables.

Example 12. Question: Find the inverse of f(x) = 2x−1x+7

.

Answer: We solve y = 2x−1x+7

for x:

y =2x− 1

x+ 7,

so

2x− 1 = xy + 7y,

so

x(2− y) = 7y + 1.

We conclude

x =7y + 1

2− y,

so

f−1(x) =7x+ 1

2− x.

Graphs. Graphs give a pictorial representation of the relationship described by a function.You should be familiar with the graphs of

1) Linear Functions: y = mx (lines)2) Quadratic Functions: y = x2 (parabolas)3) Cubic Functions: y = x3

4) Square Root Functions: y =√x

5) Absolute Value Functions: y = |x|6) Hyperbolic Functions: y = 1/x

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• Replacing x with x− a moves the graph a units to the right.• Replacing y with y − b moves the graph b units up.

Example 13. Question: Graph y = |x− 3| − 4.Answer: Rewriting as y + 4 = |x − 3| we see that this is like y = |x| but shifted 3 units

to the right and 4 units down. In class, we draw this on the board.

Points of Intersection. x-intercepts: points at which the graph crosses the x-axis.→ We find them by solving f(x) = 0.y-intercepts: points at which the graph crosses the y-axis.→ We find them by substituting x = 0.

Example 14. Question: Find the x- and y-intercepts of y = |x− 3| − 4.Answer: When y = 0, we have |x − 3| = 4, so x ∈ {−1, 7}. Thus, the y-intercepts are{−1, 7}.

When x = 0, we have y = | − 3| − 4 = −1, so the x-intercept is −1.

The break-even point is a number of units that must be sold for the total costs to be equalto the total revenue (i.e. the profit is 0).

Example 15. Producing CDs requires an initial investment of $1980 for studio time andthen $1.80 for each CD made. We sell the CDs for $9 each. Sketch the graphs of the costand revenue functions on the same axes. What is the break-even point?

Lines. A linear equation is an equation that can be written in the form y = mx+ b (slope-y-intercept form) where m is the slope of the line and b is the y-intercept.

Given two points (x1, y1) and (x2, y2) on the line we can calculate the slope.

m =y2 − y1x2 − x1

The slope is how much y changes for each 1 unit increase in x.Given a point (x1, y1) on the line and the slope m we can find the linear equation using

y − y1 = m(x− x1).

Example 16. Suppose that the relationship between the demand, x, for a product and theprice per unit, p is linear. If we can sell 1000 units at $20 each and 3000 units at $15 each,find the demand function.

Using the point-intercept form, we have

(y − 20) =15− 20

3000− 1000(x− 1000).

Trigonometric Functions. Recall the definitions of the trigonometric functions from a right-angled triangle.

cos θ =adjacent

hypotenuse

sin θ =opposite

hypotenuse

tan θ =opposite

adjacent11

sec θ =1

cos θ=hypotenuse

adjacent

csc θ =1

sin θ=hypotenuse

opposite

cot θ =1

tan θ=

cos θ

sin θ=adjacent

opposite

Or, from the unit circle, x2 + y2 = 1.

x = cos θ and y = sin θ.

The graph of sin(x) looks like:

You may recall from highschool that there are very many trig formulas. We don’t focuson them in this course, but you should be familiar with them. The most important for this

12

course are:

cos(x) = sin(x+π

2)

sin2(x) + cos2(x) = 1

sin(x+ y) = sin(x) cos(y) + cos(x) sin(y)

cos(x+ y) = sin(x) cos(y)− cos(x) sin(y)

sin2(x) =1

2(1− cos(2x))

cos2(x) =1

2(1 + cos(2x))

sin(x) + sin(y) = 2 sin(x+ y

2) cos(

x− y2

).

Calculus Preview. We’ll talk about the rate of change, which motivated the developmentof calculus. The point of studying differential calculus is to study the relationship betweenthe value of a function and how quickly that value is changing. I think this is normally easiestto understand by looking at the relationship between the position of an object and its speed.

Definition 0.1 (Average Speed). Let f(x) be some function. Its average speed over theinterval [a, b] is given by

v =f(b)− f(a)

b− a.

Example 17 (Car Travel). Question: I travel from Ottawa to Toronto, a distance of 450kilometers, in 5 hours. What was my average speed?

Answer: Let d(t) be the distance traveled after t hours. We calculate

v =d(5)− d(0)

5− 0=

450− 0

5− 0= 90.

This seems like a very sensible way to measure speed for this example.

We give another example, illustrating the fact that an object’s average speed may notalways accurately reflect how quickly it is moving at any given moment.

Example 18 (Height of a Spring). We have a weight bobbing on the end of a spring. Theheight of that spring is given by the formula

h(t) = 4 + cos(πt),

where h is measured in feet and t is measured in seconds.1

How quickly is the spring moving? For our current definition, we need to say both thestarting time and the end time. For starting t = 0, we get the following speeds:

We can see that the answer depends a huge amount on the time interval that we look at!However, it seems that small time intervals agree with our intuition pretty well: the weight”isn’t moving” when we let go of it at time t = 0, and the ”average speed” gets close to 0 asthe interval becomes small.

1Note to physicists - I’m mostly choosing constants to make the equations easy to write down. They arenormally not realistic.

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Table 1. Speed

Interval Speed

(0, 200) 0(0, 2) 0(0, 1) -2

(0, 0.75) -2.28(0, 0.5) -2(0, 0.25) -1.17(0, 0.1) -0.49(0, 0.01) -0.049(0, 0.001) -0.0049

We can try to look at the rate of change using equations:

Example 19. Question: A ball falls from a window that is 50 meters up from the ground.For the first few seconds of its fall, its height in meters is given by the formula h(s) =50−9.8s2. Give a formula for the average speed of the ball over the time interval [1.5, 1.5+h].

Answer: We plug into our earlier formula:

v =h(1.5)− h(1.5 + h)

1.5 + h− 1.5

= 9.8(1.5 + h)2 − (1.5)2

h

= 9.83h+ h2

h= 29.4 + 9.8h.

So, that was an easy algebra exercise. However, we’ll take a second to notice some thingsabout this formula that seem important:

• Our ball is accelerating, so the longer the interval h, the higher the average speed.• For h very small, the speed is basically 29.4. This is suggestive that maybe the ‘instan-

taneous’ speed should be 29.4, even though we haven’t really defined ‘instantaneous’speed yet.• In class: We draw h, then draw the line with slope 29.4 passing throughthe point (1.5, h(1.5)). We point out that this line seems to ‘kiss’ h. Sucha line is called a tangent, and will be very important.

None of these things are coincidences, and we’ll get back to this soon.

Making this careful needs some more definitions. Unfortunately, the very first definition,the limit, is also the most difficult definition in the entire course. We’ll get to this next class.

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Additional Worked Problems

Recap of Techniques and Problems. In our first class, we saw:

(1) a review of prerequisite subjects, and(2) an introduction to rate-of-change problems.

Short Quiz. Let’s try a few questions:

• A car’s position at time s is given by x(s) = 2s + 3. Calculate the average rate ofchange over the interval (1, 2).• Let f(x) = x2 − 4x+ 8. Find {x : f(x) ≥ 2}.• Find {x : |x− 1| ≥ |x+ 1|}.• Simplify x2−2x+1

x2−1 .

Further Examples.

Example 20 (Rate of Change Problem). The position of a weight on a spring is given byx(s) = 2 + cos(s)e−s. Calculate the average rate of change over the intervals (0, π

2) and

(0, 0.1).Plugging into a calculator,

v1 =x(π

2)− x(0)π2

= − 2

π

and

v1 =x(0.1)− x(0)

0.1− 0≈ −6.34.

Example 21 (Inequalities 1). Find all values of x ≥ 1 for which x2 − 30x+ 125 ≥ 0.By the quadratic formula,

x =30±

√900− 500

2= 15± 10.

Thus, the solutions to x2 − 30x+ 125 = 0 are x ∈ {5, 25}. Drawing the picture, we have

{x : x2 − 30x+ 125 ≥ 0} = (−∞, 5] ∩ [25,∞).

Thus, the solution is [1, 5] ∩ [25,∞).

Example 22 (Inequalities 1). Find all values of x ≤ 10 for which |x− 1| > |x− 5|.We break the line into three regions:

(1) x ≤ 1: In this case, |x− 1| = 1− x and |x− 5| = 5− x, so

|x− 1| − |x− 5| = (1− x)− (5− x) = −4 < 0.

This is never positive.(2) 1 < x < 5: In this case, |x− 1| = x− 1 and |x− 5| = 5− x, so

|x− 1| − |x− 5| = x− 1− (5− x)

= 2x− 6.

This is positive for 3 ≤ x < 5.15

(3) x ≥ 5: In this case, |x− 1| = x− 1 and |x− 5| = x− 5, so

|x− 1| − |x− 5| = x− 1− (x− 5) = 4 > 0.

This is always positive.

Putting these together, the solution is:

∅ ∪ [3, 5) ∪ [5,∞) = [3,∞).

Example 23 (Simplifying Functions). Simplify f(x) = x2+2x−35x2+14x+49

.We write

f(x) =(x− 5)(x+ 7)

(x+ 7)2

=x− 5

x+ 7.

Example 24 (Inverses). Find the inverse of f(x) = x+52x−1 .

We solve:

y =x+ 5

2x− 1so

x+ 5 = 2xy − yso

x(1− 2y) = −y − 5

so

x =y + 5

2y − 1.

Example 25 (Plotting). Plot f(x) =√x+ 5 and g(x) = cos2(x).

Example 26 (Lines). We have y = mx + b for some unknown m, b. You know that (2, 10)and (5, 12) are on the line. Calculate m and b.

16

Lecture 2: July 7

Review Quiz!

(1) Let f(x) = −2x2 + 45x− 1. Find the values of x for which f(x) ≥ 0.

(2) Simplify (x2)5 and x25.

(3) Simplify g(x) = 21+√x. What could the domain of this function be? The range?

(4) Find the inverse of h(x) = x+ 7.(5) Graph `(x) = |x− 1|+ 1.

Solutions will be briefly discussed in class. Since this is meant as a reviewquiz, they will not be included in the lecture notes.

Limits and Continuity. I’ll begin with a vague definition to get our intuition going. Thisis the one that you’ll probably think about the most often:

Definition 0.2 (Limit - Vague Definition). Let f be a function. If f(x) gets closer andcloser to some number L as x gets close to y, we write

limx→y

f(x) = L.

Note, we allow L to be ∞ or −∞ if f just gets bigger and bigger or smaller and smaller.When f doesn’t get close to some value, we say that the limit does not exist.

In class: We draw a picture of a continuous function, and see what the limitis.

Example 27 (Types of Limits). By drawing pictures in class, we can see the followingintuitive examples:

limx→0

x = 0

limx→0

1

x2=∞

limx→0

sin(1

x) = DNE.

Remark 0.3 (Plug-In Principle). For “nice” functions, limx→y f(x) = f(y). Functions thatsatisfy this ‘plug-in’ principle are called continuous functions. In practice, when computinglimits, you will mostly calculate them using this idea.

Example 28. Question: Calculate limx→2

√x+ 7.

Answer: We draw the function√x+ 7. It seems nice, so the plug-in principle suggests

that the limit is√

2 + 7 = 3.

Sometimes, the plug-in principle can’t give us a nice answer:In this case, there is no limit. Even worse, the sometimes principle can sometimes be

misleading. That is, it can give you an answer that is wrong. Consider:

f(x) =

{x2 + 1 x 6= 0

−1 x = 0.17

We have f(0) = −1, but by drawing a picture, limx→0 = 1. We need to write down acareful definition for the limit exactly because the plug-in principle can give you the wronganswer.

The textbook introduces some other definitions. I’ll repeat them here, but want to warnyou that they are also a little bit vague. I’ll then give a careful definition.

Definition 0.4 (Left and Right Limits). Let f be a function. If f(x) gets closer and closerto some number L as x gets close to y from the left, we say that L is the left-limit of f at xand write

limx→y−

f(x) = L.

If f(x) gets closer and closer to some number R as x gets close to y from the right, we saythat R is the right-limit of f at x and write

limx→y+

f(x) = R.

Sometimes, we can have left-limits and right-limits without having limits:

Other strange stuff can happen:18

These types of discontinuities all have special names, which you can find in your textbook.I will not use these names on exams, and in my experience they don’t show up very often inother classes.

19

Definition 0.5 (Limit - Another Vague Definition). If

limx→y−

f(x) = L = R = limx→y+

f(x),

then we say that the limit of f at x is L and write

limx→y

f(x) = L.

Example 29. Question: Consider the function

f(x) =

{x2+5x+1

0 ≤ x ≤ 5

x+ 4 x > 5.

Does f have a left limit at 5? A right limit? A limit?Answer: We can calculate the left and right limits by the plug-in principle. They are

L =52 + 5

6= 5

R = 5 + 4 = 9.

Thus, f has left and right limits at x = 5 of 5 and 9. Since these are not equal, f does nothave a limit at x = 5.

You might ask: what is wrong with this definition? Mostly, if the function f is reallycomplicated, it can be a little hard to tell what we mean by f(x) getting ‘closer and closer’to some number L. For example, does sin( 1

x) have a limit as x goes to 0? What about

x sin( 1x)? What do we do about functions that we have a hard time drawing, like 1

x2e−

1x or

(cos(x)2−1)5sin(x)

?

For this reason, we need a precise definition that makes things very careful. Unfortunately,this comes at the cost of being much harder for most people to read at first glance:

Definition 0.6 (Limit - Precise Definition). Let f be a function. We say that

limx→y

f(x) = L

if, for all ε > 0, there exists some δ > 0 so that

{|x− y| < δ} =⇒ {|f(x)− L| < ε}.Example 30 (The ε-δ game). The definition of a limit has a lot of qualifiers in it. If you’renot used to logic, the sentence can be quite hard to parse. One way to do this is to imaginethis as a game.

In the game, you are trying to prove that limx→y = L; your opponent wants to check this.Your opponent does this by choosing any value of ε; you must then produce some value of δso that the desired inequality is true. The order matters here! Your opponent chooses εfirst, and you get to respond. If you had to choose δ first, this game would be impossible!

Example 31 (An ε-δ Proof). Question: Prove that limx→2 5x = 10.Answer: First, let’s draw a picture and notice that the conclusion is basically obvious!How do we prove this? Before we actually prove the statement, let’s play the game. Imag-

ine my opponent gives me some ε > 0. What should δ be? Well, we can look at the pictureof f(x) = 5x. Algebraically, we want:

ε > |f(x)− 10|20

= |5x− 10|= 5|x− 2|.

Thus, we need |x − 2| < ε5. Thus, we should choose δ < ε

5. For example, we might choose

δ = ε10

.We now check that this choice of δ actually works. If |x− 2| < ε

10, then

|f(x)− 10| = 5|x− 2| < 5ε

10=ε

2< ε.

So, this worked!

Remarks 0.7. The proof itself is very short!I promise that there will be no ε-δ proofs on any exam or midterm in this course. However,

this is the ‘right’ definition. If you’re interested in any subjects that involve a bunch of math,like physics, statistics, computer science, or much of economics, you will have to understandwhat is going on here. If you understand the definition, this can also help you figure outthe answers to some questions in this class, where our various rules-of-thumb might giveconflicting advice.

In addition to our definition of limits, limits have a lot of nice formulas. The following areall true if we assume that all of the limits involved exist:

limx→y

(f(x) + g(x)) = limx→y

f(x) + limx→y

g(x)

limx→y

(f(x)g(x)) = (limx→y

f(x))(limx→y

g(x))

There’s another formula that we’d like to be true, but which is false (and can lead to trickyexam questions):

limx→y

f(g(x))??? =???f(limx→y

g(x)).

That formula is true for almost every function that you might write down, but is not truein general!

Let’s use these formulas to do a question that seems hard at first:

Example 32. Question: Let f(x) satisfy

limx→4

f(x)− 5

(x− 3)7= 6.

limx→3

f(x)− 5

(x− 3)7= 1.

What is limx→4 f(x)? What is limx→3 f(x)?Answer: We calculate

limx→4

f(x) = 5 + limx→4

(f(x)− 5)

= 5 + limx→4

f(x)− 5

(x− 3)7limx→4

(x− 3)7

= 5 + (6)(1) = 11.21

We calculate

limx→3

f(x) = 5 + limx→3

(f(x)− 5)

= 5 + limx→3

f(x)− 5

(x− 3)7limx→3

(x− 3)7

= 5 + (1)(0) = 5.

Note: Zero is ok!

Note that we can’t ignore the assumption that all of the limits exist when using theseformulas!

Example 33. Question: Calculate limx→0 x−2x3, limx→0 x

−3x2 and limx→0 x−2x2.

Answer: Doing this directly is easy:

limx→0

x−2x3 = limx→0

x = 0

limx→0

x−3x2 = limx→0

x−1 = DNE

limx→0

x−2x2 = limx→0

1 = 1.

Using the formulas, however, gives nonsense:

limx→0

x−2x3? = limx→0

x−2 limx→0

x3 = DNE?0?

limx→0

x−3x2? = limx→0

x−3 limx→0

x2 = DNE?0?

limx→0

x−2x2? = limx→0

x−2 limx→0

x2 = DNE?0?.

Having defined the limit, we’ve gotten past the hardest part of the course. Our nextdefinition builds on the limit:

Definition 0.8 (Continuous Function). We say that a function f is continuous at x if

limy→x

f(y) = f(x).

We say that a function f is continuous if it is continuous at x for all x in its domain.Roughly speaking, a function is continuous if it can be drawn without ever lifting your penfrom your paper.

Remember that we had some formulas for limits:

limx→y

(f(x) + g(x)) = limx→y

f(x) + limx→y

g(x)

limx→y

(f(x)g(x)) = (limx→y

f(x))(limx→y

g(x)).

We wanted this formula to be true

limx→y

f(g(x)) = f(limx→y

g(x)),

22

but it wasn’t in general. The missing ingredient is continuity : the formula is true if f and gare continuous functions.2 Let’s apply our newest formula:

Example 34. Question: Given f(5) = 3 for some continuous function f , calculatelimx→25 f(

√x).

Answer: We have

limx→25

f(√x) = f(lim

x→5

√x) = f(5) = 3.

Example 35 (Short Recap Quiz). (1) Calculate limx→2 x2 + 1.

(2) Let f(x) = x+ 1 for x ≤ 3 and f(x) = x2 − 5 for x > 3. Is f continuous? What doyou have to check?

(3) More advanced: Using the δ-ε definition, prove that limx→1 x2 = 1.

Applications and Calculations for Limits. We do some more examples, getting morecomplicated as we go.

Example 36 (Plug-In Principle). Question: Calculate:

limx→2

(x+ 4)(2x2 + 3) =?

limx→3

x2 + 1

x+ 2=?

Example 37 (Left and Right Limits). Consider the function f(x) =

1 + x x < 2

2 x = 2

x2 x > 2

.

This is an example of a piecewise defined function. Look at limits of f at x = 2.limx→2−

f(x) = limx→2−

(1 + x) = 3,

limx→2+

f(x) = limx→2+

x2 = 4,

so limx→2

f(x) does not exist (so f(x) cannot be continuous at x = 2),

while f(2) = 2,but f(x) has a jump discontinuity at x = 2.

Example 38 (Limits and Asymptotes). Question: Calculate:

limx→2+

1

x− 2=?

limx→2−

1

x− 2=?

limx→2

1

x− 2=?

Answer: In class, we draw a picture of 1x−2. By inspection of the picture,

limx→2+

1

x− 2=∞

2The textbook has many more formulas, but all of them follow from the three that I’ve written downhere.

23

limx→2−

1

x− 2= −∞

limx→2

1

x− 2= DNE.

Example 39 (Limits and Infinities). Question: Calculate:

limx→∞

x =?

limx→∞

x2 + 6

3x2 + 4x− 8=?

Answer: The first is simple: by looking at the picture, it is clearly infinity. To do thesecond, we do what is often called leading term analysis. The idea is to divide through by thebiggest power of x:

limx→∞

x2 + 6

3x2 + 4x− 8= lim

x→∞

1 + 6x−2

3 + 4x−1 − 8x−2

=1

3.

Example 40 (Limits and Indeterminate Forms). Question: Calculate

limx→2

x2 − 3x+ 2

x− 2=?

limx→0

√9 + x− 3

x=?

Answer: These questions look hard! If we try to plug in, we get 00. So, what to do? One

option is to draw the picture, but that is quite hard. Another option is to try to factor theexpression and use our rules:

limx→2

x2 − 3x+ 2

x− 2= lim

x→2

(x− 2)(x− 1)

x− 2= lim

x→2(x− 1) = 1.

In more complicated situations, we must use factorization tricks. In this case, we usea2 − b2 = (a− b)(a+ b):

limx→0

√9 + x− 3

x= lim

x→0

√9 + x− 3

x

√9 + x+ 3√9 + x+ 3

= limx→0

9 + x− 9

x

1√9 + x+ 3

= limx→0

1√9 + x+ 3

=1

6.

Note: This is a nice trick to know for now, but factorization tricks are hard! Later instudying calculus, you will learn a general technique for calculating these sorts of 0

0limits,

24

called L’hopital’s rule. It is much better, because you don’t have to think much when applyingit.

Example 41. Question: Let fC(x) be given by the formula

fC(x) =

{x2+Cx+1

0 ≤ x ≤ 5

2x x > 5.

For what value of C is fC(x) continuous?Answer: The function is clearly continuous everywhere except possibly x = 5. We calcu-

late the left and right limits at 5:

limx↑5

fC(x) =25 + C

5 + 1

limx↓5

fC(x) = 10.

These two limits must be equal if fC is continuous, so we must have

25 + C

6= 10,

so C = 35.

25

Lecture 3: July 11

Review Quiz! Take a few minutes to do the following review questions by yourself. We’lldiscuss the answers after 5 minutes.

Let f(x) = x2−2x+1x2−1 . Calculate

limx→5

f(x) =?

limx→1

f(x) =?

limx→−1+

f(x) =?

limx→−1−

f(x) =?

limx→−1

f(x) =?

In class: we discuss the answers. The key trick is to factor the numerator anddenominator of f .

The Derivative.

Definition 0.9 (Derivative). The derivative of a function f at a point x ∈ R is defined tobe the value of the limit

limh→0

f(x+ h)− f(x)

h

when that limit exists and is finite. When the limit does not exist, or is infinite, we say thatthe derivative of f does not exist at x.

There are two common ways to write the derivative:

d

dxf(x) ≡ f ′(x) ≡ lim

h→0

f(x+ h)− f(x)

h.

Remark 0.10. WARNING: The expression ddx

is NOT really a fraction! It is its ownsymbol, and it just happens to look like a fraction. For example, you CANNOT cancel thed’s and write d

dx= 1

x, or write d

dxx2 = dx2

dx= x. Until you get used to this, it can be helpful

to put a little box around the symbol to remind you that it can’t interact with anything else.Yes, this is annoying. Unfortunately, we’ve been using this notation for over 300 years

and it is probably not going anywhere any time soon.

Remark 0.11. The derivative is the limit as the time interval h goes to 0 of our formulafor the rate of change. For this reason, the derivative is often called the instantaneous rateof change.

Example 42 (Simple Direct Derivatives). Question: Let f(x) = x2 + 2x + 1. Calculatethe derivative of f(x) directly from the definition.Answer: We have

f ′(x) = limh→0

f(x+ h)− f(x)

h

= limh→0

(x+ h)2 + 2(x+ h) + 1− x2 − 2x− 1

h26

= limh→0

2xh+ h2 + 2h

h= lim

h→0(2x+ 2 + h)

= 2x+ 2

Example 43 (Complicated Direct Derivatives). Question: Let f(x) =√x+ 1 on x ∈

(−1,∞). Calculate the derivative of f(x) directly from the definition.Answer: This requires a bit of a trick! We have to remember the factorization (a2−b2) =

(a− b)(a+ b). We calculate:

f ′(x) = limh→0

f(x+ h)− f(x)

h

= limh→0

√x+ h+ 1−

√x+ 1

h

= limh→0

√x+ h+ 1−

√x+ 1

h

√x+ h+ 1 +

√x+ 1√

x+ h+ 1 +√x+ 1

= limh→0

x+ h+ 1− (x+ 1)

h

1√x+ h+ 1 +

√x+ 1

= limh→0

1√x+ h+ 1 +

√x+ 1

=1

2√x+ 1

.

The derivative is related to the plot of the function! We have:

Definition 0.12 (Tangent Line). The tangent line to the function f at the point (x, f(x))is the line that passes through (x, f(x)) and has slope f ′(x).

In pictures, the tangent line sort of ‘kisses’ f at (x, f(x)). In class, we draw somepictures here.

This notion lets us recognize that some functions probably do not have derivatives:

Example 44 (Non-Differentiable Functions). Question: Consider the function

f(x) =

{12− 2x 0 ≤ x ≤ 5

7x− 33 x > 5.

Is f continuous? Differentiable?

Answer: We plot the function. It is easy to see that the function is continuous. To seethat f isn’t differentiable, we notice that the function has slope −2 for 0 ≤ x < 5 and hasslope 7 for x > 5. The slope doesn’t seem to be unique (and there isn’t a unique tangent) atx = 5.

Since the derivative (or instantaneous rate of change or slope of the tangent line) is definedby a limit, we should be careful to realize that it does not have to exist for all x. We say

that f(x) is differentiable at x = a if f ′(a) = limh→0

f(a+ h)− f(a)

hexists. How could f ′(a)

27

fail to exist ? If a is not in the domain of f , f(a) is undefined and f ′(a) cannot exist. If f isdiscontinuous at x = a, f ′(a) will also fail to exist (we can’t draw a tangent to the curve atthat point). But, it is also possible for f(x) to be continuous at x = a and for f ′(a) to fail

to exist. Consider y = f(x) = |x| =

{x x ≥ 0

−x x < 0.

Notice that f(x) is continuous at x = 0 as f(0) = 0 and limx→0−

f(x) = 0 = limx→0+

f(x). But,

if we try to find f ′(0), f ′(0) = limh→0

f(0 + h)− f(0)

h, we’ll have to consider the two one-sided

limits separately as the definition of the function is different for positive and negative values.

limh→0−

f(0 + h)− f(0)

h= lim

h→0−

|0 + h| − |0|h

= limh→0−

|h|h

= limh→0−

−hh

= limh→0−

−1 = −1

And

limh→0+

f(0 + h)− f(0)

h= lim

h→0+

|0 + h| − |0|h

= limh→0+

|h|h

= limh→0+

h

h= lim

h→0+1 = 1.

Thus

limh→0−

f(0 + h)− f(0)

h6= lim

h→0+

f(0 + h)− f(0)

h

and so

f ′(0) = limh→0

f(0 + h)− f(0)

h

does not exist. And hence, f(x) is not differentiable at 0. The graph of y = f(x) = |x| issaid to have a corner at x = 0.

f ′(a) will also fail to exist if there is a cusp or vertical tangent at x = a.

You may have noticed that our definition of the derivative is a little impractical. Whenwe actually compute derivatives, we’ll use a bunch of special formulas. These first formulasare all on page 76 of the textbook. In each, f, g are functions and k is a constant:

d

dxk = 0

d

dxxk = kxk−1

d

dxkf(x) = k

d

dxf(x)

d

dx(f(x) + g(x)) =

d

dxf(x) +

d

dxg(x).

Let’s derive these rules! The first is easy:28

d

dxk = lim

h→0

k − kh

= limh→0

0 = 0.

Next, we derive the second rule when k is an integer (it is also true for all k ∈ R, but theproof is too hard for our class):

d

dxxk = lim

h→0

(x+ h)k − xk

h

= limh→0

h−1(k∑j=0

k!

j!(k − j)!xk−jhj − xk)

= limh→0

(kxk−1 +k∑j=2

k!

j!(k − j)!xk−jhj−1)

= kxk−1 +k∑j=2

k!

j!(k − j)!xk−j lim

h→0hj−1

= kxk−1.

We apply this:

Example 45. Question: Calculate the tanget to the curve f(x) = 2x3 at x = 3.Answer: We have

f ′(x) = 6x2.

Thus,

f ′(3) = (6)(9) = 54

f(3) = (2)(27) = 54,

so the tangent curve is

(y − 54) = 54(x− 3)

in point-slope form.

The next two formulas are easier to check:

d

dxkf(x) = lim

h→0

kf(x+ h)− kf(x)

h

= k limh→0

f(x+ h)− f(x)

h

= kd

dxf(x),

and

d

dx(f(x) + g(x)) = lim

h→0

f(x+ h) + g(x+ h)− f(x)− g(x)

h29

= limh→0

f(x+ h)− f(x)

h+ lim

h→0

g(x+ h)− g(x)

h

=d

dxf(x) +

d

dxg(x).

Let’s use these rules:

Example 46. Question: Let f(x) = 3x4 + 5x3 − 12x+ 1 and calculate f ′(1).Answer: We calculate

d

dx(3x4 + 5x3 − 12x+ 1) = 3

d

dxx4 + 5

d

dxx3 − 12

d

dxx+

d

dx1

= 12x3 + 15x2 − 12.

Thus

f ′(1) = 12 + 15− 12 = 15.

Some more examples:

Example 47. (1)d

dx

(3√x)

=d

dx

(x1/3

)=

1

3x−2/3 =

1

3x2/3

(2)d

dx

(4x2)

= 4d

dx

(x2)

= 4(2x) = 8x

(3) If f(x) = 2x4 − 3x2 + x, then f ′(x) = 2(4x3)− 3(2x) + 1 = 8x3 − 6x+ 1.

(4) If y = 3t2 −√t, then

dy

dt= 6t− 1

2√t.

(Note that the rules are independent of the names given to the variables.)

(5) Find the equation of the tangent line to the curve y = f(x) = 2x3 + 6x2 − 4x + 3 atx = −2.

The slope of the line is m = f ′(−2).f ′(x) = 6x2 + 12x− 4, so m = f ′(−2) = 6(−2)2 + 12(−2)− 4 = −4.The point on the curve has y coordinate f(−2) = 2(−2)3 + 6(−2)2− 4(−2) + 3 = 19.So the tangent line is y − 19 = (−4)(x− (−2)) or y = −4x+ 11.

(6) The height above the ground, in metres, of an object dropped from the top of a build-ing of height 60 m, after t seconds, is h(t) = 60− 4.9t2.(i) How fast is the object falling after 2 s ?(ii) How fast does the object hit the ground ?

(i) The velocity is v(t) = h′(t) = −9.8t, so after 2 s, v(2) = −19.6 m/s (the negativeindicates downward motion).(ii) The object hits the ground when h(t) = 0 or 4.9t2 = 60 or t = 3.5 s, so v(3.5) =34.3 m/s.

The next derivative rule is very powerful (although at the moment, it might look a silly)

Definition 0.13 (Product Rule). Fix any functions f, g, and let h(x) = f(x)g(x). We have30

d

dx(f(x)g(x)) = g(x)

d

dxf(x) + f(x)

d

dxg(x),

or, in our other notation,

h′(x) = f(x)g′(x) + f ′(x)g(x).

Let’s check that this seems plausible:

Example 48. Let f(x) = x3, g(x) = x2, and h(x) = f(x)g(x) = x5. We know h′(x) = 5x4.By the product rule,

h′(x) = f ′(x)g(x) + f(x)g′(x)

= (3x2)(x2) + (x3)(2x)

= 5x4.

So, this works!

Let’s apply the product rule:

Example 49. Question: Let f(x) = (x2 + 1)(x3 + 7x− 1) and calculate f ′(x).

Answer: By the product rule,

f ′(x) = (x3 + 7x− 1)d

dx(x2 + 1) + (x2 + 1)

d

dx(x3 + 7x− 1)

= (x3 + 7x− 1)(2x) + (x2 + 1)(3x2 + 7).

Some more examples:

Example 50. (1)d

dx

((x2 + 1)(x2 + 3x− 2)

)=

(d

dx

(x2 + 1

))(x2 + 3x− 2) + (x2 + 1)

(d

dx

(x2 + 3x− 2

))= (2x)(x2 + 3x− 2) + (x2 + 1)(2x+ 3)= 2x3 + 6x2 − 4x+ 2x3 + 2x+ 3x2 + 3= 4x3 + 9x2 − 2x+ 3(check this by expanding first and then differentiating).

(2) If f(t) = (t+ 1)(3t4 − t2),then f ′(t) = (1)(3t4 − t2) + (t+ 1)(12t3 − 2t)= 3t4 − t2 + 12t4 + 12t3 − 2t2 − 2t= 15t4 + 12t3 − 3t2 − 2t(check this, too).

(3) If y = (2x2 + 1)2 = (2x2 + 1)(2x2 + 1),

thendy

dx= (4x)(2x2 + 1) + (2x2 + 1)(4x)

= 2(2x2 + 1)(4x) (there is a pattern here)= (8x)(2x2 + 1)= 16x3 + 8x.

31

(4) How aboutd

dx

((2x2 + 1)3

)?

d

dx

((2x2 + 1)3

)=

d

dx

((2x2 + 1)2(2x2 + 1)

)= (2(2x2 + 1)(4x)(2x2 + 1) + (2x2 + 1)2(4x))= 3(2x2 + 1)2(4x) (can you see the pattern ?)

Remarks 0.14 (Useful Facts). We’ve already seen a lot of formulas in this class, and therewill be many more. If you’re like me, you probably don’t like memorizing formulas. Analternative is to memorize only a smaller number of more general formulas. For example,we currently know the derivative formulas:

d

dxk = 0

d

dxxk = kxk−1

d

dxkf(x) = k

d

dxf(x)

d

dx(f(x) + g(x)) =

d

dxf(x) +

d

dxg(x)

d

dx(f(x)g(x)) = g(x)

d

dxf(x) + f(x)

d

dxg(x).

However, you don’t need to remember all of them. The first one is a special case of thesecond, and the third is a special case of the fifth. There are some other relationships betweenthem, though the two I mentioned are the easiest.

If you want to understand calculus, you should keep your eye out for these sorts of rela-tionships. I’ll also point some out as we go.

Mid-Class Review Quiz. Calculate the derivatives of:

f(x) = (2x+ 5)(x2 + 3)

g(x) =x+ 4

x− 1

h(x) =√x2 − 1.

Also say the domain and range of the functions and their derivatives. Solutions will bediscussed in class.

In Section 2.3 of the textbook, they take a break to talk about velocity, acceleration andsecond derivatives. Briefly, if f(s) is the position of an object at time s, then

• The derivative g(s) ≡ f ′(s) of f is the instantaneous velocity of the object at time s,and• The second derivative h(s) ≡ g′(s) ≡ f ′′(s) is the instantaneous acceleration of the

object at time s.

We won’t talk about this too much, but I give some more notation for the second derivative:32

Definition 0.15 (Second Derivatives). The second derivative of a function f is the derivativeof its derivative. That is,

d2

dx2f(x) ≡ d

dx(d

dxf(x)).

We often write this as f ′′(x).

The next derivative rule is probably the most important one, and the one that is generallyhardest to use:

Definition 0.16 (Chain Rule). Let f, g be two functions and let h(x) = f(g(x)). Then

h′(x) = f ′(g(x))g′(x).

You will see in the textbook the following alternative way for writing down the chain rule:

Definition 0.17 (Alternative Version of Chain Rule). Let y = f(u) and u = g(x). Then

dy

dx=dy

du

du

dx.

WARNING: This version of the chain rule practically invites you to cancel the du termsand make other weird errors! For that reason, I personally suggest that you do not use thisnotation for the chain rule.

Let’s use the chain rule:

Example 51 (Immediate Application of Chain Rule). Question: Calculate the derivativeof f(x) =

√x+ 5, x ∈ R+.

Answer: We write g(x) = x+ 5 and h(x) =√x, so f(x) = h(g(x)). Thus, by the chain

rule,

f ′(x) = h′(g(x))g′(x) =1

2(x+ 5)−

12 × d

dx(x+ 5) =

1

2(x+ 5)−

12 .

Example 52 (Chain Rule: Direct Application). Question: Let f(x) = x−2, let g(x) =3x2 + 2x+ 1, and let h(x) = f(g(x)). Calculate h′(x).

Answer: By the chain rule,

h′(x) = f ′(g(x)) g′(x)

= −2(3x2 + 2x+ 1)−3(6x+ 2).

Example 53 (Chain Rule and Powers). Question: Let f(x) = (x2 + 3x−4)8 and calculatef ′(x).

Answer: This question is harder than the previous questions: we need to notice that wewant to use the chain rule, and we need to decide on the pieces of the function. How wouldyou notice that you should use the chain rule? What should the functions be?

We’ll solve the rest of this question in class.

Let’s do this again:

Example 54 (Chain Rule and Powers, Part 2). Question: Let f(x) = (2x − 1)3 andcalculate f ′(x).

33

Answer: We’ll do the same thing. Let g(x) = 2x− 1 and h(x) = x3, so f(x) = h(g(x)).By the chain rule,

f ′(x) = h′(g(x))g′(x)

= 3(2x− 1)2 × 2.

As you use the chain rule, you’ll notice a lot of patterns and shortcuts. Let’s find a simpleone, based on the previous two questions:

Example 55. Fix a function f and an integer n. Let g(x) = xn and let h(x) = g(f(x)).Then

h′(x) = g′(f(x))f ′(x) = nf(x)n−1f ′(x).

You can either memorize this formula, or remember how to do the previous two examples.More likely, you’ll do a bit of both. Let’s apply the formula:

Example 56 (Chain and Product Rules Together). Question: Let f(x) = (x2 +2)5(x+1).Calculate f ′(x).

Answer: Combining the chain and product rules,

f ′(x) = (x+ 1)(5)(x2 + 2)4 × (2x) + (x2 + 2)5 × 1.

Example 57. Question: Let f(x) =√x2 + 1. Calculate f ′(x).

Answer: This is the same as the previous examples:

f ′(x) =1

2(x2 + 1)−

12 × (2x).

We will get another important formula from the chain rule and the product rule, whichwill be called the quotient rule. First, we do an example by hand:

Example 58. Question: Let r(x) = x2+13x+4

. Calculate r′(x).Answer: Like other difficult chain rule questions, we have to decide on our functions.

We let f(x) = x2 + 1, let g(x) = 3x+ 4, and let h(x) = x−1, so that

r(x) = f(x)h(g(x)).

By the product rule and then the chain rule,

r′(x) = f(x)d

dxh(g(x)) + f ′(x)h(g(x))

= f(x)h′(g(x))g′(x) + f ′(x)h(g(x))

= (x2 + 1)(−1)(3x+ 4)−2 + (2x)(3x+ 4)−1.

Phew!

That was pretty inconvenient, wasn’t it? We can use a formula that summarizes the ideaof this example:

Definition 0.18 (Quotient Rule). Fix functions f, g and let h(x) = f(x)g(x)

. Then

h′(x) =f ′(x)g(x)− f(x)g′(x)

g(x)2.

34

We use the quotient rule:

Example 59. Question: Calculate f ′(x), g′(x) and h′(x), where

f(x) =x2 − 2x+ 1

x2 − 1

g(x) =

√x4 + 16

2x+ 1

h(x) =x+ 7

2x+ 1

Answer: Applying the quotient rule,

f ′(x) =(2x− 2)(x2 − 1)− (x2 − 2x+ 1)(2x)

(x2 − 1)2

g′(x) =12(x4 + 16)−

12 (4x3)(2x+ 1)− 2

√x4 + 16

(2x+ 1)2

h′(x) =(2x+ 1)− 2(x+ 7)

(2x+ 1)2.

Some more applications:

Example 60. (1)d

dx

(2x+ 1

3x+ 2

)=

(ddx

(2x+ 1))

(3x+ 2)− (2x+ 1)(ddx

(3x+ 2))

(3x+ 2)2

=(2)(3x+ 2)− (2x+ 1)(3)

(3x+ 2)2

=1

(3x+ 2)2

(2) If f(x) =

√x+ 1

2x2 + x, then

f ′(x) =12(x+ 1)−1/2(1)(2x2 + x)− (x+ 1)1/2(4x+ 1)

(2x2 + x)2

=12(x+ 1)−1/2 [(2x2 + x)− 2(x+ 1)(4x+ 1)]

(2x2 + x)2

=−(6x2 + 9x+ 2)

2(2x2 + x)2√x+ 1

.

Rate of Change Problems. We apply our knowledge of calculus to some problems inother areas. We’ll start with some ideas from Business and Economics. The demandor price function, p(x), is the price per unit when x units of a product are sold. Therevenue function is R(x) = xp(x), the amount of money received for selling x units at pricep(x) each. The cost function, C(x), is the total cost of producing x units of the product.

35

The profit function is P (x) = R(x)− C(x), the profit from the sale of x units.

The marginal cost function is C ′(x), the rate of change of cost.The marginal revenue function is R′(x), the rate of change of revenue.The marginal profit function is P ′(x), the rate of change of profit.All are wrt x, the number of units produced or sold.

The marginal cost at x = 100, C ′(100), for example, is an estimate of the true cost of pro-ducing the 101st unit of the product, whereas, P ′(100) is an estimate of the change in profitfor selling the 101st unit.

Example 61. The coffee shop sells 100 mochas per day at $2.75 and we saw that they havedetermined that they will sell 5 fewer per day for each 25 cent price increase.Let n be the number of 25 cent price increases, then the price is p = 2.75 + 0.25n and

the number sold will be x = 100 − 5n. Then n =100− x

5= 20− 0.2x and so the price is

p(x) = 2.75+0.25(20−0.2x) = 7.75−0.05x and the revenue is R(x) = xp(x) = 7.75x−0.05x2.So the marginal revenue is R′(x) = 7.75 − 0.1x. So if they have raised the price to $3.25,they will be selling x = 90 mochas per day and their revenue would be R(90) = $292.50 andthe marignal revenue would be R′(90) = −$1.25 and thus revenue is decreasing at this point.

Example 62. The cost of producing x units of a product is C(x) = −0.002x2 + 10x+ 4000.Compare the marginal cost at 500 units with the true cost of producing the 501st unit.

C ′(x) = −0.004x+ 10, so marginal cost is C ′(500) = 8.The true cost is ∆C = C(501)− C(500)= (−0.002(501)2 + 10(501) + 4000)− (−0.002(500)2 + 10(500) + 4000) = 7.998 (very close).

Now, let’s look at an idea from Physics. The kinetic energy, in Joules J, of an object with

mass m kg and velocity v m/s is K = 12mv2.

Example 63. A baseball with mass 150 g is thrown upward with an initial velocity of 30 m/s,so its velocity function is v(t) = 30− 9.8t.So the kinetic energy is K(t) = 1

2mv2 = 1

2(0.150)(30− 9.8t)2 = 0.075(30− 9.8t)2,

then K ′(t) = 2(0.075)(30− 9.8t)(−9.8) = −1.47(30− 9.8t).So K ′(2) = −15.288 J/s and the ball is slowing down at this time as its kinetic energy isdecreasing.

36

Additional Worked Problems

Recap of Techniques and Problems. Here are the definitions we’ve learned in the lastweek:

• Rate of change.• Limits; left-limits and right-limits.• Continuity.• Derivative.

Here are the types of problems we’ve worked on:

• Calculating limits using:– Looking at plots of functions.– The plug-in principle.– Formulas for complicated limits in terms of simple limits.– Factorization tricks.

• Checking to see if a function is continuous.• Calculating derivatives using:

– The basic definition.– Special formula for polynomials.– The product, quotient, and chain rules.

Limit Examples. There were basically five elementary types of limit questions: findinglimits by looking at pictures, plug-in questions, limits with asymptotes, limits as x goesto infinity, and limits with indeterminate forms. We’ll do one of each of these elementaryquestions: In class: We draw a picture as an example problem for the first typeof limit.

The remaining four types of questions are:

limx→5

(x2 + 2x− 7) =?

limx→5+

1

(x− 5)(x+ 2)=?

limx→∞

2x2 + 3x− 5

−x2 + 3x=?

limx→1

(x− 1)(x+ 3)

x2 − 1=?

Beyond the elementary questions, the most important tool we learned were special formulasthat limits follow:

limx→y

(f(x) + g(x)) = limx→y

f(x) + limx→y

g(x)

limx→y

(f(x)g(x)) = (limx→y

f(x))(limx→y

g(x))

These can be used for lots of reasons - for example, to calculate limits even if you are notgiven the function:

37

Example 64 (Calculating Limits Without a Function). Question: Assume that limx→0 x−2f(x) =

1. Using this information, can you calculate

limx→0

f(x) =?

limx→0

x−4f(x) =?

limx→0

x−3f(x) =?

Answer: Let’s try!

limx→0

f(x) = limx→0

x−2f(x) limx→0

x2 = (1)(0) = 0

limx→0

x−4f(x) = limx→0

x−2f(x) limx→0

x−2 = (1)(∞) =∞

limx→0

x−3f(x) = limx→0

x−2f(x) limx→0

x−1 = (1)(DNE) =???

So, we can calculate the first two. What about the third? Our strategy failed, but this doesnot imply that the limit doesn’t exist!

What should we do next? I think the obvious thing is to try some examples and see if theyagree. In this case, I think the easiest examples are f1(x) = x2 and f2(x) = x |x|. NOTE:The fact that you have to look for examples makes this question much harder than theother ones.

We have also learned about continuity. There aren’t very many questions here. Onequestion is to check when functions are continuous:

Example 65. Question: Let fC(x) be given by the formula

fC(x) =

{√x+ C 0 ≤ x ≤ 2

x+ 1 x > 2.

For what value of C is fC(x) continuous?Answer: The function is clearly continuous everywhere except possibly x = 2. We calcu-

late the left and right limits at 5:

limx↑2

fC(x) =√

2 + C

limx↓2

fC(x) = 3.

These two limits must be equal if fC is continuous, so we must have

2 + C = 9,

so C = 7.

Example 66. Question: Let f(x) satisfy

limx→1

f(x)− 2

x− 1= 2.

What is limx→1 f(x)?38

Answer: We calculate

limx→1

f(x) = 2 + limx→1

(f(x)− 2)

= 2 + limx→1

f(x)− 2

x− 1limx→1

(x− 1)

= 2 + (2)(0) = 2.

Another question is to use the formula limx→y f(g(x)) = f(limx→y g(x)), which requirescontinuity:

Example 67. Question: Let f be continuous and satisfy limx→−1 f(x) = 2, limx→2 = −1.Given this information, can you calculate:

limx→−1

f(√x2 + 1) =??

Answer: Yes, it is

limx→−1

f(√x2 + 1) = f( lim

x→−1

√x2 + 1)

= f(2) = −1.

More examples. After that brief review of all the ‘types’ of applications, we’ll look at somemore examples

Example 68 (Indeterminate Forms). Consider the following limits.

(i) limx→3

x2 − 9

x− 3

(ii) limx→1

√x+ 3− 2

x− 1

(iii) limx→2

(x− 1)2 − 1

x− 2

Substituting into all of these will yield the indeterminate form 00

(which is undefined). Wecan evaluate limits like these by doing certain manipulations.

(i) limx→3

x2 − 9

x− 3

= limx→3

(x+ 3)(x− 3)

x− 3(factor)

= limx→3

(x+ 3) (cancel common factor – allowed because x 6= 3)

= 6

(ii) limx→1

√x+ 3− 2

x− 1

= limx→1

√x+ 3− 2

x− 1×√x+ 3 + 2√x+ 3 + 2

(rationalize the numerator)

= limx→1

(x+ 3)− 4

(x− 1)(√x+ 3 + 2)

39

= limx→1

x− 1

(x− 1)(√x+ 3 + 2)

= limx→1

1√x+ 3 + 2

(cancel common factor)

= 1/4

(iii) limx→2

(x− 1)2 − 1

x− 2

= limx→2

x2 − 2x+ 1− 1

x− 2(expand)

= limx→2

x2 − 2x

x− 2

= limx→2

x(x− 2)

x− 2(factor)

= limx→2

x (cancel common factor)

= 2

Example 69. Is f(x) =x2 − 2x− 3

x2 + 5x+ 4continuous at x = −1 ? Does the limit exist at x = −1 ?

f(−1) = 00, so f(x) is not defined at x = −1 and so it cannot be continuous there.

Example 70. limx→−1

x2 − 2x− 3

x2 + 5x+ 4

limx→−1

(x− 3)(x+ 1)

(x+ 1)(x+ 4)

limx→−1

x− 3

x+ 4= −4/3so yes, lim

x→−1f(x) exists.

And this discontinuity is removable – so the graph of the function would have a hole at point(−1,−4/3).

Derivative Examples.

Example 71. Question: Calculate the derivative of f(x) =√x2 − 2x+ 4, x ∈ R+.

Answer: We have

d

dx(x2 − 2x+ 4)

12 =

1

2(x2 − 2x+ 4)−

12 × d

dx(x2 − 2x+ 4) =

1

2(x2 − 2x+ 4)−

12 (2x− 4).

Example 72 (Chain Rule: Direct Application). Question: Let f(x) = x−1, let g(x) =x2 + 2, and let h(x) = f(g(x)). Calculate h′(x).

Answer: By the chain rule,

h′(x) = f ′(g(x)) g′(x)

= −(x2 + 2)−2(2x).40

Example 73. Differentiate f(x) =√

2x4 + 5x2.

d

dx

(√2x4 + 5x2

)=

d

dx

((2x4 + 5x2)1/2

)=

1

2(2x4 + 5x2)−1/2

d

dx

(2x4 + 5x2

)=

1

2(2x4 + 5x2)−1/2(8x3 + 10x)

=8x3 + 10x

2√

2x4 + 5x2

=4x3 + 5x√2x4 + 5x2

.

Example 74 (Chain Rule: Indirect Application). Question: Let h(x) = f(g(x)) for someunknown functions f, g : R 7→ R. You know that g(0) = 1, g(1) = 4, g′(0) = 3, g′(1) = 1,f ′(0) = 2, f ′(1) = −2. Based on this information, can you calculate h′(0)? Can you calculateh′(1)? In both cases, calculate the derivative if sufficient information is given.Answer: From the chain rule,

h′(0) = f ′(g(0)) g′(0) = f ′(1)g′(0) = (−2)(3) = −6.

Similarly,

h′(1) = f ′(g(1)) g′(1) = f ′(4) g′(1).

Since we don’t know the value of f ′(4), we don’t have enough value to compute this derivative.

Example 75. Question: Calculate the tangent line to the curve f(x) = (x2 − 9)0.5 atx = 4.

Answer: We calculate the derivative of f :

f ′(x) =1

2(x2 − 9)−0.5 (2x).

Thus, we have f ′(4) = 12(4)−0.5 (8) = 2 and f(4) = 2. Thus, using the point-slope form of a

line,

y = f(1) + f ′(1)(x− 4)

= 2 + 2(x− 4).

A more serious application:

Example 76. At time t = 0, a car starts driving eastward from point P at 60 km/h. Asecond car, initially at point Q 200 km south of the first starts driving northward at 80 km/h.After t hours, what is the distance between the cars and at what rate is that distance chang-ing ?

Car 1 has travelled 60t km east of P and car 2 has travelled 80t km north of Q, which is200− 80t km south of P .So the distance between the cars (by Pythagoras) is

d(t) =√

(60t)2 + (200− 80t)2

=√

3600t2 + 40000− 32000t+ 6400t2

41

=√

10000t2 − 32000t+ 40000= 100

√t2 − 3.2t+ 4 km.

Rate of change isd′(t) = 100

(12

)(t2 − 3.2t+ 4)−1/2(2t− 3.2)

=50(2t− 3.2)√t2 − 3.2t+ 4

km/h.

So, after 1 hour, for example, d(1) ≈ 134 km and d′(1) ≈ −44.7 km/h(d′(1) < 0 means the distance between the cars is decreasing).

42

Lecture 4: July 14

Review Quiz! The basic ideas we saw last class were:

• The definition of the derivative.• The definition of the tangent.• Derivative formulas for polynomials.• Derivative formulas for sums, products and quotients.• The chain rule.

Most of the techniques we learned were just formulas to plug into - but there were a lotof formulas, and at least one of them requires some real thought to use:

• The algebra tricks required to calculate a derivative directly. We will also call thiscalculating a derivative from first principles.• Applying the derivative formulas for polynomials, sums, products and quotients cor-

rectly.• Recognizing when the chain rule is required and applying it.

Take a few minutes to calculate the derivatives of the following functions by yourself. We’lldiscuss the answers after 5 minutes.

f(x) = (2x+ 7)(3x2 − x+ 5)

g(x) =x2 + x+ 1

x5 − 3

h(x) =√x2 + x+ 1

If you have time, try to calculate the derivative of u(x) =√x+ 7 from first principles.

Curve Sketching. We’ll spend quite a lot of time using calculus to draw curves moreaccurately. These questions will involve many steps. We begin by using calculus to findextreme values.

We’ll use the following definitions:

Definition 0.19 (Global Maxima and Minima). Fix a function f . Say that x is a globalmaximum of f if, for all a, f(x) ≥ f(a). Say that x is a global minimum of f if, for all a,f(x) ≤ f(a).

Note that a function may have several global minima and maxima! In class, we sketchsin(x).

Similarly, functions can have local minima and maxima:

Definition 0.20 (Local Maxima and Minima). Fix a function f . Say that x is a localmaximum of f if there exists an interval (a, b) so that:

(1) x ∈ (a, b), and(2) for all y ∈ (a, b), f(y) ≤ f(x).

Say that x is a local minimum of f if there exists an interval (a, b) so that:

(1) x ∈ (a, b), and(2) for all y ∈ (a, b), f(y) ≥ f(x).

43

Every global maximum/minimum is also a local maximum/minimum, but the opposite isnot true. In class, we sketch some functions with local maxima/minima that are not globalmaxima/minima.

If you know the minima/maxima of a function f , you can often draw a useful picture off . We will use the following three basic ideas, which will be left up in class, to hunt formaxima/minima:

(1) What does f(x) look like for x very large or very small?(2) Where does f(x) ‘blow up’ - that is, where is there a vertical asymptote?(3) Where is f(x) increasing and decreasing?

If you know these things, you can often draw a pretty good picture of f . After we’vediscussed these three considerations, we’ll look at a more subtle property of f - the ‘concavity’- and relate it to the picture.

We’ll look at these three in order, though calculus only shows up for the third.

Remarks 0.21 (Value of f(x) for x very large or small). We would like to calculatelimx→∞ f(x) and limx→−∞ f(x). We basically have four types of functions f :

(1) f(x) = adxd + ad−1x

d−1 + . . .+ a0, with d > 0 even.(2) f(x) = adx

d + ad−1xd−1 + . . .+ a0, with d > 0 odd.

(3) f(x) = adxd + ad−1x

d−1 + . . .+ a0, with d < 0.

(4) f(x) = adxd+ad−1x

d−1+...+a0bcxc+bc−1xc−1+...+b0

.

In each case, in order to understand the limit, we just have to look at the leading orderterms. That is, when c, d > 0 we look at adx

d and bcxc and ignore the rest of the functions;

when d < 0 we look at a0:

(1) In this case,

limx→∞

adxd + ad−1x

d−1 + . . .+ a0 = ad limx→∞

xd = (sign(ad))×∞

limx→−∞

adxd + ad−1x

d−1 + . . .+ a0 = ad limx→−∞

xd = (sign(ad))×∞

(2) This case is similar, except that xd is very negative when x is very negative:

limx→∞

adxd + ad−1x

d−1 + . . .+ a0 = ad limx→∞

xd = (sign(ad))×∞

limx→−∞

adxd + ad−1x

d−1 + . . .+ a0 = ad limx→−∞

xd = −(sign(ad))×∞

(3) In this case,

limx→±∞

adxd + ad−1x

d−1 + . . .+ a0 = a0

(4) Finally, in this case when c, d > 0 we have:

limx→±∞

adxd + ad−1x

d−1 + . . .+ a0bcxc + bc−1xc−1 + . . .+ b0

=adbc

limx→±∞

xd−c,

which reduces to one of the previous situations.

These are pretty hard to memorize as individual rules and formulas - the thing to memorizeis how to find the leading terms.

44

We apply this:

Example 77 (Drawing Asymptotes as x goes to ±∞). Question: For each function,compute the limit as x goes to ±∞ and draw the corresponding ‘bits’ of the function:

f(x) =2x2 + 5x+ 3

3x2 + 4

g(x) = 3x3 + 19928x2 + 828100

h(x) =2x2 + 5x+ 3

3x7 + 4

Answer: Looking at the leading terms, we have for f :

limx→∞

2x2 + 5x+ 3

3x2 + 4= lim

x→∞

2x2

3x2=

2

3

limx→−∞

2x2 + 5x+ 3

3x2 + 4= lim

x→−∞

2x2

3x2=

2

3.

For g, we have:

limx→∞

3x3 + 19928x2 + 828100 = limx→∞

3x3 =∞

limx→−∞

3x3 + 19928x2 + 828100 = limx→−∞

3x3 = −∞

Finally, for h we have:

limx→∞

2x2 + 5x+ 3

3x7 + 4= lim

x→∞

2x2

3x7= 0

limx→−∞

2x2 + 5x+ 3

3x7 + 4= lim

x→−∞

2x2

3x7= 0

Remember, we had three steps:

(1) What does f(x) look like for x very large or very small?(2) Where does f(x) ‘blow up’ - that is, where is there a vertical asymptote?(3) Where is f(x) increasing and decreasing?

We’ve seen the first; it is often easy, and we will need to do it for every question. The secondcan be harder to check, but many functions don’t have any vertical asymptotes.

The idea: Look for all values of x for which limy→x |f(y)| =∞.The strategy: We know that polynomials, like f(x) = x12 − x4 + 2x + 1, don’t have

vertical asymptotes. We know that rational functions, like g(x) = x4−x2+2x17−3x2+1

, may havevertical asymptotes. In order to check, we follow this algorithm:

Definition 0.22 (Algorithm for Finding Vertical Asymptotes of Rational Functions). Let

f(x) = g(x)h(x)

. We look for vertical asymptotes by:

• Factor g(x) = a0∏m

i=1(x− ai), h(x) = b0∏n

i=1(x− bi).45

• Cancel all of the terms that occur in both factorizations. Let c1, . . . , ck be the remain-ing roots of h. These are the values of x where a vertical asymptote happens.• To figure out what the asymptote at x = ci looks like, check to see if f(x) is very

negative or very positive when x is a tiny bit bigger or smaller than ci.

Let’s apply this to a few examples:

Example 78 (Calculating Vertical Asymptotes). Question: For each of the following func-tions, find the vertical asymptotes and draw them:

f(x) =x+ 4

x− 1

g(x) =x2 − 2x+ 1

x2 + 6x− 7

h(x) =1

x2 − 2x+ 1

Answer: The function f is already factored. There is an asymptote at x = 1; it has

limx→1−

f(x) = −∞

limx→1+

f(x) =∞.

The function g is not yet factored. We find

g(x) =x2 − 2x+ 1

x2 + 6x− 7=

(x− 1)2

(x− 1)(x+ 7)=x− 1

x+ 7,

so there is an asymptote at x = −7. We have

limx→−7−

g(x) =∞

limx→−7+

g(x) = −∞.

The function h is not yet factored. It is h(x) = 1(x−1)2 , so there is only an asymptote at

x = 1. We find

limx→1−

h(x) =∞

limx→1+

h(x) =∞.

Note that both asymptotes are positive! This is because we have a double-root at x = 1.

Remarks 0.23. This algorithm, with lots of factoring, is special to rational functions. Formore general functions, we would look for candidate values of x by:

(1) Write f(x) as a ratio f(x) = g(x)h(x)

.

(2) Look for all of the values c1, . . . , ck for which h(ci) = 0.(3) For each of these values, calculate limx→c±i

f(x).46

Note: Since f is not a rational function, we can’t cancel out roots. Thus, some of thecandidate locations c1, . . . , ck for an asymptote may not actually have an asymptote! Youcan see this by running this algorithm for the three functions in the previous example. We’llalso come back to this later.

So, we’ve now found asymptotes as x→ ±∞ and also vertical asymptotes. The last taskis to find local minima and maxima. This is the task that we need calculus for. Recall thatthe tangent line to a curve f(x) at the point x is just the line passing through (x, f(x)) withslope f ′(x). We do this:

Example 79. Question: Calculate the tangent line to the curve f(x) = (3x2 − 2)13 atx = 1.

Answer: We calculate the derivative of f :

f ′(x) = 13(3x2 − 2)12 (6x).

Thus, we have f ′(1) = 13(1)12(6) = 78 and f(1) = 1. Thus, using the point-slope form of aline,

y = f(1) + f ′(1)(x− 1)

= 1 + 78(x− 1).

Let’s see how tangents behave around maxima/minima.

Remarks 0.24 (Local Maxima, Local Minima and Tangents). Lets draw a nice function;we do this in class. We can easily point to the minima and maxima. How do these relateto tangents?

Question: What are the slopes of the tangents at the minima and maxima?Answer: 0.

Question: If x is a maximum, what does f ′(y) look like for y a little less or larger thanx? What if x is a minimum? Answer: If x is a maximum, then f ′(y) > 0 for y a little lessthan x, and f ′(y) < 0 for y a little bigger than x. The reverse is true if x is a minimum.Question: Are there points where the slope of the tangent is 0, besides the minima and

maxima?Answer: It can certainly happen - look at f(x) = x3. In this case, we will have f ′(y) > 0for both y < x and y > x, or we will have f ′(y) < 0 for both y < x and y > x.

Definition 0.25 (Critical Points). Because minima and maxima are so important, we define

{x : f ′(x) = 0}to be the set of critical points of f .

Our observation suggests the following algorithm for finding minima and maxima:

(1) Calculate the derivative f ′(x).(2) Find all of the solutions r1, . . . , rk to f(x) = 0 - that is, find the critical points of ε.(3) Calculate f(r1 ± ε) for small values of ε to see if r1 is a local max, a local min, or

neither.

Let’s do this for a simple example:

Example 80 (Finding local minima and maxima). Question: Let f(x) = x3 − 6x2 + 12.Find and classify the local extrema.

Answer: We go through the steps:47

(1) f ′(x) = 3x2 − 12x = 3x(x− 4).(2) Factoring f ′(x), the roots are x ∈ {0, 4}.(3) When we look f(x) for x ≈ 0, we find f ′(−0.0001) > 0 while f ′(0.0001) < 0. Thus,

x = 0 is a local maximum for f .Similarly, f(4− 0.0001) < 0 and f(4 + 0.0001) > 0, so x = 4 is a local minimum

for f .

Remarks 0.26. We have only found the local minima and maxima this way. We may nothave found the global minimum or the global maximum; indeed, these may not exist! Forexample:

f(x) = x2

g(x) = 1 +1

1 + x2

h(x) =1

x

r(x) =1

x2

f has a local (and global) minimum at x = 0. However, it has no local or global maximum;it just gets bigger and bigger as x goes to ±∞.g has a local (and global) maximum at x = 0. However, it has no local or global minimum;

it gets closer and closer to 1 from above as x goes to ±∞.h has no local or global minima or maxima. However, it has vertical asympotes that go to±∞ as x goes to 0.r, like h, has no local or global minima or maxima. It has a vertical asymptote that goes

to ∞ as x goes to 0 from either direction; its small values occur as x goes to ±∞.We’ll talk about the issue of finding the smallest and largest values of a function after

we’re done with graphing.

Note that local and global maxima can also occur at points where f ′(x) doesn’t exist:

Example 81. Let f(x) = |x|. Then f ′(x) = −1 for x < 0, f ′(x) = 1 for x > 0, and f ′(x)is never 0.

Drawing the function, we can see that there is only one local extremum - a minimum atx = 0.

We’ve seen a lot of examples! Let’s recap what we know:

(1) A function f can have local minima or maxima at values x for which:(a) f ′(x) = 0 and f ′ changes sign around x.(b) f ′(x) doesn’t exist.

(2) A function f can ‘converge’ to a local or global maximum or minimum by either:(a) Looking at limx→±∞ f(x). These are the horizontal asymptotes of f .(b) Looking at limx→r f(x) for some point r at which f(x) doesn’t exist. These are

the vertical asymptotes of f .

Let’s do a full problem:48

Example 82. Consider f(x) = 2x3 − 3x2 − 12x+ 2.Then f ′(x) = 6x2 − 6x− 12 = 6(x2 − x− 2) = 6(x+ 1)(x− 2),so the critical numbers are x = −1 and x = 2.If x < −1, f ′(x) > 0 and f(x) is increasing.If −1 < x < 2, f ′(x) < 0 and f(x) is decreasing. So there is a local max at x = −1.If x > 2, f ′(x) > 0 and f(x) is increasing. So there is a local min at x = 2.

We’ve implicitly been talking about functions f that are defined on all of R. What do wedo if f is only defined on an interval [a, b]?

Answer: we do the same thing as before, but replace the calculation of limx→±∞ with thecalculation of f(a) and f(b).

Let’s do another example:

Example 83. Find the absolute extrema of f(x) = x3 + 3x2 + 4 on the interval −3 ≤ x ≤ 3.f ′(x) = 3x2 + 6x = 3x(x+ 2), so the critical numbers are x = −2 and x = 0.f(−2) = 8, f(0) = 4f(−3) = 4, f(3) = 58So the absolute max is 58 (at x = 3) and the absolute min is 4 (at x = −3 and x = 0).

Midterm Review. Remember that there is a midterm at the start of next class.

0.0.1. Midterm Overview. Remember that there is a midterm next class! You may bring:

(1) A non-programmable, non-graphing calculator.(2) A single double-sided ‘cheat sheet.’

To get perfect marks on the midterm, you should:

(1) Be familiar with our basic notation (e.g. sets, functions, domain, range, ‘simplify,’etc) and basic algebraic tricks (e.g. solving equations, tricks for simplifying rationalfunctions, factoring polynomials, etc).

(2) Know the definition of trig functions (you don’t need to know any of the trigonometricidentities on the midterm).

(3) Solve inequalities involving lines, quadratic polynomials, and absolute values.(4) Be able to do “rate of change” (or “average speed”) problems.(5) Be able to calculate limits (by looking at pictures, and also by using our three formulas

and various factorization/simplification tricks), and also understand the intuitivedefinition of limits.

(6) Understand the definition of a continuous function, and be able to check if a functionthat is defined piecewise is continuous..

(7) Calculate a simple derivative from first principles.(8) Calculate more complicated derivatives using our formulas for derivatives.(9) Use calculus to plot simple functions.

In the rest of these notes, I give a ‘super-midterm’ that covers all of the types of questionson the real midterm (as well as giving some additional questions). We probably won’t finishthis in class, so I strongly suggest that you go through the remainder of the questions byyourself!

(1) Basic Notation: There will not be any questions directly about this material.However, you will need it to be able to answer some of the other questions.

49

(2) Trigonometric Functions: There will not be any questions directly about thismaterial. However, you will need it to be able to answer some of the other questions.

(3) Solving Inequalities: There will not be any questions directly about this material.However, you will need it to be able to answer some of the other questions.

(4) Rate of Change Problems: Here are typical exam problems, with wording that isvery similar to an exam.• Let f(x) = sin(x)2. Calculate the average rate of change over the interval (0, 0.2).

Answer: We have

v =f(0.2)− f(0)

0.2− 0

=sin(0.2)2 − sin(0)2

0.2≈ 0.197.

• Let g(x) = 52x. Calculate the average rate of change over the interval (3, 4).Answer: We have

v =g(4)− g(3)

4− 3

= 58 − 56

= 375000.

(5) Limits: Here are typical exam problems, with wording that is very similar to anexam.• Calculate limx→2(x

2 + 2).Answer: We have

limx→2

(x2 + 2) = (4 + 2) = 6.

• Calculate limx→2x2−4x+4

(x−2)(x−7)2 .

Answer: We have

limx→2

x2 − 4x+ 4

(x− 2)(x− 7)2= lim

x→2

(x− 2)2

(x− 2)(x− 7)2

= limx→2

x− 2

(x− 7)2

= 0.

• Calculate limx→7x2−4x+4

(x−2)(x−7)2 .

Answer: We have

limx→7

x2 − 4x+ 4

(x− 2)(x− 7)2= lim

x→7

(x− 2)2

(x− 2)(x− 7)2

= limx→7

x− 2

(x− 7)2

=∞.

• Calculate limx→∞x2−4x+4

4x2−25x+2.

50

Answer: We have

limx→∞

x2 − 4x+ 4

(x− 7)2= lim

x→∞

x2 − 4x+ 4

4x2 − 25x+ 2

= limx→∞

1− 4x−1 + 4x−2

4− 25x−1 + 2x−2

=1 + 0 + 0

4 + 0 + 0=

1

4.

• Calculate limx→3x2−9√3x−3 .

We have

limx→3

x2 − 9√3x− 3

= limx→3

(x− 3)(x+ 3)√3x− 3

√3x+ 3√3x+ 3

= limx→3

(x− 3)(x+ 3)(√

3x+ 3)

3x− 9

= limx→3

(x+ 3)(√

3x+ 3)

3

=(6)(6)

3= 12.

(6) Continuity: A typical question is: Define fr(x) = x2 + r for 0 ≤ x ≤ 2 andfr(x) = rx for x > 2. Find all values of r ∈ R for which fr is continuous.

Answer: We observe that the only possible point of discontinuity is at x = 2. Wecalculate the left and right limits at that point:

limx→2−

fr(x) = limx→2−

(x2 + r)

= 4 + r,

and

limx→2+

fr(x) = limx→2+

rx

= 2r.

If fr(x) is continuous at x = 2, these two limits must equal each other:

4 + r = 2r,

so we conclude

r = 4.

(7) Derivatives from First Principles: Typical questions are:• Let f(x) =

√x+ 1. Using only the definition of the derivative, calculate f ′(x).

Explain for which values of x the derivative exists.Answer: We have

f ′(x) = limh→0

f(x+ h)− f(x)

h

= limh→0

√x+ 1 + h−

√x+ 1

h51

= limh→0

√x+ 1 + h−

√x+ 1

h

√x+ 1 + h+

√x+ 1√

x+ 1 + h+√x+ 1

= limh→0

(x+ 1 + h− (x+ 1))

h(√x+ 1 + h+

√x+ 1)

= limh→0

1√x+ 1 + h+

√x+ 1

=1

2√x+ 1

.

This derivative exists as long as x+ 1 > 0 (i.e. as long as x > −1).• Let g(x) = 1

x2. Using only the definition of the derivative, calculate g′(x). Ex-

plain for which values of x the derivative exists.We have

g′(x) = limh→0

g(x+ h)− g(x)

h

= limh→0

(x+ h)−2 − x−2

h

= limh→0

x2 − (x+ h)2

h(x2)((x+ h)2)

= limh→0

x2 − x2 − 2xh− h2

h(x2)((x+ h)2)

= limh→0

−2x− h(x2)((x+ h)2)

= limh→0

−2

x3.

This derivative exists as long as x 6= 0.(8) Calculating derivatives: Typical questions are:

• Let f(x) = x3 + 2x+ 1. Calculate f ′(x).Answer: Using our basic formula for derivatives,

f ′(x) = 3x2 + 2.

• Let g(x) =√x2 + 5x+ 3. Calculate g′(x).

Answer: Using the chain rule,

g′(x) =1

2√x2 + 5x+ 3

× d

dx(x2 + 5x+ 3)

=2x+ 5

2√x2 + 5x+ 3

.

• Let h(x) = 2x+13x+7

. Calculate h′(x).Answer: Using the quotient rule,

h′(x) =(2)(3x+ 7)− (2x+ 1)(3)

(3x+ 7)252

=11

(3x+ 7)2.

(9) Plotting functions: Typical questions might look like:• Let f(x) = x3 − x2 + x − 1. Calculate the critical points of f , as well as

limx→∞ f(x) and limx→−∞ f(x). Using this information (and any other informa-tion that you might find useful), plot f(x).

• Let g(x) = x2+x+1x2−5x . Calculate the critical points of g, as well as limx→∞ g(x)

and limx→−∞ g(x). Using this information (and any other information that youmight find useful), plot g(x).

We will give answers in class, where we can more easily draw pictures.

0.0.2. Miscellaneous Examples. Please let me know if you have additional questions- I’ll try to add them here.

Example 84 (Chain Rule Example). (1)d

dx

((5x2 − x)7

)= 7(5x2 − x)6(10x− 1)

(2)d

dx

((2x+ 1)4(6x2 − 2)3

)(product rule and chain rule)

= 4(2x+ 1)3(2)(6x2 − 2)3 + (2x+ 1)4(3)(6x2 − 2)2(12x)= 4(2x+ 1)3(6x2 − 2)2 [2(6x2 − 2) + 9(2x+ 1)]= 4(2x+ 1)3(6x2 − 2)2(12x2 + 18x+ 5)

(3) If p(t) = 3√t2 + 3t+ 2 = (t2 + 3t+ 2)1/3, then p′(t) = 1

3(t2 + 3t+ 2)−2/3(2t+ 3).

53

Lecture 5: July 18

Midterm! Good luck!

Continued Curve Sketching. When sketching curves last class, it was annoying to try tofigure out if a point x was a local minimum, a local maximum, or neither. It turns out thatthere is a simple test that can often answer this question. We need some more definitionsfirst.

Definition 0.27 (Concave and Convex). If f ′′(x) > 0 on the interval a < x < b, we saythat f(x) is concave up on the interval and the curve would be bending upwards as slopes oftangents would be increasing. If f ′′(x) < 0, f(x) is concave down and the curve bends down-wards. A point where the concavity changes (from up to down or down to up) is called aninflection point. The concavity changing requires f ′′(x) to change sign and hence inflectionpoints occur where f ′′(x) = 0 or is undefined. But f ′′(c) = 0 or being undefined does notmean that there must be an inflection point at x = c – see the x4 example below.

Example 85. If f(x) = x3 − 3x2 + 6, then f ′(x) = 3x2 − 6x and f ′′(x) = 6x− 6,so f ′′(x) = 0 if x = 1.If x < 1, f ′′(x) < 0 and so f(x) is concave down.If x > 1, f ′′(x) > 0 and so f(x) is concave up and there is an inflection point at (1, 4).

Example 86. If f(x) = x4, then f ′(x) = 4x3 and f ′′(x) = 12x2.Then f ′′(0) = 0, but f ′′(x) ≥ 0 for all x and the function is always concave up.There is no point of inflection as the concavity has not changed.But there is a local (and absolute) minimum at x = 0.

Consider the functions f(x) = x2 and g(x) = −x2. Then f ′(x) = 2x, f ′′(x) = 2 and so thefunction is always concave up and tangent lines lie below the curve and there is a local min,whereas g′(x) = −2x, g′′(x) = −2 and that function is always concave down and tangentlines always lie above the curve and there is a local max. (This tells us that the tangent linewould have to cross the curve at a point of inflection.)These graphs also show us the connection between concavity and local extrema.

The Second Derivative TestSuppose that f ′(c) = 0 and f ′′(c) 6= 0.(i) If f ′′(c) > 0, f(x) has a local min at x = c.

54

(ii) If f ′′(c) < 0, f(x) has a local max at x = c.

Example 87. Consider our example above, f(x) = x3 − 3x2 + 6.Then f ′(x) = 3x2 − 6x = 3x(x− 2) and f ′′(x) = 6x− 6.So f ′(x) = 0 if x = 0 or 2.f ′′(0) = −6 < 0 =⇒ local max at x = 0.f ′′(2) = 6 > 0 =⇒ local min at x = 2.

Example 88. Consider the function f(x) = x4 − 8x2 + 3(this function has even symmetry as f(−x) = f(x) for all x).

Then f ′(x) = 4x3 − 16x = 4x(x2 − 4) = 4x(x+ 2)(x− 2)and so f ′(x) = 0 if x = −2, 0 or 2.If x < −2, f ′(x) < 0 =⇒ f(x) decreasing.If −2 < x < 0, f ′(x) > 0 =⇒ f(x) increasing, so (−2,−13) is a local min (FDT).If 0 < x < 2, f ′(x) < 0 =⇒ f(x) decreasing, so (0, 3) is a local max (FDT).If x > 2, f ′(x) > 0 =⇒ f(x) increasing, so (2,−13) is a local min (FDT).And f ′′(x) = 12x2 − 16 = 4(3x2 − 4),

so f ′′(x) = 0 if x = ±√

4/3 = ±2/√

3.

If x < −2/√

3, f ′′(x) > 0 and f(x) is concave up (agrees with min at x = −2).If −2/

√3 < x < 2/

√3, f ′′(x) < 0 and f(x) is concave down (agrees with max at x = 0)

and there is an inflection point at (−2/√

3,−53/9).If x > 2/

√3, f ′′(x) > 0 and f(x) is concave up (agrees with min at x = 2) and there is an

inflection point at (2/√

3,−53/9).Or, we could have used the SDT to classify the extrema:f ′′(−2) > 0 =⇒ local min at x = −2,f ′′(0) < 0 =⇒ local max at x = 0,and f ′′(2) > 0 =⇒ local min at x = 2.See the graphs below.

We have focused so far on polynomial functions. We’ll now look at rational functions.Recall:

55

Definition 0.28 (Rational Function). A rational function has the form f(x) =P (x)

Q(x), where

both P (x) and Q(x) are polynomials. A rational function will be undefined wherever Q(x) = 0(as we cannot divide by 0) and it will be continuous everywhere on its domain, {x |Q(x) 6= 0}.

If x = a is a value where the denominator is zero and the numerator is not, ie Q(a) = 0and P (a) 6= 0, the rational function will have a vertical asymptote at x = a. More ex-actly, f(x) has a vertical asymptote x = a if f(a) is undefined and lim

x→a−f(x) = ±∞ and/or

limx→a+

f(x) = ±∞. If we graph f(x), the curve cannot cross a vertical asymptote.

Example 89. Consider f(x) =2

x+ 1.

f(x) is undefined at x = −1 (denominator is 0)

and limx→−1−

f(x) = limx→−1−

2

x+ 1= −∞

and limx→−1+

f(x) = limx→−1+

2

x+ 1=∞

and so x = −1 is a vertical asymptote.

The signs of f ′(x) and f ′′(x) can change from one side of a vertical asymptote x = a tothe other, so we must check for that when we look for intervals of increase/decrease and/orintervals of concavity. But there will be no local extremum or inflection point at x = a evenif there is a sign change in the appropriate derivative as there is no point on the curve whenx = a.

Example 90. Consider f(x) =4

x2 + x− 6=

4

(x+ 3)(x− 2).

f(x) is undefined at x = −3 and x = 2.

limx→−3−

f(x) = limx→−3−

4

(x+ 3)(x− 2)=∞

and limx→−3+

f(x) = limx→−3+

4

(x+ 3)(x− 2)= −∞,

56

so there is a vertical asymptote at x = −3.

limx→2−

f(x) = limx→2−

4

(x+ 3)(x− 2)= −∞

and limx→2+

f(x) = limx→2+

4

(x+ 3)(x− 2)=∞,

so there is a vertical asymptote at x = 2.

f(x) =4

x2 + x− 6= 4(x2 + x− 6)−1,

so f ′(x) = −4(x2 + x− 6)−2(2x+ 1) =−4(2x+ 1)

(x2 + x− 6)2.

f ′(x) = 0 if 2x+ 1 = 0 or x = −1/2 (where the numerator is 0).If x < −3, f ′(x) > 0, so f(x) is increasing.If −3 < x < −1/2, f ′(x) > 0, so f(x) is increasing.If −1/2 < x < 2, f ′(x) < 0, so f(x) is decreasing and there is a local max at (−1/2,−16/25).If x > 2, f ′(x) < 0, so f(x) is decreasing.

f ′′(x) = −4

[2(x2 + x− 6)2 − (2x+ 1)(2)(x2 + x− 6)(2x+ 1)

(x2 + x− 6)4

]= −4(2)

[(x2 + x− 6)− (2x+ 1)2

(x2 + x− 6)3

]= −8

(x2 + x− 6− (4x2 + 4x+ 1)

(x2 + x− 6)3

)=−8(−3x2 − 3x− 7)

(x2 + x− 6)3

=8(3x2 + 3x+ 7)

(x2 + x− 6)3,

so f ′′(x) = 0 if 3x3 + 3x+ 7 = 0,

or if x =−3±

√(3)2 − 4(3)(7)

2(3)=−3±

√9− 84

6, so there are no real roots.

If x < −3, f ′′(x) > 0 and f(x) is concave up.If −3 < x < 2, f ′′(x) < 0 and f(x) is concave down.If x > 2, f ′′(x) > 0 and f(x) is concave up.But there are no inflection points. Why not ?

Example 91. Consider f(x) =x

x2 + 1(this function has odd symmetry as f(−x) = −f(x) for all x).The denominator is never zero, so no vertical asymptotes (ie a rational function does nothave to have them).

f ′(x) =(1)(x2 + 1)− x(2x)

(x2 + 1)2=

1− x2

(x2 + 1)2,

so f ′(x) = 0 if x = ±1.If x < −1, f ′(x) < 0 =⇒ f(x) is decreasing.If −1 < x < 1, f ′(x) > 0 =⇒ f(x) is increasing and there is a local min at (−1,−1/2).If x > 1, f ′(x) < 0 =⇒ f(x) is decreasing and there is a local max at (1, 1/2).

f ′′(x) =−2x(x2 + 1)2 − (1− x2)(2)(x2 + 1)(2x)

(x2 + 1)457

=−2x(x2 + 1)− 2(2x)(1− x2)

(x2 + 1)3

=−2x[x2 + 1 + 2(1− x2)]

(x2 + 1)3

=−2x(3− x2)

(x2 + 1)3

=2x(x2 − 3)

(x2 + 1)3

and so f ′′(x) = 0 if x = 0 or x = ±√

3.If x < −

√3, f ′′(x) < 0 and f(x) is concave down.

If −√

3 < x < 0, f ′′(x) > 0 and f(x) is concave up and thus there is an inflection point at(−√

3,−√

3/4).If 0 < x <

√3, f ′′(x) < 0 and f(x) is concave down and thus there is an inflection point at

(0, 0).If x >

√3, f ′′(x) > 0 and f(x) is concave up and thus there is an inflection point at

(√

3,√

3/4).

We put together the techniques we’ve assembled. If we want to sketch the graph of afunction y = f(x), we can obtain information about the shape and behaviour of the curve

58

and about special points on the curve by finding the following:

(1) the domain

(2) the intercepts (where the curve crosses the x and y axes)

(3) symmetry (if the function is even f(−x) = f(x) or odd f(−x) = −f(x), though mostfunctions are neither)

(4) vertical asymptotes

(5) horizontal asymptotes (check if either limx→−∞

f(x) or limx→∞

f(x) is a constant)

(6) intervals of increase and decrease and local extrema

(7) intervals of concavity and points of inflection.

Example 92. y = f(x) = x+1

xThis function is defined for all x except x = 0, so x = 0 could be a vertical asymptote.

limx→0−

f(x) = limx→0−

x+1

x= 0−∞ = −∞

and limx→0+

f(x) = limx→0+

x+1

x= 0 +∞ =∞,

so, yes, x = 0 is a vertical asymptote.f(0) is undefined, so there is no y-intercept.

f(x) = 0 only if x+1

x= 0 or x2 + 1 = 0, which is impossible (x is real), so no x-intercepts

either.

f(−x) = (−x) +1

(−x)= −x− 1

x= −

(x+

1

x

)= −f(x), so this function is odd (and the

symmetry is about the origin as we’ll see in the graph).

limx→−∞

f(x) = limx→−∞

x+1

x= −∞− 0 = −∞

and limx→∞

f(x) = limx→∞

x+1

x=∞+ 0 =∞,

so no horizontal asymptotes.

f ′(x) = 1− 1

x2,

so f ′(x) = 0 if x = ±1.If x < −1, f ′(x) > 0 and so f(x) is increasing.If −1 < x < 0, f ′(x) < 0 and so f(x) is decreasing and there is a local max at (−1,−2).If 0 < x < 1, f ′(x) < 0 and so f(x) is decreasing.If x > 1, f ′(x) > 0 and so f(x) is increasing and there is a local min at (1, 2).Why did we use x = 0 as one of the divisions for the intervals ?

f ′′(x) =2

x36= 0 for any x.

59

If x < 0, f ′′(x) < 0 and f(x) is concave down.If x > 0, f ′′(x) > 0 and f(x) is concave up.But there is no inflection point. Why not ?Put it all together to get the graph.

Example 93. y = f(x) = x3 − 6x2 − 36xf(x) is a polynomial, so it is defined for all x and there are no vertical asymptotes.f(0) = 0 =⇒ (0, 0) is the y-intercept.f(x) = 0 if x3 − 6x2 − 36x = x(x2 − 6x− 36) = 0 or if x = 0

or if x =6±

√(−6)2 − 4(−36)

2(1)=

6±√

5(36)

2=

6± 6√

5

2= 3± 3

√5,

and thus the x-intercepts are 0, 3− 3√

5 ≈ −3.71 and 3 + 3√

5 ≈ 9.71.This function has no symmetry (it has both even and odd powered terms).lim

x→−∞f(x) = −∞ and lim

x→∞f(x) =∞, so no horizontal asymptotes.

f ′(x) = 3x2 − 12x− 36 = 3(x2 − 4x− 12) = 3(x+ 2)(x− 6),so f ′(x) = 0 if x = −2 or x = 6.If x < −2, f ′(x) > 0, so f(x) is increasing.If −2 < x < 6, f ′(x) < 0, so f(x) is decreasing and there is a local max at (−2, 40).If x > 6, f ′(x) > 0, so f(x) is increasing and there is a local min at (6,−216).f ′′(x) = 6x− 12 = 6(x− 2),so f ′′(x) = 0 if x = 2.If x < 2, f ′′(x) < 0 and f(x) is concave down.If x > 2, f ′′(x) > 0 and f(x) is concave up and there is an inflection point at (2,−88).

60

Optimization Problems. We can use our knowledge of how to find the absolute extremaof a function on a closed interval and the Derivative Tests to solve real-world optimizationproblems like minimizing costs, maximizing area, etc. When solving these problems, thereare simple things to keep in mind:

(1) we need to be sure what the question is asking for and what the given information is

(2) we need to be able to identify what function needs to be optimized and what theinterval is (if appropriate), which usually requires reduction to a single variable anddetermination of constraints

(3) a diagram often helps.

Example 94. A farmer has 800 m of fence and wishes to enclose a rectangular field. Whatdimensions will maximize the area enclosed ?

Let l and w be the length and width of the field.We are told that he has 800 m of fence, so the perimeter of the rectangle must be 800 m.So we have that 2l + 2w = 800 or l + w = 400.The area enclosed is A = lw, but w = 400− l, so A(l) = l(400− l) = 400l − l2.Clearly, l ≥ 0 and since w ≥ 0, we must have l ≤ 400 (because l+w = 400), so the intervalis 0 ≤ l ≤ 400.A′(l) = 400− 2l, so A′(l) = 0 if l = 200.A(0) = 0, A(200) = 40 000 and A(400) = 0,so the maximum area enclosed is 40 000 m2 if the field is a 200 m by 200 m square.

Example 95. What if the farmer only needed to enclose three sides with 100 m of fence tocreate a pen beside the barn ?Now l + 2w = 100, so w = (100− l)/2and the area is A(l) = l(100− 1)/2 = 50l − l2/2 and the interval is 0 ≤ l ≤ 100. Why ?Then A′(l) = 50− l, so A′(l) = 0 if l = 50.A(0) = A(100) = 0 and A(50) = 1250,

61

so the maximum area enclosed is 1250 m2 with dimensions 50 m by 25 m, with the 50 mside parallel to the barn.

Example 96. A soup can is to have a volume of 500 mL. What dimensions will minimizethe amount of tin used ?Let the radius and height of the can be r and h, respectively.The amount of tin used corresponds to the surface area of the can,which is A = 2πr2 + 2πrh (can you see that ?).

But the volume is supposed to be V = πr2h = 500 cm3, so h =500

πr2.

And so A(r) = 2πr2 + 2πr

(500

πr2

)= 2πr2 +

1000

r.

Clearly, the dimensions must be positive values.

A′(r) = 4πr − 1000

r2,

so A′(r) = 0 if 4πr3 = 1000 or r3 =1000

4π=

250

πor r =

3

√250

π≈ 4.30 cm.

A′′(r) = 4π +2000

r3> 0 for all r > 0, so this value must be a local (and absolute) min for

the function (see the graph for confirmation).

The height is then h =500

π(4.3)2≈ 8.61 cm

(notice that the height and the diameter are equal here).

Example 97. Suppose that the material for the top and bottom of the can costs twice asmuch as that for the side. What dimensions would minimize the cost of producing the can ?The (relative) cost would be C = 2(2πr2) + 2πrh = 4πr2 + 2πrh,

so C(r) = 4πr2 +1000

r.

Then C ′(r) = 8πr − 1000

r2, so C ′(r) = 0 if r =

3

√125

π≈ 3.41 cm

and then the height would be h ≈ 13.69 cm(notice that the height is now twice the diameter).

Example 98. If the cost of producing x widgets is C(x) = 0.1x2 + 30x+ 100 and the pricefunction is p(x) = 125 − 0.2x, how many widgets should be produced and sold to maximizeprofits ?The profit function isP (x) = R(x)− C(x)= xp(x)− C(x)= x(125− 0.2x)− (0.1x2 + 30x+ 100)= −0.3x2 + 95x− 100.So P ′(x) = −0.6x+ 95, and then P ′(x) = 0 if x = 95/0.6 ≈ 158 (round to nearest integer).P ′′(x) = −0.6 < 0 =⇒ this is a local (and absolute) max for the function (which is aparabola opening downwards).

62

Additional Examples.

Example 99. Question: Let f(x) = x3 − 12x + 1. Calculate f ′(x) and f ′′(x). Use thisinformation to calculate the critical points of f and the intervals on which f is concave upand concave down. Use this information to sketch the graph of f .

Answer: We calculate

f ′(x) = 3x2 − 12 = 3(x− 2)(x+ 2)

f ′′(x) = 6x.

From the first line, we find that the critical points are at x ∈ {−2, 2}, while f changesconvexity at x = 0. These 3 points split the real line into 4 pieces. In these regions, we have:

Table 2. Critical Points

x f ′(x) f ′′(x)

−∞ < x < −2 + -−2 < x < 0 - -0 < x < 2 - +2 < x <∞ + +

We then use this information to sketch the graph.

Example 100. Question: Let f(x) = x+4x−5 . Calculate f ′(x) and f ′′(x). Use this infor-

mation to calculate the critical points of f and the intervals on which f is concave up andconcave down. Use this information to sketch the graph of f .Answer: We calculate

f ′(x) =x− 5− (x+ 4)

(x− 5)2=

−9

(x− 5)2

f ′′(x) =18

(x− 5)3.

Notice, the equation −9(x−5)2 = 0 doesn’t have any solutions. From this, it is tempting to

think that f might not have any critical points. However, there is one other importantpoint: x = 5, where the denominator is 0. This is a critical point, and potentially a placewhere the function can change from concave up to down.

Checking, we find that f ′(x) < 0 for all x ∈ R, but f ′′(x) < 0 for x < −5 and f ′′(x) > 0 forx > 5. We also note that limx→∞ f(x) = limx→−∞ f(x) = 1. We then use this informationto sketch the graph.

Example 101. Question: Let f(x) = x4− 8x3 + 10x2 + 1. Calculate f ′(x) and f ′′(x). Usethis information to calculate the critical points of f and the intervals on which f is concaveup and concave down. Use this information to sketch the graph of f .Answer: We calculate

f ′(x) = 4x3 − 24x2 + 20x = 4x(x2 − 6x+ 5) = 4x(x− 1)(x− 5)

f ′′(x) = 12(x2 − 4x+5

3) = 12(x− 3.53)(x− 0.47).

63

From the first line, we find that the critical points are at x ∈ {0, 1, 5}, while f changesconvexity at roughly x ∈ {3.53, 0.47}. These 5 points split the real line into 6 pieces. Inthese regions, we have:

Table 3. Critical Points

x f ′(x) f ′′(x)

−∞ < x < 0 - +0 < x < 0.47 + +0.47 < x < 1 + -1 < x < 3.53 - -3.53 < x < 5 - +5 < x <∞ + +

We then use this information to sketch the graph.

Example 102. Question: Let f(x) = 1x+2

, g(x) = 1x2−4 , and h(x) = 1

x2+4. Find all vertical

asymptotes of all three functions.Answer: These functions are well-behaved except perhaps at the roots of the polynomials

in the denominator. The roots are:

f(x) : x = −2

g(x) : x ∈ {−2, 2}h(x) : x ∈ {−2i, 2i}.

Thus, f has a vertical asymptote at x = −2, g has vertical asymptotes at x ∈ {−2, 2}, andh does not have vertical asymptotes at all.

Important note: in this class, it is not correct to say that h has vertical asymptotes atx ∈ {−2i, 2i}. This is because all of our functions take only real numbers as inputs.

64

Lecture 6: July 21

Review Quiz! You would like to build a window frame using 4 feet of framing material.The window will consist of a rectangle, topped by a semicircle (See picture in class.).What is the largest window that you can make?

Special Functions. Recall that the graph of an exponential function y = f(x) = ax (fora > 1) would look like the one below.

The function is always positive, increasing, concave up, passes through (0, 1) and has limx→−∞

ax = 0

and limx→∞

ax =∞.

What is the derivative of this function ? Graphically, we can draw some tangents and see

thatd

dx(ax) would have the same shape as ax itself – increasing, concave up, always positive,

starts small (for negative x) and gets larger.So it appears that the derivative of an exponential is another exponential function. Is itreally ?

d

dx(ax) = lim

h→0

ax+h − ax

h

= limh→0

axah − ax

h

= limh→0

ax(ah − 1)

h

= ax limh→0

ah − 1

h(because ax is independent of the limit)

= kax,

where k = limh→0

ah − 1

his a constant.

Indeed,d

dx(ax) is an exponential function – in fact,

d

dx(ax) = kax.

But, we can do a little better, since if f(x) = ax,

f ′(0) = limh→0

f(0 + h)− f(0)

h65

= limh→0

a0+h − a0

h

= limh→0

ah − 1

hand so k = f ′(0), the slope of the tangent line to the curve at x = 0.

Let’s see what happens with 2x, 3x and 4x.

For 2x, f ′(0) ≈ 0.6931, for 3x, f ′(0) ≈ 1.0986 and for 4x, f ′(0) ≈ 1.3863. So we can see

that limh→0

ah − 1

hincreases as a increases. And we can see something else – there must be a

number between 2 and 3 (closer to 3) such that the slope of the tangent at x = 0, or thelimit, is 1. We call that number e.

So e is the number such thateh − 1

h≈ 1 for small h. Or eh − 1 ≈ h. Or eh ≈ h + 1. So

e ≈ (1 + h)1/h.

And thus e = limh→0

(1 + h)1/h

= limn→∞

(1 +

1

n

)n(n = 1/h)

≈ 2.718 281 828 459 . . . .

And we have thatd

dx(ex) = ex, ie this exponential function is everywhere equal to its own

derivative (and second derivative and third derivative, etc.).

Since we have defined the exponential function f(x) = ex, we can define its inverse function,called the natural logarithm, f−1(x) = loge x = lnx (pronounced as “lawn of x”), definedfor all x > 0.

This looks like:So we have that elnx = x = ln(ex) (since they are inverses). We now calculate the derivativeof ln(x).

66

Let y = lnx, so that x = ey,

andd

dx(x) =

d

dx(ey).

Differentiating, this gives 1 = eydy

dx(by the chain rule).

Sody

dx=

1

ey=

1

x.

And thus, we have thatd

dx(lnx) =

1

xfor x > 0.

Example 103. Suppose the population of a bacterial culture after t days is given by P (t) =1500e0.08t.Then the initial population is P (0) = 1500e0.08(0) = 1500e0 = 1500.The population after two days is P (2) = 1500e0.08(2) = 1500e0.16 = 1500(1.1735) ≈ 1760.How long would it take for the population to reach 2000 ?2000 = 1500e0.08t means that e0.08t = 2000

1500= 4

3,

so ln(e0.08t) = ln(4/3) or 0.08t = ln(4/3)

and so t = ln(4/3)0.08

≈ 3.6 days.

67

Since we have defined lnx, we can identify the constant k = f ′(0) in the derivative of ax.Let y = ax, so that ln(y) = x ln(a). Differentiating and using the chain rule, we have

1

y

dy

dx= ln(a),

so

dy

dx= y ln(a) = ln(a) ax.

Thus, ddxax = ln(a) ax.

Example 104. (i)d

dx(3x) = 3x ln 3

(ii)d

dt

(t2et)

= 2tet + t2et = (2t+ t2)et

(iii)d

dx(x5x) = 5x + x5x ln 5 = (1 + x ln 5)5x

We now look at some more complicated examples. If we have f(x) = eg(x), then the chainrule tells us that

f ′(x) =d

dx

(eg(x)

)= eg(x)

d

dx(g(x)) = eg(x)g′(x).

Let’s apply this: Note: Some of these examples use trigonometric polynomials,which we haven’t seen yet. You can come back to these for practice after nextclass - we don’t talk about them in class today.

(1) f(x) = ecosx, f ′(x) = −(sinx)ecosx

(2) y = 2et2+3,

dy

dt= 4tet

2+3

(3)d

dx

(e4x−sinx

)= (4− cosx)e4x−sinx

(4)d

dx(ex sinx) = ex sinx+ ex cosx

(5)d

dx

(x2ex

2)

= 2xex2

+ 2x3ex2

(6) If f(x) =xex

cosx,

then f ′(x) =(ex + xex) cosx− xex(− sinx)

cos2 x

=ex(cosx+ x(cosx+ sinx))

cos2 x.

(7)d

dx

(sin(ex

2+5x+1))

= cos(ex2+5x+1)

d

dx

(ex

2+5x+1)

= cos(ex2+5x+1)ex

2+5x+1 d

dx

(x2 + 5x+ 1

)68

= (2x+ 5)ex2+5x+1 cos(ex

2+5x+1)

Example 105. Analyze the function y = f(x) = e−x2

and sketch its graph.f(x) is defined for all x, so no vertical asymptotes.f(0) = 1 =⇒ (0, 1) is the y-intercpet.f(x) 6= 0 for any x, so no x-intercepts.

f(−x) = e−(−x)2

= e−x2

= f(x), so the function is even.

limx→±∞

e−x2

= 0, so y = 0 is a horizontal asymptote.

f ′(x) = −2xe−x2, so f ′(x) = 0 if x = 0.

If x < 0, f ′(x) > 0 and so f(x) is increasing.If x > 0, f ′(x) < 0 and so f(x) is decreasing and (0, 1) is a local (and abs) maximum.

f ′′(x) = −2e−x2

+ 4x2e−x2

= 2(2x2 − 1)e−x2, so f ′′(x) = 0 if x = ±1/

√2.

If x < −1/√

2, f ′′(x) > 0 and f(x) is concave up.If −1/

√2 < x < 1/

√2, f ′′(x) < 0 and f(x) is concave down (agrees with local max).

So there is an inflection point at (−1/√

2, e−1/2) ≈ (−0.707, 0.607).If x > 1/

√2, f ′′(x) > 0 and f(x) is concave up.

So there is another inflection point at (1/√

2, e−1/2) ≈ (0.707, 0.607).

Example 106. Analyze the function y = f(x) = xe−x and sketch its graph.f(x) is defined for all x =⇒ no vertical asymptotes.f(0) = 0 and so (0, 0) is the y-intercept.f(x) = 0 only if x = 0 and so (0, 0) is the only intercept.This function has no symmetry.lim

x→−∞xe−x = −∞

limx→∞

xe−x = 0 and so y = 0 is a horizontal asymptote.

f ′(x) = e−x − xe−x = (1− x)e−x, and so f ′(x) = 0 if x = 1.If x < 1, f ′(x) > 0 and f(x) is increasing.If x > 1, f ′(x) < 0 and f(x) is decreasing and thus (1, e−1) ≈ (1, 0.368) is a local (and abs)max.f ′′(x) = −e−x − e−x + xe−x = (x− 2)e−x, so f ′′(x) = 0 if x = 2.

69

If x < 2, f ′′(x) > 0 and f(x) is concave up.If x > 2, f ′′(x) < 0 and f(x) is concave down and so (2, 2e−2) ≈ (2, 0.271) is an inflectionpoint.

Example 107. Analyze the function y = f(x) = (ln x)2 and sketch its graph.f(x) is defined for all x > 0.limx→0+

(lnx)2 = (−∞)2 =∞ and so x = 0 is a vertical asymptote.

No y-intercept.There is no symmetry.f(x) = 0 if lnx = 0, or if x = 1. So there is an x-intercept at (1, 0).limx→∞

(lnx)2 =∞ so no horizontal asymptotes.

f ′(x) = 2 (lnx)1

x, so f ′(x) = 0 if x = 1.

If 0 < x < 1, f ′(x) < 0 and f(x) is decreasing.If x > 1, f ′(x) > 0 and f(x) is increasing and thus (1, 0) is a local (and abs) min.

f ′′(x) =2

x2− 2 lnx

x2=

2

x2(1− lnx) and thus f ′′(x) = 0 if lnx = 1 =⇒ x = e.

If 0 < x < e, f ′′(x) > 0 and so f(x) is concave up.If x > e, f ′′(x) < 0 and f(x) is concave down and there is an inflection point at (e, 1).

Radioactive decay can be modelled using exponential functions. The mass of the radioactiveelement at time t is given by m(t) = m0e

−kt = m0e−t(ln 2)/t1/2 where m0 is the initial mass of

the sample and t1/2 is the half-life (time for half of the sample to decay) of the element.

Example 108. Radon 222, 222Rn, has a half-life of 3.8 days. How much of a 100 mg samplewill remain in 5 days ? How long does it take for the sample to be reduced to 10 mg ?m(t) = m0e

−t(ln 2)/t1/2 = 100e−t(ln 2)/3.8,so m(5) = 100e−5(ln 2)/3.8 ≈ 40.2 mg.We want t such that m(t) = 10 mg.So 100e−t(ln 2)/3.8 = 10,or e−t(ln 2)/3.8 = 0.1,

70

or −t(ln 2)3.8

= ln(0.1),

and so t = −(ln(0.1))(3.8)ln 2

≈ 12.6 days.

Example 109. Say the amount of radiocative material after t days satisfies N(t) = N0e−0.2t,

where N0 is the amount of material that you start with. At what time T does N(t) = 13N0?

What is N ′(T )?First, we solve

N0e−0.2t =

1

3N0,

which is equivalent to

e−0.2t =1

3,

which is equivalent to

−0.2t = ln(1

3),

which gives

T = 5 ln(3).

Next,

N ′(t) = N0e−0.2t(−0.2),

so

N ′(5 ln(3)) = −N0

5e− ln(3) = −N0

15.

71

Lecture 7: July 25

Derivatives of Trigonometric Functions. In Calculus, we always use radian measurefor angles – the formulas we will find below for the derivatives of the trigonometric functionsrequire radian measure to be true.

Recall the definitions of the trigonometric functions from a right-angled triangle.

cos θ =adjacent

hypotenuse

sin θ =opposite

hypotenuse

tan θ =opposite

adjacent

sec θ =1

cos θ=hypotenuse

adjacent

csc θ =1

sin θ=hypotenuse

opposite

cot θ =1

tan θ=

cos θ

sin θ=adjacent

opposite

Or, from the unit circle, x2 + y2 = 1.

x = cos θ and y = sin θ.

The graph of sin(x) looks like:72

cos(x) looks the same, but shifted by π2. tan(x) = sin(x)

cos(x)looks like:

All of the trigonometric functions are periodic, ie they repeat themselves over and over again.Some have period 2π while the others have period π – can you tell which are which ?

What would the derivatives of sinx and cosx be like ? We can make graphical approximationsand see that they will also be periodic and look like trigonometric functions themselves.

Notice thatd

dx(sinx) has the shape of cosx and that

d

dx(cosx) has the shape of − sinx.

However, because we cannot be sure of the y values of the derivatives from our graphical73

approximations, we cannot say at this point that the derivatives are exactly these functions.But we will see that they are.

d

dx(sinx) = lim

h→0

sin(x+ h)− sinx

h

= limh→0

sinx cosh+ cosx sinh− sinx

h(using the trig identity sin(θ ± φ) = sin θ cosφ± cos θ sinφ)

= limh→0

sinx(cosh− 1) + cos x sinh

h

= limh→0

sinx(cosh− 1)

h+ lim

h→0

cosx sinh

h

= (sinx) limh→0

cosh− 1

h+ (cosx) lim

h→0

sinh

h

To see what these limits are, look at the graphs. The functions are not defined at x = 0, butthe graphs show us that the limits exist.

We can see that limh→0

sinh

h= 1 and lim

h→0

cosh− 1

h= 0

And thusd

dx(sinx) = cos x (for x in radians).

74

We can use the trig identity cosx = sin(π2− x) and the chain rule to get

d

dx(cosx) =

d

dx

(sin(π

2− x))

= cos(π

2− x) d

dx

(π2− x)

= (sinx)(−1) = − sinx.

The derivatives of the other trigonometric functions are as follows.

d

dx(tanx) =

d

dx

(sinx

cosx

)=

ddx

(sinx)(cosx)− (sinx) ddx

(cosx)

(cosx)2

=(cosx)(cosx)− (sinx)(− sinx)

(cosx)2

=cos2 x+ sin2 x

cos2 x(using notation convention (sinx)n = sinn x)

=1

cos2 x(using trig identity cos2 x+ sin2 x = 1)

= sec2 x

d

dx(secx) =

d

dx

((cosx)−1

)= −(cosx)−2

d

dx(cosx)

= −(cosx)−2(− sinx)

=sinx

cos2 x

=

(1

cosx

)(sinx

cosx

)= secx tanx

d

dx(cscx) =

d

dx

((sinx)−1

)= −(sinx)−2

d

dx(sinx)

= −(sinx)−2(cosx)75

=− cosx

sin2 x

=

(−1

sinx

)(cosx

sinx

)= − cscx cotx

d

dx(cotx) =

d

dx

(cosx

sinx

)=

ddx

(cosx)(sinx)− (cosx) ddx

(sinx)

(sinx)2

=(− sinx)(sinx)− (cosx)(cosx)

(sinx)2

=− sin2 x− cos2 x

sin2 x

=−1

sin2 x= − csc2 x

Examples:

(1)d

dx(cos(3x)) = − sin(3x)(3) = −3 sin(3x)

(2)d

dx

(sin(x2)

)= cos(x2)(2x) = 2x cos(x2)

(3)d

dθ(θ tan θ) = tan θ + θ sec2 θ

(4)d

dt

(sec(t2 + 4t)

)= sec(t2 + 4t) tan(t2 + 4t)(2t+ 4) = (2t+ 4) sec(t2 + 4t) tan(t2 + 4t)

(5)d

dx(sin(tan(x+ 1))) = cos(tan(x+ 1)) sec2(x+ 1)

(6) If y = cos2(4x),dy

dx= 2 cos(4x)(− sin(4x))(4) = −8 sin(4x) cos(4x) = −4 sin(8x)

(using the double angle trig identity sin(2θ) = 2 sin θ cos θ).

(7) If g(x) = 2 cos2 x− 3x4 sin(x2),then g′(x) = 4 cosx(− sinx)− 12x3 sin(x2)− 3x4(cos(x2)(2x))= −4 sinx cosx− 12x3 sin(x2)− 6x5 cos(x2)= −2 sin(2x)− 12x3 sin(x2)− 6x5 cos(x2).

(8) If y =t cos t

t+ sin t,

dy

dt=

(cos t− t sin t)(t+ sin t)− (t cos t)(1 + cos t)

(t+ sin t)276

=t cos t− t2 sin t+ cos t sin t− t sin2 t− t cos t− t cos2 t

(t+ sin t)2

=cos t sin t− t2 sin t− t

(t+ sin t)2.

Example 110. Find the equation of the tangent line to y = tanx at x = π.The point on the curve is (π, 0).

The slope is m =dy

dx|x=π = sec2(π) = (−1)2 = 1,

so the line is y − 0 = (1)(x− π) or y = x− π.

Example 111. Analyze the function y = f(x) = x+ sinx on [0, 2π] and sketch its graph.f(x) is defined for all x in [0, 2π], so no vertical asymptotes.We’re considering only a finite interval, so no horizontal asymptotes.There is no symmetry because the interval is not symmetric around origin (though functionis odd for any interval of form [−a, a]).f(0) = 0 and f(2π) = 2π.f(x) = 0 if x+ sinx = 0, which has no solution.f ′(x) = 1 + cosx, so f ′(x) = 0 if cosx = −1 =⇒ x = π.If 0 < x < π, f ′(x) > 0 and f(x) is increasing.If π < x < 2π, f ′(x) > 0 and f(x) is increasing.So there are no local extrema, but the graph does have a horizontal tangent at x = π.f ′′(x) = − sinx, so f ′′(x) = 0 if x = π.If 0 < x < π, f ′′(x) < 0 and f(x) is concave down.If π < x < 2π, f ′′(x) > 0 and f(x) is concave up and there is an inflection point at (π, π).

We now look at some ‘word problems’ for trigonometric functions:

Example 112. The voltage in a wall socket is given by V (t) = 170 sin(120πt) volts for tmeasured in seconds.V ′(t) = 170 cos(120πt)(120π) = 20 400π cos(120πt),so V ′(t) = 0 if cos(120πt) = 0 or if 120πt = (2k + 1)π

2for k ∈ Z.

So we have the maximum voltage of 170 V if 120πt = (4k + 1)π2

or if t = 4k+1240

s.77

And we have the minimum voltage of −170 V if 120πt = (4k + 3)π2

or if t = 4k+3240

s.

The period of oscillation is T = 2π120π

= 160

s and the frequency is f = 60 Hz.

Example 113. The standard household voltage in many other countries is given by a differ-ent function. The wall socket voltage in India, for example, is given by V (t) = 325 sin(100πt).Which delivers the maximum 325 V when t = 4k+1

200s and the minimum −325 V when

t = 4k+3200

s, with a frequency of f = 50 Hz.

Midterm Review. Remember that there is a midterm at the start of next class. You maybring:

(1) A non-programmable, non-graphing calculator.(2) A single double-sided ‘cheat sheet.’

The midterm will be extremely similar to the first midterm, with the following basicchanges:

(1) Old question types removed: There will be no questions on limits, continuity, orcalculating derivatives from first principles.

(2) New question types added: There will now be optimization problems, and thegraphing problems will include questions about inflection points.

(3) Questions are harder: All of the basic calculus questions (product/chain/quotientrule calculations; graphing; optimization; etc) will appear, but with trig functionsand exponential functions as well as polynomials and rational functions.

(4) Questions are harder: There will be slightly more variety in the word problems.

To get perfect marks on the midterm, you should:

(1) Be familiar with all of the material from the first midterm.(2) Be able to find the inflection points of functions, and use this information while

plotting functions.(3) Be able to solve optimization problems.(4) Be able to do basic algebra and word problems that involve ax, ln(x), cos(x) and

sin(x).(5) Be able to do derivatives that involve ax, ln(x), cos(x) and sin(x), as well as poly-

nomials and rational functions. Be able to incorporate this knowledge while solvingthe above problem types (plotting, optimization, etc).

78

In the rest of these notes, I give a ‘super-midterm’ that covers all of the types of questionson the real midterm (as well as giving some additional questions). We probably won’t finishthis in class, so I strongly suggest that you go through the remainder of the questions byyourself!

(1) First midterm questions: See that review section!(2) Plot, using inflection points: Let f(x) = 2

(x−1)(x+1). Find the domain of f , the

points at which y = f(x) intersects with the Cartesian coordinate axes, the verticaland horizontal asymptotes of f(x), the critical points of f(x), the critical points off ′(x), and the regions on which f, f ′ are positive. Finally, plot f(x).

Answer:• The domain of f is (−∞,−1) ∪ (−1, 1) ∪ (1,∞).• f(0) = 2

−1 = −2, so f intersects that y-axis at y = −2. If f(x) = 0, then2

(x−1)(x+1)= 0. This doesn’t have any real solutions, so f does not intersect the

x-axis.• f has vertical asymptotes:

limx→−1−

f(x) =∞

limx→−1+

f(x) = −∞

limx→1−

f(x) = −∞

limx→1+

f(x) =∞.

f has horizontal asymptotes:

limx→∞

f(x) = limx→−∞

f(x) = 0.

• We have

f ′(x) =4x

(x2 − 1)2.

Thus, the only critical point of f is 0.• We have

f ′′(x) =4(x2 − 1)2 − 16x2(x2 − 1)

(x2 − 1)4

=4x((x2 − 1)− 4x)

(x2 − 1)3

=4x(x− 2−

√5)(x− 2 +

√5)

(x− 1)3.

Thus, the critical points of f ′ are {0, 2 +√

5, 2−√

5}.• Using the information about the critical points,

{x : f(x) > 0} = (−∞,−1) ∪ (1,∞)

{x : f ′(x) > 0} = (0,∞)

79

(3) Optimization Problems: Find the area of the largest rectangle that can be in-scribed inside a circle of radius 1.

Answer: Let (x, y) be the coordinates of the vertex of the rectangle in the firstquadrant. This rectangle has area A = 4xy, and the coordinates must satisfy x2+y2 =1. Thus, we have

A(x) = 4x√

1− x2,and we have 0 ≤ x ≤ 1. Taking derivatives,

A′(x) = 4√

1− x2 +4x

2√

1− x2× (−2x)

=4− 4x2 − 4x2√

1− x2

=4(1− 2x2)√

1− x2.

The only critical point in [0, 1] is x = 1√2; it has no vertical asymptotes in this interval.

Evaluating A at the boundary and the critical point, we have

A(0) = A(1) = 0

A(1√2

) = 1 > 0.

Thus, the largest possible area is 1.(4) Word problems involving trigonometry or exponentials: Let f(x) = sin(x) +

cos(x). Find the smallest value X of x > 0 for which f(x) = 0, and calculate f ′(x).Next, let g(s) = e2s. Find the smallest value S of s > 0 for which g(s) = 10g(0), andcalculate g′(s).

For the first: we need | sin(x)| = | cos(x)|, which means | sin(x)| = 1√2. Looking at

the sign of sin(x), cos(x) in the four quadrants, we have x = 3π4

. We then have

f ′(x) = cos(x)− sin(x),

so

f ′(3π

4) = − 1√

2− 1√

2= −√

2.

For the second: we need

e2s = 10,

so

2s = ln(10),

so

s =1

2ln(10).

We calculate

g′(s) = 2e2s,80

so

g′(1

2ln(10)) = 2eln(10) = 20.

(5) Derivative, graphing and optimization problems involving trigonometryand exponentials: We give a list.• Calculate the tangent to y = ex + e2x at x = 2.

We have dydx

= ex + 2e2x, so the equation for the tangent at x = 2 is:

(y − e2 − e4) = (e2 + 2e4)(x− 2).

• Differentiate f(x) cos(x2).By the chain rule,

f ′(x) = − sin(x2) (2x).

• Differentiate g(x) = ln(1 + ex).We have

g′(x) =1

1 + exex.

• Differentiate h(x) = x ln(cos(x)).By the chain and product rules,

h′(x) = ln(cos(x)) + xd

dxln(cos(x))

= ln(cos(x)) +x

cos(x)(− sin(x)).

81

Lecture 8: July 28

Midterm! Good luck!

Introduction to Vectors. The first thing we need to do is distinguish vectors from scalars.

A scalar is a quantity that describes magnitude or size only, like numbers (eg√

2, 1/3, −7),temperature (eg 12◦C), area (eg the area of a rectangle with sides of length 2 cm and 4 cmis 8 cm2), speed (eg 80 km/h) or distance (eg my friend’s house is 600 m away from mine).

A vector is a quantity that has both magnitude and direction, like velocity (eg 80 km/h East),displacement (eg my friend’s house is 600 m North of mine) or force (eg 20 N downward).

Example 114. See if you can distinguish whether the following are scalars or vectors:(i) my cat’s mass is 4.6 kg(ii) the acceleration of gravity is g = 9.8 m/s2 downward(iii) a boat is sailing at 10 knots westward(iv) my nephew has 37 video games

(i) mass is a scalar – but be careful, if I had said my cat weighs 45 N, that would be a forceand hence a vector (the direction would be downward)(ii) a vector since there is magnitude and direction(iii) a vector since there is magnitude and direction(iv) this is just a number, so a scalar

We can represent a vector in different ways:(i) in words, like 5 km East(ii) in a diagram as a geometric vector(iii) in a symbolic way as ~v (the arrow above the letter denotes that v is a vector, not a

scalar)

If we mean a directed line segment from point A (called the starting or initial point or tail)to point B (called the end or terminal point or tip or head) like in the diagram above, we

write ~AB.

The magnitude or size or length of a vector is written using absolute value bars, so themagnitude of vector ~v is |~v|.

Example 115. The magnitude of the vector 2 cm at an angle of 20◦ to the horiziontal is itslength, 2 cm.

82

There are many reasonable ways to represent the direction. The most common in mathclass is to measure the angle with respect to the horizontal:

Example:

This is 5 m/s at 215◦ to the horizontal (or 5 m/s at −145◦ to the horizontal) or 5 m/s at abearing of 235◦ or 5 m/s S55◦W.Can you see where these angles are coming from ?

Two vectors are said to be parallel if they have the same or opposite directions (but notnecessarily the same magnitude)

Example:In trapezoid ABCD,

~AB ‖ ~DC (and ~AB ‖ ~CD).

Two vectors that have the same direction and magnitude are said to be equivalent or equal(their actual locations in space do not matter).

Example:In parallelogram PQRS,

vectors ~PS and ~QR are equivalent, which we write as ~PS = ~QR.We also have that ~SP = ~RQ, ~PQ = ~SR and ~QP = ~RS.

83

Opposite vectors have the same magnitude but opposite direction (again, locations do notmatter). The opposite of vector ~v is written as −~v.

So, in the parallelogram above, ~PS = − ~RQ, ~PQ = − ~RS, ~QR = − ~SP and ~SR = − ~QP .We can now draw vectors. We now talk about doing math with vectors. We begin with

addition and subtraction of vectors. To add two vectors, we just put the arrows one afteranother and draw the line from the tail of the first to the tip of the second. When we addtwo (or more) vectors, we obtain a single vector, often called the resultant.

Let’s see this through an example.

Example:Suppose you walked 400 m East and then 500 m South.

tan θ =500

400=

5

4, so θ = arctan(5/4) ≈ 51.3◦.

Then, with the help of Pythagoras and trigonometry, we know that this is equivalent tohaving walked

√(400)2 + (500)2 ≈ 640.3 m in direction S38.7◦E.

This shows us how we can add vectors – by joining them head to tail. Note that drawingthe sum of two vectors is extremely easy! You only have to think when you want to writedown a formula for the angles.

Suppose we have vectors ~a and ~b

to add them and produce the vector ~a+~b, we move ~b so that its tail touches the head of ~a

84

and then the vector ~a+~b is the vector that joins the tail of ~a to the tip of ~b.Can you see how this is exactly what we did in the walking example above ?

Also, notice this

ie ~a+~b = ~b+ ~a and so vector addition is commutative.

Vector addition is also associative, ie (~a+~b) + ~c = ~a+ (~b+ ~c) and so the order in which weadd vectors does not matter (but it will when we subtract vectors below).

Do we do anything special if the vectors are parallel ? No, we do exactly the same thing.

To subtract vectors, we simply recognize that subtraction is adding the negative, which is

the opposite when dealing with vectors, so ~a−~b = ~a+ (−~b).

Notice that we could also do this

we can find ~a−~b by joining the vectors tail to tail and joining the head of ~b to the head of ~a.

But then

the vector ~b− ~a is the opposite of ~a−~b and so order in subtraction does matter (just as itdoes with scalars).

Suppose we take any vector ~v and subtract it from itself, the resultant is ~v − ~v = ~0, calledthe zero vector, which has length or magnitude equal to 0 and points in no specific direction.

85

The identity property for vector addition should then be clear, ~v +~0 = ~v = ~0 + ~v.

Can you see how the properties of vector addition seem to be very similar to those of scalar(number) addition ?

Example:Consider the parallelogram ABCD with diagonals AC and BD that intersect at E.

We can write vectors in the figure as sums and differences of others, like ~DB = ~DA + ~AB,~DC = ~DE + ~EC, ~AE = ~DE − ~DA, and so on...

Try writing some others this way.

Example 116. We can simplify expressions like the following:((~u+ ~v)− ~u)− ~v= ((~v + ~u)− ~u)− ~v (commutativity)= ((~v + ~u) + (−~u))− ~v (subtraction is adding opposite)= (~v + (~u+ (−~u))− ~v (associativity)

= (~v +~0)− ~v (opposites add to zero vector)= ~v − ~v (identity property)= ~v + (−~v)

= ~0

We can also multiply a vector by a scalar. If we take a vector ~v and multiply it by ascalar k (any real number), we are performing scalar multiplication and have produced thescalar multiple k~v. The vector k~v will be |k| times as long as ~v and it will be parallel to ~v.

The magnitude of k~v is |k~v| = |k||~v|.If k > 0, k~v is in the same direction as ~v.If k < 0, k~v is in the opposite direction to ~v.

86

If |k| > 1, k~v is longer than ~v.If |k| < 1, k~v is shorter than ~v.

Example:

Vectors that are parallel are also said to be collinear because they would lie on a straightline when arranged tail to head. But then they would also be scalar multiples of each other.

ie there is some k ∈ R such that ~u = k~v.

The properties of scalar multiplication are, for any vectors ~u and ~v and scalars k and c,(i) k(~u+ ~v) = k~u+ k~v (distributivity)(ii) k(c~v) = (kc)~v (associativity)(iii) 1~v = ~v (identity)

Example 117. 2(3~u− ~v) + 4~v= 2(3)~u+ 2(−~v) + 4~v= 6~u− 2~v + 4~v= 6~u+ (−2 + 4)~v= 6~u+ 2~v

A vector of the form s~u + t~v (where s and t are scalars) is called a linear combination ofvectors ~u and ~v.Like we have in the examples above and below.

Example:In the trapezoid ABCD, ~AB ‖ ~DC and DC = 4AB.

87

If we let ~DA = ~u and ~AB = ~v, then we can write ~DC = 4~v, ~AC = ~AD + ~DC = −~u + 4~vand ~BC = ~BA+ ~AD + ~DC = −~v + (−~u) + 4~v = −~u+ 3~v

The idea of linear combinations of vectors is central to linear algebra.Two vectors that are perpendicular to each other and add together to give a vector ~v are

called rectangular vector components of ~v.

Example:A cannonball of mass 100 kg is fired horizontally out of a cannon with a force of 2500 N.Gravity will act vertically (downward) with a force of (9.8 m/s2)(100 kg) = 980 kgm/s2 =980 N.So we have

and the horizontal force of 2500 N and the vertical of 980 N are rectangular components ofthe resultant force ~F .The magnitude of the resultant force is |~F | =

√(2500)2 + (980)2 ≈ 2685 N.

The direction is at angle θ below the horizontal, where tan θ =980

2500=

98

250=

49

125,

so θ = arctan(49/125) ≈ 21.4◦.

88

Municipal Holiday: August 1

See you on Thursday!

89

Lecture 9: August 4

Review Quiz! None for today - last class had few calculations.

Physics Examples. An equilibrant vector ~E is a vector that balances another vector orcombination of vectors and hence is equal in magnitude and opposite in direction to theresultant ~R.

We need to understand this idea to solve tension problems.

Tension is the equilibrant force in a rope or chain that keeps an object in place (or station-nary).

Example:A picture that weighs 60 N is hanging from a wire (attached to the picture frame) placed ona hook on the wall such that the hook is in the centre of the wire and the two segments ofwire have an angle of 120◦ between them.So we have

by symmetry, the tensions in the two wire segments will be equal in magnitude, ie | ~T1| = | ~T2|and the angles they make with the horizontal must also be equal.In fact, θ = 180◦ − 90◦ − 60◦ = 30◦.The 60 N weight of the picture is the resultant of the system, so we must have that~T1 + ~T2 = 60 N upward.We can redraw the system as

we have an isosceles triangle where α = 60◦ and hence the triangle must actually be equilat-eral. Thus | ~T1| = | ~T2| = 60 N.

90

Any vector can be resolved into rectangular (perpendicular) components, typically hori-zontal and vertical.

Suppose we have a vector ~v at angle θ to the horizontal

it can be resolved into its horizontal component ~vh and vertical component ~vv, where~v = ~vh + ~vv (and ~vh ⊥ ~vv).

then trigonometry tells us that |~vh| = |~v| cos θ and |~vv| = |~v| sin θ.

Example 118. A child pulls a wagon with a force of 40 N at an angle of 25◦ to the horizontal.Then the horizontal component has magnitude | ~Fh| = |~F | cos θ = (40)(cos(25◦)) ≈ 36.3 N

and the vertical component has magnitude | ~Fv| = |~F | sin θ = (40)(sin(25◦)) ≈ 16.9 N.

Example:An object that weighs 75 N is resting on an inclined plane that makes an angle of 15◦ withthe horizontal.

Since the object is at rest, there is a frictional force ~f acting parallel to the ramp and anormal force ~n acting perpendicularly to the ramp that balances the weight (ie the force ofgravity).So we have

and so |~f | = (75)(sin(15◦)) ≈ 19.4 N and |~n| = (75)(cos(15◦)) ≈ 72.4 N.91

Continuing Study of Vectors. Consider any vector ~v in the plane – its initial and terminalpoints, Q and R, can be defined using Cartesian coordinates. If we translate ~v so that itstail is at the origin O, then its head will be at some point P , with coordinates (a, b). This

representation of ~v (ie ~OP ) is called the position vector [a, b].

And any vector with the same magnitude and direction can be represented as [a, b].92

If we resolve ~v = [a, b] into horizontal and vertical vector components, we’ll have a vector oflength a along the x-axis and one of length b along the y-axis (which will add up to ~v).

We define the unit vectors ı and to be vectors of length 1 (hence the use of the word unit)that point in the positive x and y directions, respectively.So ı has position vector [1, 0] and it’s [0, 1] for and so we can write that ı = [1, 0] and = [0, 1].

Then the horizontal and vertical components of ~v are ~vh = aı = a[1, 0] = [a, 0] and~vv = b = b[0, 1] = [0, b]. And thus ~v = ~vh + ~vv = [a, 0] + [0, b] = [a, b] or ~v = aı+ b.And we also see that |~v| =

√a2 + b2 (by Pythagoras).

Example:

The vector [2, 3] has horizontal component of length 2, vertical component of length 3 and

length or magnitude√

(2)2 + (3)2 =√

13. It can be written as [2, 3] = [2, 0] + [0, 3] =2[1, 0] + 3[0, 1] = 2ı+ 3.

Suppose we add vectors ~u = [u1, u2] and ~v = [v1, v2]:~u+ ~v = [u1, u2] + [v1, v2]= [u1, 0] + [0, u2] + [v1, 0] + [0, v2]= u1ı+ u2+ v1ı+ v2= (u1 + v1)ı+ (u2 + v2)= (u1 + v1)[1, 0] + (u2 + v2)[0, 1]= [u1 + v1, u2 + v2]

Once we understand how we add vectors in this manner, we do not have to write out allthese details.

And scalar multiplication is k~v = k[v1, v2] = k(v1ı+ v2) = kv1ı+ kv2 = [kv1, kv2].93

So the opposite of ~v is −~v = [−v1,−v2].

Example 119. Suppose ~u = [2, 7] and ~v = [1, 3], then(i) ~u+ ~v = [2 + 1, 7 + 3] = [3, 10] = 3ı+ 10(ii) ~u− ~v = [2− 1, 7− 3] = [1, 4] = ı+ 4(iii) −3~v = [−3(1),−3(3)] = [−3,−9] = −3ı− 9(iv) 2~u+ 4~v = [4, 14] + [4, 12] = [8, 26] = 8ı+ 26

Given any vector ~v = [v1, v2] = v1ı + v2, we can always find a unit vector in the same

direction by dividing by the magnitude, ie v =~v

|~v|=

v1ı+ v2√v21 + v22

and one in the opposite

direction by multiplying by −1, ie −v.

Example 120. If ~v = [−2, 3], then v =[−2, 3]√

(−2)2 + (3)2= 1√

13[−2, 3] = 1√

13(−2ı+ 3).

Suppose we have two points P1 = (x1, y1) and P2 = (x2, y2),

then the vector that joins P1 to P2 is ~P1P2 = ~OP2− ~OP1 = [x2, y2]−[x1, y1] = [x2−x1, y2−y1]and | ~P1P2| =

√(x2 − x1)2 + (y2 − y1)2 (which is the distance between the points).

Example 121. The vector that joins point A = (−1, 0) to point B = (5, 2) is ~AB =

[5− (−1), 2− 0] = [6, 2] and its length is√

(6)2 + (2)2 =√

40 = 2√

10.

We have also seen that the horizontal and vertical vector components are expressible as|~vh| = |~v| cos θ and |~vv| = |~v| sin θ, where θ is the angle (measured counterclockwise) thevector makes with the positive x-axis.

~v = [v1, v2] = [|~vh|, |~vv|] = [|~v| cos θ, |~v| sin θ] = |~v| cos θı+ |~v| sin θ = |~v|(cos θı+ sin θ)

Can you then see that v = cos θı+ sin θ ?

94

Example 122. A force of 150 N at 15◦ to the horizontal is then~F = [|~F | cos θ, |~F | sin θ] = [(150) cos(15◦), (150) sin(15◦)] ≈ [144.9, 38.8] N = 144.9ı+38.8 N.

We have seen how to multiply vectors by scalars. It is much less obvious what it shouldmean to multiply two vectors. It turns out that we will actually have a few ways to do this!

The first method is called the dot product. The dot product of two vectors ~u and ~v isdefined to be ~u · ~v = |~u||~v| cos θ (read as ~u dot ~v), where θ is the acute (0 ≤ θ ≤ 180◦) anglebetween the vectors ~u and ~v when arranged tail to tail. The dot product will clearly be

commutative, ie ~u ·~v = ~v ·~u. Also, the dot product produces a scalar (not vector) result andrepresents a measurement of the projection of one vector onto the other, or of the tendencyof the vectors to point in the same direction.

Example:

~u · ~v = (75)(92) cos(40◦) ≈ 5286

We can notice some of the properties of the dot product immediately:

(1) if ~u or ~v is ~0, then ~u · ~v = 0

(2) also if θ = 90◦, then ~u · ~v = 0 (vectors are orthogonal)

(3) if 0 < θ < 90◦, then ~u · ~v > 0 (vectors point in same general direction)

(4) but if 90◦ < θ < 180◦, then ~u · ~v < 0 (vectors point in generally opposite directions)

(5) if θ = 0, then ~u · ~v = |~u||~v| (and so ~v · ~v = |~v|2)

(6) if θ = 180◦, then ~u · ~v = −|~u||~v|

(7) ı · = · ı = 095

The dot product is used to calculate work, which is the product of the magnitude of dis-placement of an object with the magnitude of the force applied in the direction of motion.

Example:A child pulls a sled 40 m up a hill with a force of 125 N at an angle of 20◦ to the surface ofthe hill.

The work done isW = (displacement)(horizontal component of force)

= | ~AB||~F | cos(20◦)

= ~AB · ~F= (40 m)(125 N) cos(20◦)≈ 4698 J.

Other properties of the dot product are:(i) if ~u 6= ~0 and ~v 6= ~0, then ~u · ~v = 0 if and only if ~u ⊥ ~v(ii) for any scalar k, (k~u) · ~v = k(~u · ~v) = ~u · (k~v) (associative)(iii) ~u · (~v + ~w) = ~u · ~v + ~u · ~w and (~u+ ~v) · ~w = ~u · ~w + ~v · ~w (distributive)

There is a nice way to calculate ~u · ~v using the Cartesian representations of the vectors.

By the cosine law, |~w|2 = |~u|2 + |~v|2 − 2|~u||~v| cos θ = |~u|2 + |~v|2 − 2~u · ~v.But ~w = ~u− ~v, so we have~u · ~v = 1

2(|~u|2 + |~v|2 − |~u− ~v|2)

= 12(u21 + u22 + v21 + v22 − ((u1 − v1)2 + (u2 − v2)2))

= 12(u21 + u22 + v21 + v22 − u21 + 2u1v1 − v21 − u22 + 2u2v2 − v22)

= u1v1 + u2v2

ie ~u · ~v = [u1, u2] · [v1, v2] = u1v1 + u2v2

Example 123. (1) [−2, 4] · [3, 2] = (−2)(3) + (4)(2) = −6 + 8 = 2

(2) Suppose ~u = [0, 2] and ~v = [7, 3],then ~u · ~v = (0)(7) + (2)(3) = 6and (3~u− ~v) · (~u+ 5~v) = [−7, 3] · [35, 17] = (−7)(35) + (3)(17) = −245 + 51 = −194.

96

Let’s look at what the dot product can do for us. We have already seen that the workdone by a force ~F in direction of a displacement ~s = ~AB is W = ~F · ~s.Example 124. A force of 30 N is acting in the direction of the vector ~v = [3, 2] and isexerted on an object moving from point (0, 2) to point (2, 7) (distances measured in metres).What is the work done ?

A unit vector in the direction of ~v is v =~v

|~v|=

[3, 2]√(3)2 + (2)2

= 1√13

[3, 2].

So the force is ~F = (30 N)v = 30√13

[3, 2] = 30√13

(3ı+ 2) N.

The displacement is ~s = (2− 0)ı+ (7− 2) = 2ı+ 5 m.So the work done isW = ~F · ~s = 30√

13(3ı+ 2) · (2ı+ 5) J = 30√

13((3)(2) + (2)(5)) J = 480√

13J ≈ 133 J.

Since the dot product is ~u·~v = |~u||~v| cos θ, we have that cos θ =~u · ~v|~u||~v|

and so θ = arccos

(~u · ~v|~u||~v|

)and hence we can find the angle between the vectors.

Example:The angle between the vectors ~u = [6,−2] and ~v = [−1, 4] is

θ = arccos

(~u · ~v|~u||~v|

)= arccos

([6,−2] · [−1, 4]√

(6)2 + (−2)2√

(−1)2 + (4)2

)= arccos

((6)(−1) + (−2)(4)√

40√

17

)= arccos

(−14√40√

17

)≈ 122.5◦.

The shadow of one vector ~v onto another ~u is called the projection of ~v on ~u and is writtenproj~u~v.

If the angle between the vectors, θ, is less than 90◦, then proj~u~v is the component of ~v in

the direction of ~u and thus proj~u~v = |~v| cos θ u = |~v| cos θ

(~u

|~u|

)=

(|~v||~u|

cos θ

)~u.

If 90◦ < θ < 180◦, proj~u~v is in the direction opposite to ~u and thus

proj~u~v = |~v| cos(180◦ − θ)(−u) = |~v|(− cos θ)

(−~u|~u|

)=

(|~v||~u|

cos θ

)~u

(the same formula as above).97

And so proj~u~v =

(|~v||~u|

cos θ

)~u =

(|~v||~u| cos θ

|~u||~u|

)~u =

(~v · ~u~u · ~u

)~u.

Example:The projection of ~v = 2ı− 4 on ~u = ı+ 3 is

proj~u~v =

(~v · ~u~u · ~u

)~u

=

((2ı− 4) · (ı+ 3)

(ı+ 3) · (ı+ 3)

)(ı+ 3)

=

((2)(1) + (−4)(3)

(1)(1) + (3)(3)

)(ı+ 3)

= −1010

(ı+ 3)= (−1)(ı+ 3)= −ı− 3 = −~u.

Notice that if θ = 90◦, proj~u~v = ~0 as there is no shadow of ~v on ~u or, equivalently, there isno component of ~v in the direction of ~u.

We can use the dot product to calculate other quantities. For example, suppose the CandyStore sold 40 bags of jelly beans and 25 giant jawbreakers one week. Then we could representthese sales with the vector [40, 25]. If a bag of jelly beans sells for $1.25 and a giant jawbreakerfor 50 cents, we could represent these prices with the vector [1.25, 0.50]. The dot product ofthese vectors [40, 25] · [1.25, 0.50] = (40)(1.25) + (25)(0.50) = $62.50 would be the revenuefrom these sales.

So far, we have only looked at vectors in two dimensions. We will now look at vectors inthree dimensions. This will be very similar to two dimensions. The biggest difference is thatit is much harder to draw pictures - we will need to rely more on calculations.

The three-dimensional Cartesian coordinate system isthis system is called right-handed since, if we were to curl the fingers of our right hand fromthe x-axis towards the y-axis, our thumb would point in the direction of the z-axis.

Points are specified by ordered triples P = (x, y, z). The axes divide space into 8 octants,the one where all three coordinates are positive is called the first octant (but there is no

98

agreement on how to label the remaining seven). The origin O = (0, 0, 0) is the point wherethe three axes intersect.

Example:The point (2, 3,−1) is 2 units in front, 3 units to the right and 1 unit below the origin.

The unit vectors in 3-space are ı = [1, 0, 0], = [0, 1, 0] and k = [0, 0, 1].

The position vector of the point P = (x, y, z) is ~v = ~OP (from the origin O to the point P ).~OP = [x, y, z] = xı+ y+ zk.

The magnitude (or length) of the vector would be |~v| =√x2 + y2 + z2 (from Pythagoras).

Example 125. The vector ~v = [2, 4,−1] = 2ı+4−k has magnitude |~v| =√

(2)2 + (4)2 + (−1)2 =√21.

A scalar multiple of vector ~v = [v1, v2, v3] is k~v = [kv1, kv2, kv3], where k ∈ R and which is avector collinear with ~v.

Example 126. For what value of c is ~u = [2, c, 1] collinear with ~v = [4, 6, 2] ?For ~u to be collinear with ~v, it must be a scalar multiple of it, ie ~u = k~vor [2, c, 1] = k[4, 6, 2] = [4k, 6k, 2k] or 2 = 4k, c = 6k and 1 = 2k.So clearly k = 1/2 and thus c = 3.

For two vectors ~u = [u1, u2, u3] and ~v = [v1, v2, v3], ~u + ~v = [u1 + v1, u2 + v2, u3 + v3] and~u− ~v = [u1 − v1, u2 − v2, u3 − v3].

And the vector from point P1 = (x1, y1, z1) to point P2 = (x2, y2, z2) is~P1P2 = [x2 − x1, y2 − y1, z2 − z1].

Example 127. Suppose ~u = 2ı− + 3k and ~v = ı+ 3− 4k, then

(1) ~u+ ~v = (2 + 1)ı+ (−1 + 3)+ (3− 4)k = 3ı+ 2− k

(2) ~v − ~u = (1− 2)ı+ (3− (−1))+ (−4− 3)k = −ı+ 4− 7k

(3) 2~u−3~v = 2(2ı− +3k)−3(ı+3−4k) = (4ı−2+6k)−(3ı+9−12k) = ı−11+18k

99

(4) A unit vector in the direction of ~u is u =~u

|~u|=

2ı− + 3k√(2)2 + (−1)2 + (3)2

= 1√14

(2ı− +

3k).

The dot product of ~u = [u1, u2, u3] and ~v = [v1, v2, v3] is ~u·~v = |~u||~v| cos θ = u1v1+u2v2+u3v3.

Example 128. (1) If ~u = [−1, 5, 0] and ~v = [2, 6,−1], then ~u · ~v = (−1)(2) + (5)(6) +(0)(−1) = 28.

(2) What is the angle bewteen ~u = ı− 2+ 3k and ~v = 4ı+ 7+ k ?

θ = arccos

(~u · ~v|~u||~v|

)= arccos

((1)(4) + (−2)(7) + (3)(1)√

(1)2 + (−2)2 + (3)2√

(4)2 + (7)2 + (1)2

)= arccos

(−7√

14√

66

)≈ 103.3◦.

Are you comfortable with the two notations for vectors ?

Two nonzero vectors ~u and ~v are orthogonal if and only if ~u · ~v = 0. So if we want to find avector orthogonal (perpendicular) to a given one, we use the dot product to help us.

Example 129. Find a vector orthogonal to ~p = [2,−3, 5].Let ~v = [x, y, z] be a vector orthogonal to ~p, then ~p · ~v = 0, which means that[2,−3, 5] · [x, y, z] = 0 or 2x− 3y + 5z = 0. So there are infinitely many vectors orthogonalto ~p (all of which will lie in the same plane) and we can pick any solution to the equation,say ~v = [1,−1,−1].

Now, we’ll define a second type of multiplication. This time, the product of two vectors willbe a vector. The cross product of two vectors ~u and ~v is defined to be ~u× ~v = |~u||~v| sin θ n(which is read as ~u cross ~v) where θ is the acute angle between ~u and ~v and n is a unit vectorperpendicular to both ~u and ~v such that ~u, ~v and n form a right-handed system (curling thefingers of the right hand from ~u towards ~v gives the thumb pointing in the direction of n).

100

Immediately, we have that ~v× ~u = |~v||~u| sin θ (−n) = −~u×~v and hence the cross product isnot commutative.

If ~u = [u1, u2, u3] and ~v = [v1, v2, v3], we calculate the components of ~u× ~v this way~u× ~v = [u2v3 − u3v2, u3v1 − u1v3, u1v2 − u2v1].

Example 130. If ~u = 2ı− + 3k and ~v = ı+ 3+ 4k, then~u× ~v = ((−1)(4)− (3)(3))ı+ ((3)(1)− (2)(4))+ ((2)(3)− (−1)(1))k = −13ı− 5+ 7k,also (~u× ~v) · ~u = [−13,−5, 7] · [2,−1, 3] = (−13)(2) + (−5)(−1) + (7)(3) = 0,and (~u× ~v) · ~v = [−13,−5, 7] · [1, 3, 4] = (−13)(1) + (−5)(3) + (7)(4) = 0and so ~u× ~v is orthogonal to both ~u and ~v (as it should be).

The magnitude of the cross product |~u× ~v| = |~u||~v| sin θ is equal to the area of the parallel-ogram formed by ~u and ~v.

The area is A = |~v|h = |~v||~u| sin θ.

So the area of the parallelogram formed by ~u and ~v in our example above is√(−13)2 + (−5)2 + (7)2 =

√243 ≈ 15.6.

Suppose that ~u× ~v = ~0, what would have to be true ?Either ~u = ~0 or ~v = ~0 or ~u and ~v are parallel (ie θ = 0) and there is k ∈ R such that ~u = k~v.

Other properties of the cross product are:(i) ~u× (~v + ~w) = ~u× ~v + ~u× ~w (distributive)(ii) (~u+ ~v)× ~w = ~u× ~w + ~v × ~w (distributive)(iii) k(~u× ~v) = (k~u)× ~v = ~u× (k~v) (associative)

We can use the cross product to find the angle between vectors (though using the dot product

would be easier) as |~u×~v| = |~u||~v| sin θ means that sin θ =|~u× ~v||~u||~v|

and so θ = arcsin

(|~u× ~v||~u||~v|

).

For our example, |~u| =√

(2)2 + (−1)2 + (3)2 =√

14 and |~v| =√

(1)2 + (3)2 + (4)2 =√

26,

so θ = arcsin

(15.6√14√

26

)≈ 54.9◦.

Since ı = [1, 0, 0] and = [0, 1, 0],ı× = [(0)(0)− (0)(1), (0)(0)− (1)(0), (1)(1)− (0)(0)]

= [0, 0, 1] = k.

Can you see this visually from the coordinate system ?

101

We now do some physics problems related to dot-products and cross-products. The for-mulas for work and projection are the same in 3-space as they are in 2-space.

Example 131. A force ~F = [20, 100, 75] N acts on an object with displacement ~d =

[2, 4, 5] m, so the work done is W = ~F · ~d = [20, 100, 75] · [2, 4, 5] = (20)(2) + (100)(4) +(75)(5) = 815 J.

Gravity acts in direction −k (ie downward), so the work done against gravity is[0, 0, 75] · [0, 0, 5] = 375 J.

Example 132. The projection of ~v = [2,−1, 3] on ~u = [3, 2, 7] is

proj~u~v =

(~v · ~u~u · ~u

)~u =

([2,−1, 3] · [3, 2, 7]

[3, 2, 7] · [3, 2, 7]

)[3, 2, 7]

=

((2)(3) + (−1)(2) + (3)(7)

(3)2 + (2)2 + (7)2

)[3, 2, 7]

= 2562

[3, 2, 7]

= 7562ı+ 50

62+ 175

62k.

Torque is a measure of the force ~F acting on an object that causes it to rotate and is given

by ~τ = ~r× ~F , where ~r is the torque arm, the vector from the pivot point to the point where

the force ~F acts.

Example:A wrench of length 30 cm is used to tighten a bolt. A force of 50 N is applied in a clockwisedirection at an angle of 75◦ to the handle.

or

so the torque is ~τ = ~r× ~F = |~r||~F | sin θ n = (0.3 m)(50 N) sin(75◦) n ≈ 14.5 Nm inward (thebolt is being tightened).

We use Nm as the units of the vector torque and not J which is used for the scalar energy.102

If we have the vectors ~u, ~v and ~w, we can define the scalar triple product to be ~w ·~u×~v (thecross product has to be done first in order for the product to make sense). This representsthe volume of the parallelopiped formed by the vectors. We use absolute value bars if theresult is negative.

Example:The volume of the parallelopiped formed by ~u = [1, 1, 1], ~v = [2, 2, 4] and ~w = [3, 1, 6] is~w · ~u× ~v = [3, 1, 6] · [1, 1, 1]× [2, 2, 4]= [3, 1, 6] · [(1)(4)− (1)(2), (1)(2)− (1)(4), (1)(2)− (1)(2)]= [3, 1, 6] · [2,−2, 0]= (3)(2) + (1)(−2) + (6)(0)4

103

Lecture 10: August 8

Review Quiz! Having studied vectors, we now study lines. In two-space, a line can begiven in slope-intercept form y = mx+ b or with the scalar equation Ax+By + C = 0.

Example:

Example 133 (h).

The line y = 2x + 6 has slope m = 2 and y-intercept b = 6. Its scalar equation form wouldbe 2x− y + 6 = 0.

Example 134 (h).

We can represent this line with a vector equation. A direction vector parallel to the line is~m = [1, 2]. A position vector that has its tip on the line would be ~r0 = [1, 8] (since the point(1, 8) is on the line). Another position vector ~r = [3, 12] also touches the line (since (3, 12)is on the line). Then, if we let ~s be the vector joining (1, 8) to (3, 12), we would have that~r = ~r0+~s. But since ~s is parallel to ~m, we have that ~r = ~r0+t~m. In this particular case, t = 2– can you see that ? If we were to let t vary, we could get any other point on the line, so wehave that the vector equation of the line is ~r = ~r0 + t~m, t ∈ R or [x, y] = [x0, y0] + t[m1,m2].Notice that this is not unique – we can use any point (x0, y0) on the line and any directionvector ~m parallel to it.

104

If we separate the components, we have the parametric equations of the line, x = x0 + tm1,y = y0 + tm2, t ∈ R.

So for our example above, x = 1 + t and y = 8 + 2t.

Example 135. Is the point (−1, 4) on the line ? How about the point (2, 9) ?If (−1, 4) is on the line, then we must have that −1 = 1 + t and 4 = 8 + 2t for the samevalue of t. So, yes, (−1, 4) is on the line since t = −2 satisfies both equations.For the other point, 2 = 1 + t requires t = 1 but 9 = 8 + 2t requires that t = 1/2 (ie not thesame value) so the point (2, 9) is not on the line.

Example 136. Find the vector equation of the line passing through the points P1 = (−1, 2)and P2 = (2,−4).

We need a direction vector, so we use ~m = ~P1P2 = [3,−6]. We can use either point, so take~r0 = [−1, 2] and thus ~r = ~r0 + t~m is [x, y] = [−1, 2] + t[3,−6] (and then P2 corresponds tot = 1 ).

Two lines L1 and L2 are parallel if they have the same slope and are coincident if they arethe same line. For lines in vector equation form, they will be parallel if the direction vectorsare parallel.

So the lines in our two examples above are not parallel, whereas the linesL1 ~r = (2 + 3t)ı+ (1− 2t) and L2 ~r = (6− 6t)ı+ (4 + 4t) are parallel.(The notation has changed – can you see what you need to see to see that the lines are par-allel ?)

Let’s go back to our example y = 2x + 6. We know that ~m = [1, 2] is a direction vectorfor the line, so any vector ~n that is perpendicular to ~m (ie ~m · ~n = 0 ) will be normal orperpendicular to the line, like ~n = [2,−1]. But notice the scalar equation 2x− y+ 6 = 0. Inother words, given the scalar equation Ax+ By + C = 0 for a line, the vector ~n = [A,B] isnormal to it.

What do we do in three-space ?First of all, a scalar equation of three variables Ax + By + Cz + D = 0 describes a planein three-space, not a line, so we do not have a scalar equation, nor do we have a slope-intercept form. But we do have a vector equation and parametric equations. The line pass-ing through P0 = (x0, y0, z0) with direction vector ~m = [m1,m2,m3] is ~r = ~r0 + t~m, t ∈ Ror [x, y, z] = [x0, y0, z0]+t[m1,m2,m3] or x = x0+tm1, y = y0+tm2 and z = z0+tm3 for t ∈ R.

Example 137. The line passing through P1 = (1, 0, 5) and P2 = (4, 2, 1) has ~m = [3, 2,−4],so~r = [4, 2, 1] + t[3, 2,−4] (and then P1 corresponds to t = −1 ).Does (10, 5,−7) lie on the line ?4 + 3t = 10 =⇒ t = 2

105

2 + 2t = 5 =⇒ t = 3/21− 4t = −7 =⇒ t = 2 so no, it does not.

All of the vectors normal to a line in two-space are parallel, but this is not the case in three-space.

In class, we will be able to more easily see this by rotating a piece of chalk around theplane that is perpendicular to a pen.

Having discussed lines, we now discuss planes.Example:Consider the scalar equation x+y−z−2 = 0. if we set two of the coordinates equal to 0, wecan see where the graph of the object would cross the axes. If y = z = 0, x+ (0)− (0)− 2 =0 =⇒ x = 2, so the x-intercept is 2. If x = z = 0, (0) + y − (0)− 2 = 0 =⇒ y = 2, so they-intercept is 2. If x = y = 0, (0) + (0)− z − 2 = 0 =⇒ z = −2, so the z-intercept is −2.

106

This object is certainly not a line. The graph of this equation is a plane, a flat, two-dimensional surface in three-space that extends off to infinity in every direction.

Other points on the plane are P = (2, 2, 2), Q = (0, 7, 5), R = (6, 3, 7) and S = (4, 0, 2).There are infinitely many points on the plane – can you see that these points satisfy theequation ?However, the point (10, 5, 7) is not on the plane as (10) + (5)− (7)− 2 6= 0 (ie the point doesnot satisfy the equation).

Using the points on the plane, we can find direction vectors that are parallel to the plane(and not to each other). For example, ~PQ = [−2, 5, 3] and ~RS = [−2,−3,−5].

Suppose we have two non-parallel direction vectors ~a and ~b on a plane.

Then, the vector from any point P0 = (x0, y0, z0) to any other point P = (x, y, z) on the

plane can be represented by a linear combination of the direction vectors, ie ~P0P = t~a+ s~b,where t, s ∈ R. But since ~P0P = [x, y, z]− [x0, y0, z0], we can rearrange the vector equation

to write [x, y, z] = [x0, y0, z0]+t~a+s~b = [x0, y0, z0]+t[a1, a2, a3]+s[b1, b2, b3] or ~r = ~r0+t~a+s~band we have the vector equation of a plane.

So for our example above, we can take P0 = (2, 2, 2), ~a = [−2, 5, 3] and ~b = [−2,−3,−5], so[x, y, z] = [2, 2, 2] + t[−2, 5, 3] + s[−2,−3,−5]. Notice that if t = −1/4 and s = 1/4, we have[x, y, z] = [2, 0, 0], the position vector of the x-intercept.

The parametric equations of a plane would be x = x0 + ta1 + sb1, y = y0 + ta2 + sb2 andz = z0 + ta3 + sb3 where t, s ∈ R.

So for our example, x = 2− 2t− 2s, y = 2 + 5t− 3s and z = 2 + 3t− 5s. Picking values fort and s will give us other points on the plane, like if t = 1 and s = −2, we’ll have (4, 13, 15).Can you see that this does satisfy the original scalar equation ?

Example 138. Let’s find the equation of the plane that contains the line [x, y, z] = [2, 1, 4]+t[2, 3, 4] and is parallel to the line [x, y, z] = [7, 4, 2] + s[−1, 0, 6].Since the first line lies on the plane, we can use (2, 1, 4) as the point on the plane and [2, 3, 4]as a direction vector. Since the plane is parallel to the second line and [−1, 0, 6] is not parallelto [2, 3, 4] (not a scalar multiple of it), we can use [−1, 0, 6] as the second direction vector. A

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vector equation for the plane would be [x, y, z] = [2, 1, 4] + p[2, 3, 4] + q[−1, 0, 6] for p, q ∈ R.Or, in parametric form, x = 2 + 2p− q, y = 1 + 3p and z = 4 + 4p+ 6q.

We now discuss some general properties of planes. Suppose we know a point P0 =(x0, y0, z0) on a plane and that the vector ~n = [A,B,C] = Aı+B+ Ck is normal (perpen-dicular or orthogonal) to the plane.

If we take any point P = (x, y, z) on the plane, then the vector ~P0P = [x− x0, y− y0, z− z0]is parallel to the plane and perpendicular to ~n. Thus ~P0P · ~n = 0.So [x− x0, y − y0, z − z0] · [A,B,C] = 0or A(x− x0) +B(y − y0) + C(z − z0) = 0or Ax+By + Cz + (−Ax0 −By0 − Cz0) = 0or Ax + By + Cz + D = 0 where D = −Ax0 − By0 − Cz0 is a constant. This is thescalar equation of the plane.

Example 139. The plane containing the point P0 = (4,−2, 3) and having normal vector~n = [1,−2, 1] has scalar equation (1)x + (−2)y + (1)z + (−(1)(4)− (−2)(−2)− (1)(3)) = 0or x− 2y + z − 11 = 0, which is also written as x− 2y + z = 11.

Example 140. Find the scalar equation of the plane containing the points P = (2, 1, 4),Q = (4, 0, 3) and R = (3, 4,−2).

The vectors ~PQ = [2,−1,−1] and ~PR = [1, 3,−6] are vectors on the plane. Then ~PQ× ~PRwill be normal to the plane, so~n = [2,−1,−1]× [1, 3,−6]= [(−1)(−6)− (−1)(3), (−1)(1)− (2)(−6), (2)(3)− (−1)(1)]= [9, 11, 7].The scalar equation has the form 9x+11y+7z+D = 0. We can plug any of the three pointsin to find D. Let’s use Q = (4, 0,−3), then 9(4) + 11(0) + 7(3) +D = 0 =⇒ D = −57 andthe scalar eqaution of the plane is 9x+ 11y + 7z = 57.

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Lecture 11: August 11

Review Quiz.

Intersections of Lines and Planes. We finally get to the most important subject in thishalf of the course: how to find intersections of lines and plans. For two lines L1 and L2 intwo-space, there are three possibilities.

(i) The lines intersect at a single point.

(ii) The lines are coincident (and hence intersect at infinitely many points).

(iii) The lines are parallel and distinct (and hence do not intersect).

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There is a direct connection to linear systems of two equations in two unknowns.

(a) Consider2x+ 3y = 43x+ 4y = 9

we can rewrite the system as an augmented matrix[2 3 43 4 9

]where the first row represents the first equation and the second row represents the secondequation. The first column represents x and the second column represents y. The verticalbar represents the equalities and the last column represents the constants on the right-handsides of the equations. This is simply a shorthand notation for the system of equationswhere we keep track of the coefficients. We perform elementary row operations like mul-tiplying or dividing a row by a nonzero scalar constant, interchanging rows or adding orsubtracting multiples of one row from another. The goal is to change the matrix into itsreduced row echelon form (RREF), where the first nonzero entry in each row is a 1, called aleading 1 or pivot. Each leading 1 is strictly to the right of any leading 1 in a row above it andall other elements in a column that contains a leading 1 are zero (so both above and below theleading 1). Any zero rows must appear at the bottom of the matrix. When we have reducedrow echelon form, we will be able to read the solution of the linear system right out of thematrix. The procedure to reduce a matrix to this form is called Gauss-Jordan elimination.[

2 3 43 4 9

]R1/2 (create leading 1 in first row)[

1 3/2 23 4 9

]R2 − 3R1 (clear column under leading 1)[

1 3/2 20 −1/2 3

]R2 ×−2 (make leading 1 in second row)[

1 3/2 20 1 −6

]R1 − (3/2)R2 (clear column above leading 1)[

1 0 110 1 −6

]this is RREF, the solution is x = 11 and y = −6.

These lines intersect at a single point, which is a unique solution.

(b) Consider2x+ 3y = 46x+ 9y = 12

as a matrix

110

[2 3 46 9 12

]R1/2[

1 3/2 26 9 12

]R2 − 6R1[

1 3/2 20 0 0

](RREF)

we have a row of zeros and the solution is x+(3/2)y = 2, which means that there are infinitelymany solutions (these lines are coincident). Let y = t be a parameter, then x + (3/2)t = 2or x = 2 − (3/2)t. So the parametric equations of the line of solutions are x = 2 − (3/2)tand y = t and the vector equation would be [x, y] = [2, 0] + t[−3/2, 1].

(c) Consider2x+ 3y = 46x+ 9y = 16

as a matrix[2 3 46 9 16

]R1/2[

1 3/2 26 9 16

]R2 − 6R1[

1 3/2 20 0 4

]this system is inconsistent. The last row says 0x + 0y = 4, which is impossible, so there isno solution here (these lines are parallel).

If we have two lines L1 and L2 in three-space, there are four possibilities.

(i) The lines intersect at a single point (unique solution).

(ii) The lines are coincident (there are infinitely many solutions).

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(iii) The lines are parallel and distinct (no solution).

(iv) The lines are skew – they are distinct, not parallel and do not intersect (no solution).This is the ‘usual’ situation.

Example 141. The lines L1 [x, y, z] = [2, 3, 4]+t[1, 1, 3] and L2 [x, y, z] = [1, 0,−1]+s[1, 1, 3]are parallel (same direction vector). Are they coincident or distinct ?If they are coincident, every point on L1 would be on L2, so all we have to do is check if(2, 3, 4) is on L2. The parametric equations for L2 are x = 1 + s, y = s and z = −1 + 3s. Ifwe substitute the point (2, 3, 4) in, 2 = 1 + s =⇒ s = 13 = s =⇒ s = 3and 4 = −1 + 3s =⇒ s = 5/3 and so the point is not on L2 and the lines are distinct.

112

Example 142. Find the point of intersection of L1 ~r = [2, 1, 5] + t[3, 1,−2] and L2 ~r =[0, 3, 1] + s[5,−1, 2]These lines are not parallel because their direction vectors are not parallel.We write the lines in parametric form x = 2 + 3t, y = 1 + t and z = 5 − 2t; and x = 5s,y = 3 − s and z = 1 + 2s. We look for a point of intersection by equating the coordinates2 + 3t = 5s, 1 + t = 3− s and 5− 2t = 1 + 2s. This system has a unique solution s = t = 1.Plugging these values for the parameters into the equations for the lines yields the point ofintersection (5, 2, 3) (which is a unique solution).

Example 143. Are the lines L1 [x, y, z] = [2, 1, 5] + t[3, 1,−2] and L2 [x, y, z] = [1, 3, 2] +s[5,−1, 2] skew ?The lines are not parallel, so we check for a point of intersection.2 + 3t = 1 + 5s =⇒ 3t− 5s = −11 + t = 3− s =⇒ t+ s = 2and 5−2t = 2 + 2s =⇒ 2t+ 2s = 3, which contradicts the previous equation, so this systemis inconsistent and there is no point of intersection. ∴ the lines are skew.

The shortest distance between two skew lines L1 and L2 would be the length of the commonperpendicular | ~AB|.

Suppose we have points P1 and P2 on the lines, then if ~n is any vector in the direction of~AB, we would have that proj~n ~P1P2 = ~AB. But | ~AB| is the length we want, so

| ~AB| = |proj~n ~P1P2| =| ~P1P2 · ~n||~n|

. We can find a normal vector ~n that is perpendicular to

both lines by taking the cross product of the direction vectors of the lines, ie ~n = ~m1 × ~m2.

Example 144. For our skew lines above, ~m1 = [3, 1,−2] and ~m2 = [5,−1, 2].So ~n = ~m1 × ~m2

= [3, 1,−2]× [5,−1, 2]= [(1)(2)− (−2)(−1), (−2)(5)− (3)(2), (3)(−1)− (1)(5)]= [0,−16,−8].We can divide by −8 and use ~n = [0, 2, 1] for convenience.

Use the points on the lines P1 = (2, 1, 5) and P2 = (1, 3, 2) to get ~P1P2 = [−1, 2,−3].

So | ~AB| = |~P1P2 · ~n||~n|

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=|[−1, 2,−3] · [0, 2, 1]|√

(0)2 + (2)2 + (1)2

=|(−1)(0) + (2)(2) + (−3)(1)|√

5= 1/

√5 ≈ 0.45 (so the lines are close).

If we have a line and a plane in three-space, there are three possibilities.

(i) The line intersects the plane at a single point (unique solution).

(ii) The line lies on the plane (so they intersect at infinitely many points and there are in-finitely many solutions).

(iii) The line is parallel to the plane (but not on it), so the line does not intersect the plane(no solution).

Example 145. Dose the line x = 2 + t, y = 6 − t and z = 3 + 2t intersect the plane2x+ 3y + 5z = 55 ? If yes, at what point ?We plug the parametric equations of the line into the scalar equation of the plane to see ifthere is a solution.

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2(2 + t) + 3(6− t) + 5(3 + 2t) = 554 + 2t+ 18− 3t+ 15 + 10t = 559t = 18 =⇒ t = 2Since there is a single solution for t, the line and plane intersect at the single point (4, 4, 7).

Example 146. Is the line ~r = [2, 3, 1] + t[1, 5,−2] parallel to the plane 3x + y + 4z = 13 ?If yes, is the line in the plane or not ?The direction vector of the line is ~m = [1, 5,−2] = ı + 5 − 2k and a normal vector of the

plane is ~n = [3, 1, 4] = 3ı+ + 4k.Let’s check ~m · ~n = [1, 5,−2] · [3, 1, 4] = (1)(3) + (5)(1) + (−2)(4) = 0.Since ~m · ~n = 0, ~m ⊥ ~n and hence the line is parallel to the plane.Let’s check for points of intersection.3(2 + t) + (3 + 5t) + 4(1− 2t) = 136 + 3t+ 3 + 5t+ 4− 8t = 1313 + 0t = 130t = 0Since all values of t satisfy this equation, there are infinitely many points of intersection andhence the line does lie on the plane.

The line would be parallel to the plane 3x+ y + 4z = 10 as well, but it would not lie on it.3(2 + t) + (3 + 5t) + 4(1− 2t) = 1013 + 0t = 100t = −3Which has no solution.

If a line is parallel to a plane, but not on it, how far apart are they ? Suppose we have apoint P on the line and a point Q on the plane.

If ~n is a normal vector to the plane, then the projection of ~PQ onto ~n would be the perpen-

dicular distance d from P to the plane. So we have d =| ~PQ · ~n||~n|

(which is similar to what

we did for the skew lines).

Example 147. How far is the line ~r = [2, 3, 1]+ t[1, 5,−2] from the plane 3x+y+4z = 10 ?

Take P = (2, 3, 1) and Q = (2, 0, 1), then ~PQ = [0,−3, 0] and we know ~n = [3, 1, 4].115

So d =|[0,−3, 0] · [3, 1, 4]|√

(3)2 + (1)2 + (4)2

=|(0)(3) + (−3)(1) + (0)(4)|√

26= 3/

√26 ≈ 0.59 (so pretty close).

We now look at intersections of planes. If we have two planes in three-space, there are threepossibilities.

(i) They intersect in a line (infinitely many solutions with one parameter).

(ii) They are coincident (infinitely many solutions with two parameters).

(iii) They are parallel and distinct (no solution).

A system of two equations in three unknowns is either inconsistent (has no solution) or hasinfinitely many solutions – a unique solution is not possible (two planes cannot intersect ata single point).

116

Example 148. Consider the planes 2x+ 2y + z = 7 and x+ y + z = 6.The normals for the planes are not parallel, so the planes are not parallel and hence theymust intersect.Let’s solve for the line of intersection. We’ll rewrite the system as a matrix.[

1 1 1 62 2 1 7

]R2 − 2R1[

1 1 1 60 0 −1 −5

]R2 ×−1[

1 1 1 60 0 1 5

]R1 −R2[

1 1 0 10 0 1 5

](RREF)

The solution is x+ y = 1, z = 5.Let y = t, then x = 1− t, y = t and z = 5 are the parametric equations of the line.The vector equation is [x, y, z] = [1, 0, 5] + t[−1, 1, 0].

Example 149. Consider the planes x− 2y + 3z = 10 and 2x− 4y + 6z = 12.The normals are parallel, so the planes are parallel. But are they coincident or distinct ?[

1 −2 3 102 −4 6 12

]R2 − 2R1[

1 −2 3 100 0 0 −8

]This is an inconsistent system, so there is no solution and the planes are distinct.

The planes x−2y+3z = 10 and 2x−4y+6z = 20 are coincident – can you see that ? Noticethat one equation is just a multiple of the other, unlike the previous example!

For three planes, there are six possibilities. There are three where the linear system isconsistent (has solutions).

(1) The planes intersect at a single point (unique solution).

(2) The planes intersect in a line (infinitely many solutions with one parameter).

(3) The planes are coincident (infinitely many solutions with two parameters).

And there are three where the system is inconsistent (no solutions).

(1) The planes are parallel (and at least two are distinct).

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(2) Two of the planes are parallel (and distinct), but the third is not parallel.

(3) The planes intersect in pairs.

The normals in the situations would have to be as follows. When there are solutions:

(1) not parallel, nor coplanar

(2) coplanar, but not parallel

(3) parallel

When there are not solutions:

(1) parallel

(2) two are parallel

(3) coplanar, but not parallel.

Example 150. Consider the planes x+2y+3z = 10, 2x+3y−z = 4 and 3x+4y+5z = 12. 1 2 3 102 3 −1 43 4 5 12

R2 − 2R1, R3 − 3R1 1 2 3 100 −1 −7 −160 −2 −4 −18

R2 ×−1 1 2 3 100 1 7 160 −2 −4 −18

R3 + 2R2 1 2 3 100 1 7 160 0 10 14

R3/10 1 2 3 100 1 7 160 0 1 7/5

R1 − 3R3, R2 − 7R3 1 2 0 29/50 1 0 31/50 0 1 7/5

R1 − 2R2 1 0 0 −33/50 1 0 31/50 0 1 7/5

(RREF)

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These planes intersect at a single point (−33/5, 31/5, 7/5).

Example 151. Consider the planes x+ y−2z = 4, 2x+ 2y−4z = 6 and 3x+ 5y+ 2z = 10. 1 1 −2 42 2 −4 63 5 2 10

R2 − 2R1, R3 − 3R1 1 1 −2 40 0 0 −20 2 8 −2

This is an inconsistent system, so there is no solution.Can you see that two of the planes are parallel but distinct ?

Example 152. Consider the planes x−5y+2z = 10, x+7y−2z = −6 and 8x+5y+z = 20. 1 −5 2 101 7 −2 −68 5 1 20

R2 −R1, R3 − 8R1 1 −5 2 100 12 −4 −160 45 −15 −60

R2/12 1 −5 2 100 1 −1/3 −4/30 45 −15 −60

R3 − 45R2 1 −5 2 100 1 −1/3 −4/30 0 0 0

R1 + 5R2 1 0 1/3 10/30 1 −1/3 −4/30 0 0 0

(RREF)

So we have x+ (1/3)z = 10/3 and y − (1/3)z = −4/3.Let z = t be the parameter, then we have x = 10/3− (1/3)t, y = −4/3 + (1/3)t and z = tor ~r = [10/3,−4/3, 0] + t[−1/3, 1/3, 1] and the planes intersect in a line.

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Lecture 12: August 15

Final Exam Review. We’ll spend all of today reviewing for the final exam. The finalexam will consist of 10-15 multiple choice questions, each worth 1 point, and 6-8 long answerquestions. You will be allowed to bring an approved basic calculator, but will not be allowedto bring and books or notes. The exam will cover the entire course.

Briefly, you should be able to do all of the following:

(1) Solve all questions related to the pre-calculus review. This includes understanding ournotation (e.g. sets, domains, etc), doing basic algebraic tricks (e.g. solving equations,tricks for simplifying rational functions, factoring polynomials, etc), manipulatingtrig functions and solving related equations, and solving inequalities involving lines,quadratic polynomials, and absolute values.

(2) Be able to do “rate of change” (or “average speed”) problems, as well as ‘tangent’problems.

(3) Be able to calculate limits (by looking at pictures, and also by using our three formulasand various factorization/simplification tricks), and also understand the intuitivedefinition of limits.

(4) Understand the definition of a continuous function, and be able to find discontinuitiesof a piecewise-defined function.

(5) Calculate derivatives using our formulas for derivatives (this includes knowing specialformulas for the derivatives of polynomials, exponential, logarithmic and trigonomet-ric functions, as well as being able to use the chain, product and quotient rule tocombine these formulas when calculating more complicated derivatives).

(6) Plot functions, using our large collection of techniques (this includes finding inter-cepts with the Cartesian coordinate axes, finding vertical and horizontal asymptotes,finding critical points of f and f ′, finding inflection points, finding the regions inwhich f, f ′ and f ′′ are positive, and classifying extrema).

(7) New: Be able to answer questions about the function f if given a picture of f (e.g.does f have a local max? Is f ′(x) always positive?).

(8) Be able to solve optimization problems.(9) Be able to do basic algebra and word problems that involve ax, ln(x), cos(x) and

sin(x).(10) Be able to do simple vector operations (adding, subtracting, multiplying by scalar,

resolving into ‘standard’ coordinates i and j).(11) Know how to calculate the formulas for lines and planes given various pieces of

information (e.g. calculate a line given two points that lie on it, or calculate a planegiven three points that lie on it) and their intersections. Calculate distance betweena point and a line.

(12) Know how to calculate the formulas for lines and planes given various pieces ofinformation (e.g. calculate a line given two points that lie on it, or calculate a planegiven three points that lie on it) and their intersections.

(13) Overall: be able to answer questions about definitions (e.g. continuity, parallel,perpendicular, etc).

(14) Overall: be able to answer ‘tricky’ or ‘non-quantitative’ or ‘find-the-example’ ques-tions (e.g. a function f satisfies f ′(0) = 1 and limx→∞ f(x) = 5. Is it possible thatf ′′(x) > 0 for all x? Is it possible that f(1) = 10?).

120

Note that there will not be any questions:

(1) involving the ‘precise’ definition of a limit, or(2) asking you to calculate a derivative from first principles.

Since we may not be able to get through everything, we’ll start with the three new typesof questions (‘tricky’ questions, ‘definition’ questions and ‘reading plots’ questions), move onto the questions about vectors, and finally do some calculus review at the end. As with theprevious midterm reviews, I strongly suggest that you work through all of these problemsbefore the final exam.

New Questions.

(1) Tricky Questions:(a) Let f be a function with global maximum at x = 2 and with no global minimum.

Is it possible that f ′′(2) > 0? Is it possible that f ′′(2) = 0? Is it possible thatlimx→∞ f(x) = limx→∞ f(x) = 0?If f ′′(2) > 0, then x = 2 is a local minimum by the second-derivative test, whichmeans it cannot be a global maximum.On the other hand, we can have f ′′(2) = 0. Consider, for example, f(x) =−(x− 2)10.Finally, we can have limx→∞ f(x) = limx→∞ f(x) = 0 if there are vertical asymp-totes somewhere. For example, we could have a function that looks like f(x) = 1

x2

with a little ‘bump’ at x = 2. Important note: I don’t know a formula for afunction f that satisfies these properties - but it is very easy to draw one! Thatis good enough to answer the question!

(b) Let f be a function satisfying f ′(0) = 1 and f ′′(x) > 0 for all x ∈ R. Is itpossible that limx→∞ f(x) = 2? Is it possible that f(0) = −1? is it possible thatlimx→∞ f

′(x) = 2?Since f ′(0) = 1 and f ′′(x) > 0 for all x ∈ R, we have f ′(x) ≥ 1 for all x ≥ 0.Thus, the instantaneous rate of change of f is at least 1 for all x ≥ 0. Drawingthis sort of function, we find that limx→∞ f(x) = ∞. Thus, we can’t havelimx→∞ f(x) = 2.To answer the next part of the question, note that we can add or subtract anynumber to f without changing f ′ or f ′′. Thus, f(0) can have any value, including−1.To answer the last part of the question, we try writing g = f ′(x). Then ourconditions become g(0) = 1 and g′(x) > 0 for all x ∈ R. We can clearly geta function that looks like g, and has finite limits, by flipping tan around theline y = x. Thus, there exists a function g satisfying these rules that satsfieslimx→∞ g(x) = 2.

(c) I have two non-parallel lines in R3. Is it possible that they don’t intersect?Yes!

(2) Definition Questions:(a) Let f(x) = x2 + 2x+ 1 so that f ′(x) = 2(x+ 1). Is x = −1 a critical point of f?

Yes!(b) Are the lines y = 2x+ 4 and y = 2x+ 7 parallel?

Yes!121

(c) Is the projection of u onto v parallel to v?Yes!

(3) Plot Reading Questions: We will discuss these in class. There is really only onequestion here:

The question will give you the plot of some unknown function f , and will ask: isf(0) positive? Is f ′(0) positive? Is f ′′(0) positive?

Vector Questions.

(1) Basic Vector Operations:(a) Let u = (1, 3, 5) and v = (−1, 2, 2). Find u+ 3v.

We have:

u+ 3v = (1, 3, 5) + 3 (−1, 2, 2)

= (1, 3, 5) + (−3, 6, 6)

= (−2, 9, 11).

(b) Let |u| = 4, and let u be at angle π4

with respect to the x-axis. Write u instandard form.We have

u = (|u| cos(π

4), |u| sin(

π

4))

= (41√2, 4

1√2

)

≈ (2.82, 2.82).

(c)(2) Dot Products, Cross Products and Formulas:

(a) Let u = (2, 3) and v = (−1, 1). Calculate u · (2u+ v).We calculate

u · (2u+ v) = (2, 3) · ((4, 6) + (−1, 1))

= (2, 3) · (3, 7)

= 6 + 21 = 27.

(b) Let u = (1,−1, 3) and v = (−1,−1, 2). Calculate the projection of u onto v.We have

projv(u) =u · vv · v

v

=(1,−1, 3) · (−1,−1, 2)

(−1,−1, 2) · (−1,−1, 2)(−1,−1, 2)

=−1 + 1 + 6

1 + 1 + 4(−1,−1, 2)

= (−1,−1, 2).

(c) Let u = (1, 3, 1) and v = (1, 1, 1). Calculate u× v.Using the formula for cross-produces,

u× v = ((3)(1)− (1)(1), (1)(1)− (1)(1), (1)(1)− (3)(1))122

= (2, 0,−2).

(d) Let u = (1,−1, 0) and v = (2, 3, 1). Find a vector perpendicular to both u andv.We recall that u× v is perpendicular to both u and v, so

w = u× v= ((−1)(1)− (0)(3), (0)(2)− (1)(1), (1)(3)− (−1)(2))

= (−1,−1, 5).

(e) Let u = (1,−1, 0) and v = (1, 1, 3). Is u parallel to v? Perpendicular to v?By inspection, u and v are not parallel. Next,

u · v = (1,−1, 0) · (1, 1, 3)

= 1− 1 + 0

= 0,

so u and v are perpendicular.(f) Determine the area of the parallelogram determined by the vectors u = (1, 3, 1)

and v = (1, 1, 1).We use the formula for the area:

A = ‖u× v‖= ‖(1, 3, 1)× (1, 1, 1)‖= ‖(2, 0,−2)‖=√

22 + 0 + (−2)2

≈ 2.8.

(g) In class, we will do a physics problem.(3) Lines and Planes:

(a) Find both vector and parametric equations for the line passing through thepoints A = (1, 2) and B = (−1, 1).The vector from A to B is (−1, 1)− (1, 2) = (−2,−1). Thus, the vector form is:

L = {(1, 2) + t (2, 1) : t ∈ R}

or, equivalently,

x = 1 + 2t

y = 2 + t,

for t ∈ R.(b) Find both vector and parametric equations for the plane passing through the

points A = (1, 1, 2), B = (1,−1, 2) and C = (1, 2, 3).The differences between pairs of points are:

u = B − A = (0,−2, 0)

v = C − A = (0, 1, 1),123

so the plane is:

P = {(1, 1, 2) + s u+ t v : s, t ∈ R},or equivalently

x = 1

y = 1− 2s+ t

z = 2 + t.

(c) Find the equation of the line that contains the point (1, 2) and is perpendicularto (1, 3).We find a vector u = (1, a) perpendicular to (1, 3). We have

1 + 3a = 0,

so a = −13. Thus, the line is

{(1, 2) + t (1,−1

3) : t ∈ R}.

(d) Find the intersection of the lines 2y + 3x = 1 and 3y − x = 5.Solving the second equation for x, we have

x = 3y − 5;

plugging into the first equation, we have

2y + 3(3y − 5) = 1,

so

11y = 14,

so

y =14

11

x = 3y − 5 =42

11− 5.

(e) Does the line {(0, 0, 1) + s (2,−1, 1) : s ∈ R} intersect the plane with normalvector n = (−1, 1, 2)?We note that the vector (2,−1, 1) is not perpendicular to n, so the line and planemust intersect.

(f) How far is the line {(1, 3, 2) + s (1, 3, 1) : s ∈ R} from the plane x− y+ 2z = 4?We choose P = (1, 3, 2) on the line and Q = (1, 1, 2) on the plane. We then usethe formula

d =|PQ · n||n|

=|((1, 1, 2)− (1, 3, 2)) · (1,−1, 2)|

|(1,−1, 2)|

=|1− 1 + 4− 1− 3 + 4|√

1 + 1 + 4124

=4√6.

(g) Find the intersection of the planes x+ y + z = 2 and x+ 2y + 2x = 5.Putting the system into RREF,[

1 1 1 21 2 2 5

]R2 − 2R1[

1 1 1 20 1 1 3

]R1 −R2[

1 0 0 30 1 1 3

]R1 −R2

So, we have:

x = 3

y = t+ 3

z = t,

where t ∈ R is free.

Calculus Questions. See the Midterm 1 and Midterm 2 reviews. We will do additional reviewquestions in class, time permitting.

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Miscellaneous

Last updated on August 3, 2016.

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