A brief Overview, Review and Summary of the basic Calculus

needed for AP Physics C

Animation borrowed from: http://www.calculusinmotion.com/gsp4.html Joe Stieve 2010

3

An Overview of Calculus

The Calculus involves the solution of Three types of mathematical problems.

I. The slope problem.

II. The “un-slope” problem.

III. The area problem.

Differential Calculus

Integral Calculus

Here we have a parabolic curve.

The first problem of calculus is to assign a quantity at each point to describe the rate of change of the function at that point.

For a graph, that can be interpreted as “find the slope”.

If Charlie Brown starts a trip up the hill clearly he will find the slope becoming steeper as he climbs.

We can think of the slope at each point as if it were the slope of a line that is aligned parallel to the direction of his motion.

x

y

You can also think of the instantaneous slope as the direction of travel of a car or motorcycle as it drives along the curve as indicated by the headlight beams.

x

y

But how do we get a mathematical value for the slope, say at the point (x,y) for a specific curve? Let’s look at a relatively simple example.

y = x2

x

y

The idea is that we write a general expression for the slope of a straight line drawn from the given point let us call it (x,y) to any other point (x2,y2) on the line.

(x2,y2)

(x,y)For our specific function this becomes:

Slope = 2x +Δx

x

y (x2,y2)

(x,y)In either case mathematically observe what happens for our function if the second x gets closer to the first, that is if Δx gets small.

The slope gets close to being simply 2x. We say the slope at x is 2x

There is no reason the second x must be to the right of the first x.

Slope = 2x + Δx

Slope = 2x + Δx

(x2,y2)

As the (x2, y2) moves toward the point (x,y) the slope of the red line connecting the two points becomes less.

As the (x2, y2) moves toward the point (x,y) the slope of the blue line connecting the two points becomes greater. (note that ∆x is negative.)

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So for y = x 2 we found a general expression for the slope = 2x + Δx

x

y (x2,y2)

(x,y)

(x2,y2)

We have derived from the original function, x2, a new function 2x. We say that the second function is the derivative of the first and we call the process “taking the derivative”.

Slope = 2x

The slope of the two lines we have draw get closer to being the same as we approach from above or below x.

The line drawn through the point (x,y) with the slope equal to the derivative at that point is said to be tangent to the curve at that point and represents the instantaneous slope at that point.

x

y

(1.5, 2.25)

Slope = 2x = 3

0 1 2

6

5

4

3

2

1

0

To put numbers on our given example we can choose the point (1.5,2.25) and the slope comes out 3 at that point.

For the next slide I have borrowed an animated gif to illustrate how the slope of chords, or secant lines approach the slope of a tangent line as the two points come together.

Here the tangent at the point x = 0.5 is shown in green, and the secant line draw to the yellow moving dot is shown in purple.

Notice this is NOT the same function y = x2 that we used earlier but rather y = 0.75 x2, but the principle is the same.

Source of animation http://www.ima.umn.edu/~arnold/graphics.html

Just a word about notation before we go on to generalize what we have learned.

Frequently, instead of the y notation we substitute the function notation and write f(x) instead of y, that is f(x) = x2 instead of y =x2 .

Because the derivative represents the instantaneous slope we frequently use the notation:

€

dy

dx

Or to remind us of its ancestry we might just put a prime on the old symbol and write f’(x)

So for the example we have been working with we can write y = f(x)=x2 and

= f’(x) = 2x .

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dy

dx

Now that the idea and notation of the derived function, or derivative, is known, we find that we “luck out” and that the rules for finding the derivatives of most functions are relatively easy. As an example we have been using a monomial.

For the general monomial f(x) = axn the derivative is found by multiplying the coefficient by the exponent and reducing the exponent by one. The derivative is then f’(x) =naxn-1

What if we have a polynomial? It turns out that the derivative of a sum is simply the sum of the derivatives.

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If f (x) = 4x 3 − 2x 2 + 3x + 2 we get f '(x) = 12x 2 − 4 x + 3

Although I have not listed it as a rule ,you can see that the derivative of a constant term is zero because it doesn’t change.

Caution to students: Though this rule for sums (and differences) works for all derivatives it cannot be extended to things like products and quotients. The derivative of a product/quotient is NOT the product/quotient of the derivatives.

Some important applications of the derivative to kinematics.

Because velocity is the rate of change of position and acceleration is the rate of change of velocity we can use the calculus notation to write:

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rv =

dr s

dt where "s" is a generalized space coordinate. In a particular problem

it might be x,y,z, r etc., and

r a =

dr v

dt

If then the position of an object is given as x = 25t2 -16t +4, (where the units: 25 m/s2, -16 m/s and 4 m for the numbers have been left off for convenience) we can find its velocity as a function of time by taking the derivative…v = 50t -16, and the acceleration by taking the derivative again to get…a = 50 m/s2. Note that in the case of a quadratic position function the acceleration will always come out constant.

(The vector notation is not absolutely necessary and can be eliminated for one dimensional equations.)

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A note about notation. The derivative of a derivative is referred to as

the second derivative and is often written as f"(x) or d2x

dt 2 for d dx

dt( )dt

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

Other examples and applications will appear in the lectures for the C exam.

Part II: the “unslope” problem

Now to address the second “problem” of calculus: The “unslope”, usually referred to as the anti-derivative. The situations is that you are given the slope at all points and you are to find the original function. Inspection of the following graphs will illustrate the problem.

At any value of x the slopes for all three curves shown, and indeed for a whole family of curves, have the same value. How do we distinguish which curve?

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Here are the three equations for the graphs and

their derivatives.

f (x) = x 2 − 2 , f '(x) = 2x

f (x) = x 2 , f '(x) = 2x

f (x) = x 2 + 2 , f '(x) = 2x

To find the function from its derivative you must be given some more information.

x

y

First lets use the rules for the derivative in reverse for f’(x) = 2x. We now increase the exponent by one and divide by the new exponent. Here is this rule for taking the anti-derivative of a monomial.

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If f '(x) = ax n we have f (x) =ax n +1

n +1+ C

We need to add the “C” because we don’t know what constant might have vanished when the derivative was taken.

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In the other notation we have :

If dy

dx= ax n we have y =

ax n +1

n +1+ C

The “hint” we need is just one point on the graph. Here for example we found the equation is y = x2+C. If we are told that y =2 when x = 0 we can substitute in and solve for C= 2 and we have our answer: y = x2 +2.

y

x

Because, in the symbol dy/dx, the dy and dx are called differentials, equations that contain them are called differential equations.

Near the earth’s surface the acceleration of gravity is about 9.8 m/s2. This information could be written as the differential equation:

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dv

dt= 9.8 m/s2

Taking the anti-derivative gives us v = 9.8 m/s2 t + C

So for any free object near the earth’s surface we can calculate the velocity at any time if we are given some initial information such as at 1s the object is going 20 m/s the we can do the arithmetic and find the general equation for all times is v = 9.8 m/s2 t + 10.2 m/s

This difficulty of being stuck with a constant might seem like a major pitfall, however it is a wonder in disguise.

With a simple algebraic equation we get one solution. With a differential equation we get a family of related solutions. For this reason many of the laws and principles of classical physics and engineering are written as differential equations. These equations then apply to the solution of an infinite family of related problems. We need only insert the specific parameters. In this way we get general laws. By putting in specific data we can “pick” our curve and predict the future behavior as the path follows that curve or rule.

This technique will be applied many time over in AP C and we will illustrate that as we move through the course.

Part III: The “Area” Problem

x

y

21

Here is our old friend the parabola f(x) = x2

Suppose, for some unknown reason, we want to find the area bounded by the curve on top, the x axis on the bottom and the vertical lines at x = 1 and x = 2.

Its height would be the value of the function where it intersects the curve, f(x), and its width would be the difference in the x coordinate at its left side and that at its right side, Δx.

Its area would be length times width or simply f(x)Δx.

Stretch your imagination one more time and image that Δx is on the verge of being “infinitely” small. We represent such a quantity by a differential symbol dx to distinguish it from the finite Δx

x

y

21

f(x) = x2 The impossibly minute area of this little section can be written as dA and we have dA = f(x)dx

Now all we have to do is add up the infinite number of little rectangles that it would take to cover the area we described. That is…

The Area from x = 1 to 2 is equal to the sum of the areas dA from x = 1 to 2.

Using some notation to “spell this out” we have…

Eventually, and officially the Sum began to be written as a distorted S.

S becomes becomes becomes

And we write:

x

y

21

f(x) = x2

This notation is all well and good but how in the world do we evaluate such a thing?

We have to leave the details of the derivation to Calculus but the great thing is we can do this without knowing those details.

A “miracle” occurs.

It turns out that the solution of our problem connects with the solution of the “unslope” or anti-derivative problem.

Here the F(x) is simply any ant-derivative for the function and you need not worry about the constant since it would subtract out.

In our example the anti-derivative of x2 is x3/3 + C. Dropping the constant then our area comes out exactly (2)3/3 - (1)3/3 = 7/3

In general then for any function f(x) that has a primitive (anti-derivative) F(x) you can find the area by using the expression:

There are some “tricks”. If b>a and the area or part of an area is below the x-axis that area or part of the area will come out negative. So for example the area from x = 0 to x = 2 for f(x)=x -1 comes out zero.

x

y

21+0.5

-0.5

1

-1

The graph makes the reason for a total of zero area clear. In other words you need to be concerned about signed areas.

Now that we know the integral symbol means sum we can do a brief review of the line and surface integrals of a vector field.

For our purposes in this presentation I will simply use the gravitational field of the earth. For the AP the ideas are more heavily used for the electric and magnetic fields.

The work done in moving an object from a to b is the force times the displacement and could be represented by any of the following expressions.

Consider the gravitational field near the earth’s surface.

(The scalar or “dot” product notation is not required on the AP)

This is the line integral of the force vector taken along the line a to b.

Now that we know the integral symbol means sum we can do a brief review of the line and surface integrals of a vector field.

In the more general case the notation will be a little different but the result is the same. Here ds is a minute displacement vector and at each point we are taking the product of the magnitude of the force with the component of the displacement in that direction, that is we are calculating mgdscosθ.

The work done in moving an object from a to b is the force times the displacement that could be represented as:

If we start and end in the same spot since gravitation is a conservative field we know the work will be zero. That is not always the case, but the important thing for the future is the notation.

ds

Here the circle on the integral sign merely means we have integrated around a closed loop and it is called reasonably enough a closed integral.

Lets just “jump in” to the surface integral. In physics we most are often interested to the flow of something through a surface.

For example the flow of water in a pipe.

v

A

If you multiply the velocity of an incompressible fluid by the cross sectional area you get the volume flow rate. For example 3 m/s times an area of 4 m3 gives a flow rate of 12 m3/s.

Now if the flow in general is not perpendicular to the area in question such as water passing through the frame of a net you need to take the component of the velocity that is perpendicular to the area. This is most easily done and represented if you invent a vector of magnitude equal to the area and drawn perpendicular to the surface. This is called surface vector.

If the angle between and the net were say 30o then the flow rate for an area of 0.50 m2 with a velocity of water at 0.20 m/s would give a flow rate of (0.2m/s)(0.5m2)cos 30o =0.087 m3/s

For a complex surface with different velocities at different points clearly the problem becomes very complicated and we don’t have to do it. The notation is however very important.

The rate of flow becomes:

(Here the “S” beneath the integral just means to take the sum over the entire surface.)

Lastly we extend this notation to a closed surface (indicated by the circle on the integral sign):

The convention is that the direction of dA is taken outward for a closed surface, that is from the finite volume out into the infinite volume of space.

Thus for an incompressible fluid flowing through a volume of space the flow into the volume (-) will be equal to the flow out of the volume (+) by the conservation of the incompressible fluid. In that case we can state categorically that the total surface integral is zero. As would be indicated by a net flow of zero.

Closed line and surface integrals will be revisited when we study Electricity and Magnetism, but don’t be concerned. The notation and the associated concepts are what will be important not the calculus solution of these integrals.

When we get there it will be “Look Ma, no calculus.”