Applications of Differentiation

by M. Bourne

Introduction to Applications of Differentiation

In Isaac Newton's day, one of the biggest problems was poor navigation at sea.

Before calculus was developed, the stars were vital for navigation.

Shipwrecks occured because the ship was not where the captain thought it should be. There was not a good enough understanding of how the Earth, stars and planets moved with respect to each other.

Calculus (differentiation and integration) was developed to improve this understanding.

Differentiation and integration can help us solve many types of real-world problems.

We use the derivative to determine the maximum and minimum values of particular functions (e.g. cost, strength, amount of material used in a building, profit, loss, etc.).

Derivatives are met in many engineering and science problems, especially when modelling the behaviour of moving objects.

Our discussion begins with some general applications which we can then apply to specific problems.

1. Tangents and Normals

by M. Bourne

We often need to find tangents and normals to curves when we are analysing forces acting on a moving body.

A tangent to a curve is a line that touches the curve at one point and has the same slope as the curve at that point.

A normal to a curve is a line perpendicular to a tangent to the curve.

Note 1: As we discussed before (in Slope of a Tangent to a Curve), we can

find the slope of a tangent at any point (x, y) using dydx.

Note 2: To find the equation of a normal, recall the condition for two lines

with slopes m1and m2 to be perpendicular (see Perpendicular Lines):

m1 × m2 = −1

Applications

A car has skidded while rounding a corner, tangent to the double yellow lines curve.

Tangent:

1. If we are traveling in a car around a corner and we hit something slippery on the road (like oil, ice, water or loose gravel) and our car starts to skid, it will continue in a direction tangent to the curve.

2. Likewise, if we hold a ball and swing it around in a circular motion then let go, it will fly off in a tangent to the circle of motion.

Normal:

The spokes of a bicycle wheel arenormal to the rim.

1. When you are going fast around a circular track in a car, the force that you feel pushing you outwards is normal to the curve of the road. Interestingly, the force that is making you go around that corner is actually directed towards the center of the circle, normal to the circle.

2. The spokes of a wheel are placed normal to the circular shape of the wheel at each point where the spoke connects with the center.

Examples

1. Find the gradient of

(i) the tangent (ii) the normal

to the curve y = x3 − 2x2 + 5 at the point (2,5).

Answer

dydx=3x2−4x

The slope of the tangent is

m1=dydx∣∣∣x=2=3(2)2−4(2)=12−8=4

The slope of the normal is found using m1 × m2 = −1

m2=−14

2. Find the equation of the normal in the above example.

Answer

We use y − y1 = m(x − x1)

with x1 = 2, y1 = 5, m=−14

So

y−5=−14(x−2)

gives

y=−14x+514

or

x + 4y − 22 = 0

3. Sketch the curve and the normal in the above example.

2. Newton's Method for Solving Equations

by M. Bourne

Computers use iterative methods to solve equations. The process involves making a guess at the true solution and then applying a formula to get a better guess and so on until we arrive at an acceptable approximation for the solution.

If we wish to find x so that f(x)=0 (a common type of problem), then we

guess some value x1 which is close to the desired solution and then we get

a better approximation using Newton's Method:

x2=x1−f(x1)f′(x1)

[This is just based on the point-slope form of a straight line].

Example 1

Find the root of

2x2 − x − 2 = 0

between 1 and 2.

Answer

The graph is as follows:

Try x1 = 1.5

Then

x2=x1−f(x1)f′(x1)

=1.5−f(1.5)f′(1.5)

Now f(1.5) = 2(1.5)2 − 1.5 − 2 = 1

f '(x) = 4x − 1 and f '(1.5) = 6 − 1 = 5

So

x2=1.5−15=1.3

So 1.3 is a better approximation.

Continuing the process,

x3=x2−f(x2)f′(x2)

=1.3−f(1.3)f′(1.3)

=1.3−0.084.2=1.2809524

We can continue for as many steps as necessary to give the required accuracy.

Check: Using some Computer Algebra Systems, (eg Mathcad) we

also need to give an initial guess(say, x:=2) and the result is:

root(2x2 − x − 2, x) = 1.2807764064044

Functions with Multiple Roots

Many functions have multiple roots, so you need tounderstand what is going on and give the computer a guess close to your desired answer.

Example 2

Solve 1− t2 + 2t = 0

[Scientific Notebook cannot find the solution directly for us. We need to know how to properly use the tool to get the solution, either with graphs or

setting up Newton's Method. This could involve giving an initial estimate for the root.]

Answer

Let y = 1− t2 + 2t

The graph of y(t):

There appear to be 2 roots, one near t = -1 and the other near t = 3.

However, if we look more carefully in the region near t = 3 (by zooming in),

we see that there is more than one root there.

By simple substitution, we can see that one of the roots is exactly t = 3:

1 − (3)2 + 23 = 0

Now for the root near t = 3.4.

We will use Newton's Method to approximate the root. We need to

differentiate y = 1− t2 + 2t. Since we have t as an exponent in our

expression, we need to use logarithms when differentiating. [See how to differentiate logarithms in Derivative of the Logarithmic Function].

Let's differentiate 2t by itself first.

Let h = 2t.

Take natural log of both sides:

ln h=t ln 2

1hdhdt=ln 2

dhdt=h ln 2=2t ln 2

So

dydt=f′(t)=−2t+2t ln 2

So for Newton's Method in this example, we would have:

f(t)f′(t)=1−t2+2t−2t+2tln 2

Initial guess: t1 = 3.4

t2=t1−f(t1)f′(t1)

t2=3.4−f(t)f′(t)−1−3.42+23.4−2(3.4)+23.4ln 2=3.407615918

We can write this more conveniently (for later steps) showing the substitution as:

t2=3.4−[1−t2+2t−2t+2tln 2]t=3.4=3.407615918

Now, doing another step of Newton's Method:

t3=t2−f(t2)f′(t2)

t3=3.407615918−[1−t2+2t−2t+2tln 2]t=3.407615918=3.407450596

And doing another step:

t4=3.407450596−[1−t2+2t−2t+2tln 2]t=3.407450596=3.407450522

And another:

t5=3.407450522−[1−t2+2t−2t+2tln 2]t=3.407450522=3.407450505

We can conclude that correct to 7 decimal places, t = 3.4074505.

Using Graphs Instead

Using Scientific Notebook, we can zoom into the root and we can see (from

where the graph cuts the y-axis) that t is indeed close to 3.40745.

Now for the negative case. Let t1 = -1 be our initial guess.

t2=−1−[1−t2+2t−2t+2tln 2]t=−1=−1.213076633

t3=−1.213076633−[1−t2+2t−2t+2tln 2]t=−1.213076633=−1.198322474

t4=−1.198322474−[1−t2+2t−2t+2tln 2]t=−1.198322474=−1.198250199

We could continue until we obtained the required accuracy.

Comparing this to the zoomed in graph, we can see that the solution is t =

-1.198250197, correct to 9 decimal places.

Conclusion

So the solutions for 1− t2 + 2t = 0 are

t = −1.19825,

t = 3, or

t = 3.40745,

correct to 5 decimal places.

3. Curvilinear Motion

by M. Bourne

In the last Chapter (in The Derivative as an Instantaneous Rate of Change), we found out how to find the velocity from the displacement function using:

v=dsdt

and the acceleration from the velocity function (or displacement function), using:

a=dvdt=d2sdt2

These formulae are only appropriate for rectilinear motion (i.e. velocity and acceleration in a straight line). This is inadequate for most real situations, so we introduce here the concept of curvilinear motion, where an object is moving in a plane along a specified curved path.

We generally express the x and y components of the motion as functions

of time. This form is calledparametric form. (See another example using parametric form in Lissajous Figures.)

Example 1: Parametric Equations

Draw the curve

y(t) = cos t,

x(t) = sin t

for t = 0 to 2π in 0.5 intervals.

First, we need to set up a table of values which we obtain by substituting

various values of t:

t 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0

x(t) 0 .48 .84 1.0 .91 .60 .14 -.35 -.76 -.98 -.96 -.71 -.28

y(t) 1 .88 .54 .07 -.42 -.80 -.99 -.94 -.65 -.21 .28 .71 .96

Answer

We plot our 13 points, starting at (0,1) as follows (we move clockwise in

this example):

We see that we have formed a circle, centre (0,0), radius 1 unit.

Notice that the variable t does not appear in this graph, just the

variables x and y.

Horizontal and Vertical Components of Velocity

The horizontal component of the velocity is written:

vx=dxdt

and the vertical component is written:

vy=dydt

We want to find the magnitude of the resultant velocity v once we know

the horizontal and vertical components. We use:

v=(vx)2+(vy)2−−−−−−−−−−√

The direction θ that the object is moving in, is found using:

tan θv=vyvx

Example 2

If x = 5t3 and y = 4t2 at time t, find the magnitude and direction of the

velocity when t = 10.

Answer

When t=10, the particle is at (5000,400).

Here is the graph of the motion.

Note:

The axes are x and y (and do not involve t). The particle is accelerating as time goes on (the red dots are at one

second intervals)

We are told that

x = 5t3

So

dxdt=15t2

At t = 10, the velocity in the x-direction is given by:

dxdt=vx=15(10)2=1500 ms−1

Also, y = 4t2 so the velocity in the y-direction is:

dydt=8t

When t = 10, the velocity in the y-direction is:

dydt=vy=8(10)=80 ms−1

So the magnitude of the velocity will be:

v=(vx)2+(vy)2−−−−−−−−−−√=15002+802−−−−−−−−−−√=1502.1 ms−1

Now for the direction of the velocity (it is an angle, relative to the positive x-

axis):

tan θv=vyvx=801500

So θv=0.053 radians =3.05∘.

Example 3

If

x=20t2t+1

and

y=0.1(t2+t)

at time t, find the magnitude and direction of the velocity when t = 2. Plot

the curve.

Answer

When t=2, the particle is at (8,0.6).

x=20t2t+1

so

dxdt=(2t+1)20−20t(2)(2t+1)2=20(2t+1)2

At t=2,

dxdt=vx=2025=0.8 ms−1.

Also, y = 0.1(t2 + t) so

dydt=0.1(2t+1)

When t=2,

dydt=vy=0.5 ms−1.

So

v=(vx)2+(vy)2−−−−−−−−−−√=0.82+0.52−−−−−−−−−√=0.943 ms−1

Now for the direction:

tan θv=vyvx=0.50.8

So θv=0.625 radians

Acceleration of a Body in Curvilinear Motion

The expressions for acceleration are very similar to those for velocity:

Horizontal component of acceleration:

ax=dvxdt

Vertical component of acceleration:

ay=dvydt

Magnitude of acceleration:

a=(ax)2+(ay)2−−−−−−−−−−−√

Direction of acceleration:

tan θa=ayax

Example 4

A car on a test track goes into a turn described by x = 20 + 0.2t3, y =

20t − 2t2, where x and y are measured in metres and t in seconds.

(i) Sketch the curve for 0 ≤ t ≤ 8.

(ii) Find the acceleration of the car at t=3.0 seconds.

Answer

(i) Sketch:

Note the axes are x- and y-based, and do not involve t.

(ii) Acceleration:

Horizontal acceleration:

x = 20 + 0.2t3

vx=dxdt=0.6t2

ax=d2xdt2=1.2t

At t = 3.0, ax = 3.6

Vertical acceleration:

y = 20t − 2t2

vy=dydt=20−4t

ay=d2ydt2=−4

At t = 3.0, ay = -4

Now

a=(ax)2+(ay)2−−−−−−−−−−√=3.62+(−4)2−−−−−−−−−−√=5.38

and

θa=arctan(ayax)=arctan (−43.6)=312o [4th quadrant]

So the car's acceleration has magnitude 5.38 ms-2, and

direction 312∘ from the positive x-axis.

What if x and y are NOT given as functions of t?

Example 5

A particle moves along the path y = x2 + 4x + 2 where units are in

centimetres. If the horizontal velocity vxis constant at 3 cm s-1, find the

magnitude and direction of the velocity of the particle at the point (-1, -1).

Answer

This is a different situation to the other examples. This time we have y in

terms of x, and there are no expressions given in terms of "t" at all.

To be able to find magnitude and direction of velocity, we will need to know

vx=dxdt

and

vy=dydt

But the question already gives us

vx=dxdt=3

so all we need to find is dydt.

To find this, we differentiate the given function with respect to t throughout

using the techniques we learned back in implicit differentiation:

y = x2 + 4x + 2

dydt=2xdxdt+4dxdt+0

Since

dxdt=3

and we want to know the velocity at x=−1, we substitute these two values

and get:

dydt=2(−1)(3)+4(3)=6

So now we have vy = 6 cm s-1.

So the magnitude of the velocity is given by:

v=(vx)2+(vy)2−−−−−−−−−−√=32+62−−−−−−√=6.7082

The direction of the velocity is given by:

θv=arctan(vyvx)=arctan(63)=63.432o

So the velocity is 6.7 cm s-1, in the direction 63.4∘.

Example 6

A rocket follows a path given by (distances in km):

y=x−x390

If the horizontal velocity is given by V(x) = x, find the magnitude and

direction of the velocity when the rocket hits the ground (assume level terrain) if time is in minutes.

Answer

Let's first see a graph of the motion, to better understand what is going on.

We can see that the rocket hits the ground again somewhere around x =

9.5 km. At this point, the horizontal velocity is positive (the rocket is going

left to right) and the vertical velocity is negative (the rocket is going down).

"V(x) = x" means that as x increases, the horizontal velocity also

increases with the same number (different units, of course). So for

example, at x = 2 km, the horizontal speed is 2 km/min, and atx = 7

km, the horizontal speed is 7 km/min, and so on.

To calculate the magnitude of the velocity when the rocket hits the ground, we need to know the vertical and horizontal components of the velocity at that point.

(1) Horizontal velocity. We just need to solve the following equation to find the exact point the rocket hits the ground:

x−x390=0

Factoring gives:

x−x390=x(1−x290)

And solving for 0 gives us x=0, x=−310−−√, x=310−−√

We only need the last value, x=310−−√≈9.4868 km (This value is

consistent with the graph above).

So the horizontal speed when the rocket hits the ground is 9.4868

km/min (since V(x) = x).

(2) Vertical velocity. We now need to use implicit differentiation with

respect to t (not x!) to find the vertical velocity.

y=x−x390

dydt=dxdt−130x2dxdt

But we already know dxdt and x at impact, so we simply substitute:

dydt=(9.48683298)−130(9.48683298)2(9.48683298)

This gives us a negative velocity, as we expected before:

dydt=−18.97366596

So now we need to find the magnitude of the velocity. This takes into account both the horizontal and vertical components.

Magnitude=(dxdt)2+(dydt)2−−−−−−−−−−−−−−√

Substituting, we have:

(9.48683298)2+(−18.97366596)2−−−−−−−−−−−−−−−−−−−−−−−−−−√=21.2

1320344

Velocity has magnitude and direction. Now for the direction part.

angle of motion=arctan dy/dtdx/dt

Substituting our vertical and horizontal components, we have:

arctan −18.973665969.48683298=−1.107148718

In degrees, this is equivalent to

−1.107148718×57.25578=−63.3907∘

We can see that this answer is reasonable by zooming in on the portion of the graph where the rocket hits the ground (with equal-axis scaling):

So in summary, the velocity of the rocket when it hits the ground is 21.2

km/min in the direction 63.4∘below the horizontal.

Now let's return to the example we saw at the beginning of the calculus section. We have not seen how to

4. Related Rates

by M. Bourne

If 2 variables both vary with respect to time and have a relation between them, we can express the rate of change of one in terms of the other.

We need to differentiate both sides w.r.t. (with respect to) time.

That is, we'll be finding dfdt for some function f(t).

Important!

Recall from implicit differentiation the following for some function x of t:

ddtx2=2xdxdt

ddtx3=3x2dxdt

ddtx4=4x3dxdt

ddtx5=5x4dxdt

We use this concept throughout this section on related rates.

Example 1

A 20 m ladder leans against a wall. The top slides down at a rate of 4 ms-1.

How fast is the bottom of the ladder moving when it is 16 m from the wall?

Steps:

1. Make a sketch of the problem 2. Identify constant and variable quantities 3. Establish relationship between quantities. 4. Differentiate w.r.t time. 5. Evaluate at point of interest.

Following is a demonstration of how to solve this problem in LiveMath.

You obviously still need to understand the mathematics!

Answer

Now the relation between x and y is:

x2 + y2 = 202

Differentiating throughout with respect to time (since the value

of x and y depends on t):

ddtx2+ddty2=0

2xdxdt+2ydydt=0

That is:

xdxdt+ydydt=0

Now, we know

dydt=−4

and we need to know the horizontal velocity (dxdt) when

x=16.

The only other unknown is y, which we obtain using Pythagoras' Theorem:

y=202−162−−−−−−−−√=144−−−√=12

So

(16)dxdt+(12)(−4)=0

gives

dxdt=3 ms−1.

Example 2

A stone is dropped into a pond, the ripples forming concentric circles which expand. At what rate is the area of one of these circles increasing when the

radius is 4 m and increasing at the rate of 0.5 ms-1?

Answer

Relation:

A=πr2

Differentiate w.r.t. time, and then substitute known values:

dAdtdAdt=ddt(πr2)=2πrdrdt=2π(4)(0.5)≈12.56m2s−1

Example 3

An earth satellite moves in a path that can be described by

x272.5+y271.5=1

where x and y are in thousands of kilometres.

If dxdt=12900 km/h for x=3200 km and y>0, find dydt.

Answer

We differentiate the expression with respect to t:

x272.5+y271.5=1

2xdxdt72.5+2ydydt71.5=0

We need to find y. We do this by substituting x=3.2 into the original

expression:

3.2272.5+y271.5=1

Solving this gives:

(71.5)(3.2)2+72.5y2=(71.5)(72.5)

732.16+72.5y2=5183.75

y2=5183.75−732.1672.5

y2=61.401

y=±7.836

The question tells us to take the positive value only. Substituting our known values gives:

2(3.2)(12.9)72.5+2(7.836)dydt71.5=0(71.5)(2)(3.2)(12.9)+(72.5)(2)(7.836)dyd

t=0dydt=−5.195

This means the velocity in the y-direction is −5195 km/h.

Example 4

The tuning frequency f of an electronic tuner is inversely proportional to the

square root of the capacitance C in the circuit.

If f = 920 kHz for C = 3.5 pF, find how fast f is changing at this

frequency if dCdt=0.3 pF/s.

Answer

Now

f=kC−−√

and substituting our given values, we have that

920000=k3.5×10−12−−−−−−−−−√

and this gives k=1.721.

So

f=1.721C−−√=1.721C−1/2

We need to find dfdt.

dfdt=−1.7212C−3/2dCdt

We are told that C=3.5 pF and dCdt=0.3 pF/s.

So

dfdt=−1.7212(3.5×10−12)−3/2(0.3×10−12)

=−39424.8 Hz s−1

So the frequency of the electronic tuner is decreasing at the rate of 39.4

kHz s-1.

5. Curve Sketching

by M. Bourne

NOTES:

There are now many tools for sketching functions (Mathcad, Scientific Notebook, graphics calculators, etc). It is important in this section to learn the basic shapes of each curve that you meet. Anunderstanding of the nature of each function is important for your future learning. Most mathematical modelling starts with a sketch.

You need to be able to sketch the curve, showing important features.

Avoid drawing x-y boxes and just joining the dots.

We will be using calculus to help find important points on the curve.

The kinds of things we will be searching for in this section are:

x-intercepts

Use y=0

NOTE: In many cases, finding x-intercepts is not so easy. If so, delete

this step.

y-intercepts Use x=0

local maxima Use dydx=0, sign of first derivative changes +→−

local minima Use dydx=0, sign of first derivative changes −→+

points of

inflection Use d2ydx2=0, and sign of d2ydx2 changes

Finding Maxima and Minima

Local maximum

A local maximum occurs when y′=0 and y′ changes sign from positive to

negative (as we go left to right).

Local minimum

A local minimum occurs when y′=0andy′ changes sign from negative to

positive.

The Second Derivative

The second derivative can tell us the shape of a curve at any point.

Concave up

If d2ydx2>0, the curve will have a minimum-type shape (called concave

up)

Example 1

The curve y = x2 + 3x − 2 has dydx=2x+3.

Now d2ydx2=2 and of course, this is >0 for all values of x.

So it has a concave up shape for all x.

Concave down

If d2ydx2<0, the curve will have a maximum-type shape (called concave

down)

Example 2

The curve y = x3 − 2x + 5 has dydx=3x2−2. The second derivative

is d2ydx2=6x and this is <0 for all values of x<0.

So the curve has a concave down shape for all x<0 (and it is concave

up if x>0).

Finding Points of Inflection

A point of inflection is a point where the shape of the curve changes from

a maximum-type shape d2ydx2<0 to a minimum-type shape d2ydx2>0.

Clearly, the point of inflection will occur when

d2ydx2=0 and when there is a change in sign

(from plus → minus or minus → plus) of d2ydx2.

Example 3

Sketch the following curve by finding intercepts, maxima and minima and points of inflection:

y=x3−9x

Answer

The basic shape of a cubic is:

Keeping this in mind helps with the sketching process.

1. x-intercepts:

y=x3−9x=x(x2−9)=x(x+3)(x−3)=0

when

x=0, x=−3 and x=3

2. y-intercepts:

When x=0, y=0.

3. maxima and minima?

dydx=3x2−9=3(x2−3)=3(x+3√)(x−3√)=0

when

x=−3√ or x=3√

So we have max or min at approximately (−1.7,10.4) and (1.7,−10.4).

[We could check which is which by trying some points

near −1.7 and +1.7 to determine what the sign changes are. But we need to

find the second derivative anyway for points of inflection, so we use that to determine max or min.]

4. Second derivative:

d2ydx2=6x

If x=−1.7, y''<0, so MAX at (−1.7,10.4)

If x=+1.7, y''>0, so MIN at (1.7,−10.4)

5. Point of inflection:

d2ydx2=6x

Now

d2ydx2=0 when x=0

and

d2ydx2

changes sign from negative (concave down) to positive (concave up)

as x passes through 0.

So we are ready to sketch the curve:

General Shapes

If we learn the general shapes of these curves, sketching becomes much easier. Of course, the following are "ideal" shapes, and there are many other possibilities. But at least this helps get us started.

Quadratic Cubic

Highest power of x: 2 Highest power of x: 3

1 minimum, no maximum

[if it has a positive x2 term] 1 minimum, 1 maximum

no points of inflection 1 point of inflection

Quartic Pentic

Highest power of x: 4 Highest power of x: 5

2 minimums, 1 maximum

[if it has a positive x4 term] 2 minimums, 2 maximums

2 points of inflection 3 points of inflection

Example 4

Sketch the curve and show intercepts, maxima and minima and points of inflection:

y=x4−6x2

Answer

1. x-intercepts

y=x4−6x2=x2(x2−6)=x2(x+6√)(x−6√)=0

when

x=0, x=−6√ and x=6√

2. y-intercepts:

When x=0, y=0.

3. maxima and minima?

dydx=4x3−12x=4x(x2−3)=4x(x+3√)(x−3√)=0

Now dydx=0 when x=0 or x=−3√ and x=3√

So we have max or min at (0,0) and (−3√,−9) and (3√,−9).

4. Second derivative:

d2ydx2=12x2−12

Now y''>0 for x=−3√ so (−3√,−9) is a local MIN

Now y''<0 for x=0 so (0,0) is a local MAX

Now y''>0 for x=3√ so (3√,−9) is a local MIN

5. Points of inflection:

We now use the second derivative to find points of inflection:

d2ydx2=12x2−12

=12(x+1)(x−1)

=0

when x=−1 or x=1

If x<−1, y''>0, and for −1<x<1, we have y''<0.

So the sign of y'' has changed, so (−1,−5) is a point of inflection.

If x>−1, y''>0,

So the sign of y'' has changed, so (1,−5) is a point of inflection.

So we are ready to sketch the curve:

Example 5

Sketch the curve and show intercepts, maxima and minima and points of inflection:

y=x5−5x4

Answer

NOTE: This question is fairly sophisticated and is here to show you some of the complications that can occur. If you don't fully understand it now, don't worry!

1. x-intercepts

y=x5−5x4

=x4(x−5)

=0

when x=0, x=5

2. y-intercepts:

When x=0, y=0.

3. maxima and minima?

dydx=5x4−20x3

=5x3(x−4)

=0

when x=0 or x=4

So we have max or min at (0,0) and (4,−256).

4. Second derivative:

d2ydx2=20x3−60x2

Now y''=0 for x=0 so (0,0) is ???

Now y''>0 for x=4 so (4,−256) is a local MIN

5. Points of inflection

We now use the second derivative to find points of inflection:

d2ydx2=20x3−60x2

=20x2(x−3)

=0

when x=0 or x=3

If x<0, y''<0:

If 0<x<3, y''<0:

There is no sign change, so at x=0, there is NO point of inflection.

If x>3, y''>0:

So the sign of y'' has changed, so (3,−162) is a point of inflection.

Actually, at x=0, we have a FLAT SPOT. It is not a local maximum, even

though it may appear so.

So we are ready to sketch the curve:

6. More on Curve Sketching Using Differentiation

by M. Bourne

This section deals with curves which are NOT polynomials. They have discontinuities or other unusual behaviour. It is important to understand these types of graphs, since they arise out of real-life situations. Also, we need to be able to interpret error messages or other unexpected behaviour when we are using computers to draw them.

We use all the techniques applied in Section 5 Curve Sketching and also examine the behaviour of the function as

x → −∞

x → +∞

x → left side of the discontinuity

x → right side of the discontinuity

Symmetry

We can use symmetry about the y-axis to help us sketch the curve (it will

be a mirror image about the y-axis).

Domain and range

The domain (all possible x-values) and range (all resulting y-values) is

important when graphing certain types of questions (e.g. those involving square root).

Our method

Find the following first:

1. x-intercepts

2. y-intercepts

3. Limit as x approaches infinity

4. Domain and Range

5. Maxima and minima

6. Second derivative

7. Behaviour near discontinuity

Example 1

Sketch y=x+4x2

Answer

1. x-intercepts

x+4x2=0

x3+4=0

x=−4−−−√3≈−1.6

2. y-intercepts:

We cannot have x=0, so there is NO y-intercept, and there must be an

asymptote at x=0.

3. Limits

As x → −∞,

4x2→0 ,

so y → −∞

(In fact, the curve approaches the line y = x on the negative side.)

As x → +∞, y → ∞

(In fact, the curve approaches the line y = x on the positive side.)

4. Domain all real x, except 0

Range all real y

5. Maxima and minima?

y=x+4x2=x+4x−2

dydx=1−8x−3

=1−8x3

Now

1−8x3=0

when

x3−8=0

That is, when x=2

So we have a max or min at (2,3).

Now, as x → -∞, dydx→1,

and as x → ∞, dydx→1.

6. Second derivative:

d2ydx2=24x4>0 for all x.

So concave up for all x, except 0. So (2,3) is a MIN.

7. Near discontinuity:

As x→0−, (which means x approaches 0 from the negative side), y → ∞

[Try x=−1,−0.5,−0.1,−0.01,−0.001 etc to see this].

As x→0+, (0 from the positive side), y → ∞

[Try x=1,0.5,0.1,0.01,0.001 etc to see this].

So we are ready to sketch the curve:

Example 2

Sketch y=9xx2+9

Answer

1. x-intercepts

9xx2+9=0

when

9x=0

That is, when x=0

2. y-intercepts:

When x=0, y=0.

3. Limit as x approaches infinity:

Dividing top and bottom by x2 gives:

9xx2+9÷x2x2=9x1+9x2

As x→−∞,

y=9xx2+9=9x1+9x2→01+0=0

Also, as x→+∞, y→0

4. Domain all real x

Range [see later, in the next section].

5. Maxima and minima?

y=9xx2+9

dydx=(x2+9)(9)−(9x)(2x)(x2+9)2

=−9x2+81(x2+9)2

=−9(x2−9)(x2+9)2

Now,

dydx=0 when x=±3

So we have a max or min at (−3,−1.5) or (3,1.5).

Some consideration of the expression for dydx shows that it is

POSITIVE for –3<x<3 and

NEGATIVE for x<−3 and x>3.

We conclude that (−3,−1.5) is the MINIMUM and (3,1.5) is the MAXIMUM.

So the Range is −1.5≤y≤1.5.

Also, the slope at x=0 is

dydx=−9x2+81(x2+9)2∣∣∣∣x=0=0+8192=1

6. Second derivative:

In this case, the second derivative may be too cumbersome to calculate quickly, and the benefits are doubtful.

So we are ready to sketch the curve:

7. Applied Maximum and Minimum Problems

by M. Bourne

The process of finding maximum or minimum values is called optimisation. We are trying to do things like maximise the profit in a company, or minimise the costs, or find the least amount of material to make a particular object.

These are very important in the world of industry.

Example 1

The daily profit, P, of an oil refinery is given by

P = 8x − 0.02x2,

where x is the number of barrels of oil refined. How many barrels will give

maximum profit and what is the maximum profit?

Answer

The profit is a max (or min) if dPdx=0.

dPdx=8−0.04x

=0

when

x=80.04=200

Is it a maximum?

d2Pdx2=−0.04<0 for all x, so we have a maximum.

When x=200, P=$800.

So if the company refines 200 barrels per day, the maximum profit

of $800 is reached.

[Go here to see another way to find the maximum or minimum value of a parabola.]

Example 2

A rectangular storage area is to be constructed along the side of a tall building. A security fence is required along the remaining 3 sides of the

area. What is the maximum area that can be enclosed with 800 m of

fencing?

Answer

The area is A=xy

We know 2x+y=800 so y=800−2x

So the area is A = x(800 − 2x) = 800x − 2x2

To maximise the area, find when dAdx=0

dAdx=800−4x=0

when

x=200

Is it a maximum?

d2Adx2=−4<0 for all x

So it is a maximum.

So the maximum area occurs when x=200, y=400 and that area is:

A = 200 × 400 = 80 000 m2 = 8 ha

Example 3

[This problem was presented for discussion earlier in the Differentiation introduction.]

A box with a square base has no top. If 64 cm2 of material is used, what is

the maximum possible volume for the box?

Answer

The net for this box would be:

The volume of the box is V = x2y

We are told that the surface area of the box is 64 cm2. The area of the

base of the box is x2 and the area of each side is xy, so the area of the

base plus the area of the 4 sides is given by:

x2 + 4xy = 64 cm2

Solving for y gives:

y=64−x24x=16x−x4

So the volume can be rewritten:

V=x2y=x2(16x−x4)=16x−x34

Now

dVdx=16−3x24

and this is zero when

x=±83√≈4.62

(Note: The negative case has no practical meaning.)

Is it a maximum?

d2Vdx2=−3x2

and this is negative when x is positive. So it is a MAX.

So the dimensions of the box are:

Base 4.62 cm × 4.62 cm and sides 2.31 cm.

The maximum possible volume is

V = 4.62 × 4.62 × 2.31 ≈ 49.3 cm3

Check: Area of material:

x2 + 4xy = 21.3 + 4 × 4.62 × 2.31 = 64

Checks OK.

8. Radius of Curvature

By M. Bourne

We can draw a circle that closely fits nearby points on a local section of a curve, as follows.

Application - Radius of Curvature

When engineers design train tracks, they need to ensure the curvature of the track will be safe and provide a comfortable ride for the given speed of the trains. [Imagesource].

The radius of curvature of the curve is defined as the radius of the approximating circle. This radius changes as we move along the curve. How do we find this changing radius of curvature?

The formula for the radius of curvature at any point x for the curve y = f(x)

is given by:

Radius of curvature=[1+(dydx)2]3/2∣∣d2ydx2∣∣

Proof

The curvature of a given curve at a particular point is the curvature of the approximating circle at that point.

The curvature depends on the radius - the smaller the radius, the greater the curvature (approaching a point at the extreme) and the larger the radius, the smaller the curvature. (A very large approximating circle means the curve is almost a straight line at that point.)

The radius of curvature R is simply the reciprocal of the curvature, K. That

is,

R=1K

So we'll proceed to find the curvature first, then the radius will just be the reciprocal of that curvature.

Let P and P1 be 2 points on a curve, "very close" together, as shown.

Δs is the length of the arc PP1.

Δθ is the angle turned by the tangent line as it moves from P to P1.

The curvature of the arc from P to P1 is given by

ΔθΔs

Now, the curvature K at point P is given by:

K=limΔθ→0ΔθΔs=dθds

We now need to find dθds and we use the Chain Rule: dθds=dθdxdxds

Note that tan θ=dydx, so θ=arctan(dydx)

Returning to our formula, dθdx=ddxarctan(dydx)

In the Differentiation of Transcendental Functions chapter we'll learn the

derivative of y=arctan u, where u=f(x), is given by dydx=dudx1+u2

With u=arctan(dydx) we differentiate as follows:

dθdx=ddxarctan(dydx)

=ddx(dydx)1+(dydx)2

=d2ydx21+(dydx)2

We also need dxds, which is given by:

dxds=1dsdx=11+(dydx)2−−−−−−−−√

Putting it all together gives us the formula for curvature, K:

K=dθds=d2ydx2[1+(dydx)2]32

Now the radius of curvature is just the reciprocal of this expression, that is:

R=[1+(dydx)2]3/2d2ydx2

Of course, the radius needs to be positive, so we take the absolute value of the denominator (bottom) of the fraction.

R=[1+(dydx)2]3/2∣∣d2ydx2∣∣

Example 1

Find the radius of curvature for the cubic

y = 2x3 − x + 3

at the point x = 1.

Answer

First, let's draw the graph and see what the question means.

y = 2x3 − x + 3

I have used equal scaling along the 2 axes (so that later, when I draw the circle, it will not have an elliptical shape).

Now, to find the radius of curvature, we need:

dydx=6x2−1

And then

(dydx)2=(6x2−1)2=36x4−12x2+1

So now we are ready to substitute into the formula to give us the radius

at any point x:

R=[1+(dydx)2]3/2∣∣d2ydx2∣∣=[1+36x4−12x2+1]3/2|12x|=[36x4−12x2+2]3/2|12x|

Now to find the radius of curvature at the required point x = 1, we

substitute:

⎡⎣[36x4−12x2+2]3/2|12x|⎤⎦x=1=11.04787562

To show what we have done, let's look at the graph of the curve (blue) with the approximating circle (dark red) overlaid. The circle is a good

approximation for the curve at (1, 4).

We can show that the center of the approximating circle is (−9.8,6.17).

How did I find that center?

We know the length of the radius shown in the diagram (11.05 units).

We know 1 point on that radius line, (1,4), and we need to find the one at

the other end, the center. Let's call it (x1, y1).

We're going to set up 2 equations using these 2 unknowns.

We know that

dydx=6x2−1

So the slope of the tangent at x=1 is

6(1)2 − 1 = 5.

Now, the slope of the normal (the line at right angles to the tangent at the point of contact) is −1/5. (SeeTangents and Normals.)

So we can use the formula y − y1 = m(x − x1) and the known

slope −15 and point (1,4) to find the equation of the line containing the

radius as follows:

y − 4 = (−1/5) (x − 1), which gives:

y=−x5+215

The unknown point (x1, y1) lies on this line, so we can say

y1=−x15+215

Next, we use the distance formula:

(x2−x1)2+(y2−y1)2−−−−−−−−−−−−−−−−−√=r

We know the distance r (since it is the radius), and one of our points is

(1,4), and using our expression connecting x1 and y1, we can solve these

simultaneously as:

(1−x1)2+[4−(215−15x1)]2−−−−−−−−−−−−−−−−−−−−−−−−−−√=11.04787562

This is one equation in one unknown, which after some algebra gives us

x1 = − 9.833 (and another positive solution which doesn't apply here)

Substituting this back into y1=−x15+215 gives

y1 = 6.167

So the center of the circle is (−9.8,6.17).

Exploration

In the following interactive graph you can explore what "changing radius of curvature" means.

Slowly drag the point "P" around the curve to see the changing radius of curvature (segment CP). It works best if you use a left-right motion - don't worry about following the up-down of the graph.

2.5

5

0

7.5

0

2.5

-2.5

– o + ← ↓ ↑ →

P

C

You'll notice at the point of inflexion there is interesting behavior. The circle changes from being below the curve to above (when moving left to right). When we are right on the point of inflexion, what does the circle become?

Example 2

[This example was supplied by a reader.]

We have a curve which is defined by data points and we don't know the function for this data. How can we find the radius of curvature?

We take any 3 data points to illustrate ways of solving this. I chose the

points (1, 1), (2, 3) and (3,8).

We'll do this in 3 different ways, just for fun (and for learning about how different math approaches can be used)!

Method 1: Approximation Using a Parabolic Fit and Calculus Methods

Answer

The graph connecting the generic data points (1,1), (2,3) and (3,8) is as

follows.

One way of finding the radius of curvature is to find the (unique) parabola passing through these 3 points. Parabolas are excellent for approximating curves in a local region. (See more on Parabolas.)

We proceed as follows. In general, a parabola has the form:

y = ax2 + bx + c

Substituting our given points (1, 1), (2, 3) and (3, 8) into the general form of the parabola gives us 3 equations in 3 unknowns:

1 = a + b + c

3 = 4a + 2b + c

8 = 9a + 3b + c

Solving this set of 3 simultaneous equations gives us:

a = 1.5

b = −2.5

c = 2

So the parabola passing through the points (1,1),(2,3) and (3,8) is

y = 1.5x2 − 2.5x + 2

We see this parabola passes through each of the 3 points:

Using this as our function (which we can only do in this local region for the original data points), we apply the formula for the radius of curvature:

Radius of curvature=[1+(dydx)2]3/2∣∣d2ydx2∣∣

We need to find the first and second derivatives and evaluate them at the

center point (2,3).

dydx=3x−2.5

At x=2 (the middle of our 3 points), dydx=3.5

Now for the second derivative:

d2ydx2=3

So the radius of curvature at the middle point (2, 3) is:

[1+(dydx)2]3/2∣∣d2ydx2∣∣=[1+(3.5)2]3/23=16.08

So we've found a parabola that approximates our function for the local area near our data points. Then we found the radius of curvature - that is, the radius of a circle that "fits" our curve near our data points.

Here are the 3 data points, the parabola we found, and the circle indicating

the curvature for the given 3 points. It has radius 16.08.

Method 2: Using Linear Approximations and Calculus Methods

Answer

We can approximate the value of dydx at the middle point (2,3) of our 3

points as follows.

The slope of the line joining (1,1) and (2,3) is given by:

m1=ΔyΔx=21=2

The slope of the line joining (2,3) and (3,8) is given by:

m2=ΔyΔx=51=5

We "average" these slopes to find a crude value for dydx:

Average slope=2+52=3.5≈dydx

Now for the slope of the slope (i.e. the second derivative). We find the

change in slope m divided by the change in x for the interval x = 1.5 to x =

2.5 (which are the mid-points of our 2 lines joining the 3 given points):

Slope of slope=ΔmΔx=5−21=3≈d2ydx2

Interestingly, these approximate values are exactly the same as our parabola approximation above.

Substituting into our radius of curvature formula, we obtain the same value as for Method 1:

[1+(dydx)2]3/2∣∣d2ydx2∣∣=[1+(3.5)2]3/23=16.08

Checking our answer, we can see that the (dark red) approximating circle

(center D, with radius 16.08) passes quite closely through our data points. It

may be possible to improve on this by taking weighted averages to get more appropriate values for the first and second derivatives.

Of course, my reader would need to do the same process for every set of 3 adjacent data points. He was writing a computer program to do this, so it would not be so tedious.

Method 3: Finding the Radius of the Circle through our 3 Points

Answer

This is an exact method for finding the required radius of curvature. We'll actually find the equation of the circle passing through the 3 points.

In general, the x-value for the center of the circle passing through 3

points A (x1, y1), B (x2, y2),C (x3, y3), joined by lines with

slopes m1 and m2 is given by:

xc=m1m2(y1−y3)+m2(x1+x2)−m1(x2+x3)2(m2−m1)

(The above formula is based on finding the intersection of the perpendicular bisectors of the 2 lines joining the 3 points, as follows.)

So for our given data points, we have:

xc=(2)(5)(1−8)+(5)(1+2)−(2)(2+3)2(5−2)=−10.833

(We found the values for m1 and m2 in Method 2, above.)

Now to find the y-value for center of the circle. The formula for the first

perpendicular bisector is given by:

yPerp=−1m1(x−x1+x22)+y1+y22

So for our data points, we can obtain the y-value for the center of the circle

by substituting the known values, as follows:

yc=−12(−10.833−1+22)+1+32=8.1665

So the center of the circle passing through the points (1,1),(2,3) and (3,8) is

given by (−10.83,8.17).

Finally, we can find the radius by simply finding the distance between the center of the circle and any one of the points on the circle. I have

chosen (1,1):

Radius=(x2−x1)2+(y2−y1)2−−−−−−−−−−−−−−−−−√

Radius=(1−(−10.833))2+(1+8.1665)2−−−−−−−−−−−−−−−−−−−−−−−−−√=13

.834

[See more on Distance Formula.]

So the radius of curvature for the 3 points (1,1),(2,3) and (3,8) is 13.83,

when finding the actual circle passing through the 3 points.

Our answer is slightly different to the answers obtained by using a parabolic model and linear approximations.

Integration

Why do we need to study Integration?

by M. Bourne

The Petronas Towers

of Kuala Lumpur

Often we know the relationship involving the rate of change of two variables, but we may need to know the direct relationship between the two variables. For example, we may know the velocity of an object at a particular time, but we may want to know the position of the object at that time.

To find this direct relationship, we need to use the process which is opposite to differentiation. This is

called integration (or antidifferentiation).

The processes of integration are used in many applications.

The Petronas Towers in Kuala Lumpur experience high forces due to winds.Integration was used to design the building for strength.

Sydney Opera House

The Sydney Opera House is a very unusual design based on slices out of a ball. Many differential equations (one type of integration) were solved in the design of this building.

Wine cask

Historically, one of the first uses of integration was in finding the volumes of wine-casks (which have a curved surface).

1. The Differential

Earlier in the differentiation chapter, we wrote dydx and f′(x) to mean the

same thing. We used ddx as anoperator.

We now see a different way to write, and to think about, the derivative.

Definition

Mini-Lecture

See the mini-lecture on differentials.

The differential of y=f(x) is written:

dy=f′(x)dx

Note: We are now treating dydx as a fraction, rather than as an operator.

Example

Find the differential of y=3x5−x.

Answer

dy=f′(x)dx

dy=(15x4−1)dx

We just need to find the derivative and write it with dy on the left side,

and dx on the right.

Flash demonstration

Note

We could use the differential to estimate the real change in value of a

function (Δy) caused by a small change in x(written as Δx). Many text

books do this, but it is pretty silly, since we can easily find the exact change - why approximate it?

We are introducing differentials here as an introduction to the notation used in integration.

How are dy, dx and Δy and Δx related?

As Δx gets smaller, the ratio ΔyΔx becomes closer to the "instantaneous"

ratio dydx.

That is, limΔx→0ΔyΔx=dydx

See Slope of a tangent for some background on this.

We now go on to see how the differential is used to perform the opposite process of differentiation, which first we'll call antidifferentiation, and later integration.

2. Antiderivatives and The Indefinite Integral

by M. Bourne

Mini-Lecture

We wish to perform the opposite process to differentiation. This is called "antidifferentiation" and later, we will call it "integration".

Example 1

If we know that

dydx=3x2

and we need to know the function this derivative came from, then we "undo" the differentiation process. (Think: "What would I have to differentiate to get this result?")

y=x3 is ONE antiderivative of dydx=3x2

There are infinitely many other antiderivatives which would also work, for example:

y=x3+4

y=x3+π

y=x3+27.3

In general, we say y=x3+K is the indefinite integral of 3x2. The

number K is called the constant of integration.

Note: Most math text books use C for the constant of integration, but for

questions involving electrical engineering, we prefer to write "+K",

since C is normally used for capacitance and it can get confusing.

Notation for the Indefinite Integral

We write: ∫3x2dx=x3+K and say in words:

"The integral of 3x2 with respect to x equals x3 + K."

Here is a template you can play with to get the idea.

LIVEMath

The Integral Sign

The ∫ sign is an elongated "S", standing for "sum". Later we will see that the

integral is the sum of the areas of infinitely thin rectangles.

Other Notation for Integrals

Note: Sometimes we write a capital letter to signify integration. For

example, we write F(x) to mean the integral of f(x). So we have:

F(x)=∫f(x)dx

Example 2

Find ∫(x2−5)dx

Answer

We now learn some important general rules for integration.

A. Integral of a Constant

∫k dx=kx+K

(k and K are constants.)

The integral of a constant is that constant times x.

Example 3

Find ∫4 dx

Answer

B. Integral of a Power of x

∫xndx=xn+1n+1+K (This is true as long as n≠−1)

For the integral of a power of x: add 1 to the power and divide by the new

number.

Example 4

Integrate ∫x5dx

Answer

We use our new rule, and obtain:

∫x5dx=x5+15+1+K

=x66+K

The Constant of Integration

Don't forget the "+ K" (or, alternatively, "+ C"). This constant of integration

is vital in later applications of the indefinite integral.

The next 2 examples use Flash to show the steps:

Example 5: How to do basic integration

Example 6

Example 7

∫7x6dx

Answer

∫7x6dx=7x77+K

=x7+K

Example 8

∫(3x2+x√−5x3)dx

Answer

Our first step in this question is to re-write the exponents so it is easier to integrate:

∫(3x2+x√−5x3)dx

=∫(3x2+x12−5x−3)dx

=x3+x3232−5x−2−2+K

=x3+2x323+5x−22+K

Example 9

∫x2−−√3dx

Answer

Once again, we re-write the exponents in amore convenient form.

∫x2−−√3dx=∫x2/3dx=⎛⎝x23+123+1⎞⎠+K=x5/35/3+K=3x5−−√35+K

Notice the constant of integration, +K, in each of these examples.

Example 10

A particular curve has its derivative given by dydx=3x2−2x.

We are told that the curve passes through the point (2, 5). Find the equation of the curve.

Answer

The first step for this problem is to integrate the expression (i.e. find

the antiderivative). This will give us the expression for y.

∫(3x2−2x)dx=x3−x2+K

So we have y=x3−x2+K

This represents a family of curves, and depends on the value of K for

the y-intercept.

We must now find the value of K from the information given in the question.

Since the curve passes through (2,5), we substitute these values into

y=x3−x2+K

to give:

5=(2)3−(2)2+K

5=8−4+K

So K=1.

So the required curve is y=x3−x2+1

Example 11

Consider this integration:

∫(2x4−5)6x3dx

This is different to the other exercises above!

The expression we have to integrate containts (2x4−5)6, which is a function

of a function, and we have thatx3 at the end. We cannot do this integration

using the rules we have learned so far.

In this case, we have to do the reverse of the Chain Rule, which we met in the section on differentiation.

We introduce a new rule for integrating cases like these.

C. Power Formula for Integration

∫undu=un+1n+1+K

(This is true if n≠−1)

Mini-Lecture

This requires a substitution step, where u(x) is some function of x.

Now back to the problem to see how to apply this formula.

Integrate ∫(2x4−5)6x3dx.

Answer

We use, as a starting point, the substitution

u=2x4−5.

Now du=8x3 dx

Our question has only one x3 dx so we need to divide both sides by 8:

18du=x3 dx

We can now rewrite our question as:

∫(2x4−5)6x3 dx

=18∫u6 du

=18×u77+K

=u756+K

=(2x4−5)756+K

Let's see how it works in Flash:

More examples

Example 12

∫(x3−2)6(3x2)dx

Answer

We use, as a starting point, u=x3−2.

Now du=3x2 dx

This time, this is exactly what we have in the question, so there is no need to divide both sides.

We can now write:

∫(x3−2)6(3x2) dx

=∫u6 du

=u77+K

=(x3−2)77+K

Example 13

Find ∫xx2+9−−−−−√dx using a substitution.

Answer

Put u=x2+9

Then du=2x dx

The question has just one x dx so we divide both sides by 2:

du2=x dx

Now

∫xx2+9−−−−−√dx

=12∫duu√

=12∫u−1/2du

=12u1/21/2+K

=u1/2+K

=x2+9−−−−−√+K

Example 14

Given y′=2x+1−−−−−√, find the function y=f(x) which passes through the

point (0,2).

Answer

NOTE: y′ means the derivative of y, that is dydx.

y′=2x+1−−−−−√

So

y=∫2x+1−−−−−√ dx

This question requires us to integrate, and in the process, to find the constant of integration.

Put u=2x+1.

Now du=2 dx

So

12du=dx

So

y=∫2x+1−−−−−√ dx

=12∫u1/2du

=12×23×u3/2+K

=13(2x+1)3/2+K

Now if x=0 then y=2 (from the question).

2=13/23+K

and this gives K=53.

So

y=(2x+1)3/23+53

is the required function.

Note: You will see "+K" and "+C" in this work. Most textbooks use + C.

It's a good idea to always use +K if you are answering electrical problems.

3. The Area under a Curve

by M. Bourne

A building has parabolic archways and we need to supply glass to close in the archways. How much glass is needed?

We need to know the area under the curve.

We'll see howto do this in 2 ways on this page:

1. Using an approximation (finding areas of rectangles) 2. Using integration

Before integration was developed, they could onlyapproximate the answer by dividing the space into rectangles and adding the areas of those rectangles:

The height of each rectangle is found by calculating the function values, as

shown for the typical case x = c. We get a better result if we take more

and more rectangles.

In the above diagram, we are approxcimating the area using inner rectangles (each rectangle is inside the curve). We could also find the area using the outer rectangles.

[This method was known to the Ancient Greeks. See Archimedes and the area of a parabolic segment.]

Example 1: Approximation using rectangles

(a) Find the area under the curve y = 1 − x2 between x = 0.5 and x = 1,

for n = 5, using the sum of areas of rectangles method.

Answer

The area we are trying to find is shaded magenta in this graph:

Since n=5, the width of each rectangle will be:

h=Δx=b−an=1−0.55=0.1

We aim to find the sum of the areas of the following 5 rectangles:

Now the height of each rectangle is given by the function value for that

particular x-value.

For example, since y=f(x)=1−x2, the first rectangle has height given by:

f(0.5) = 1 − (0.5)2 = 0.75

It has area given by:

Area1=0.75×0.1=0.075

The second rectangle has height:

f(0.6) = 1 − (0.6)2 = 0.64

The 5th rectangle has height

f(0.9) = 1 − (0.9)2 = 0.19

Adding the areas together gives us the following. (We are writing it using summation notation, which just means the sum of the 5 rectangles. Also, we are adding the heights first then multiplying by the width, which is the same for each rectangle.)

A=Σ5i=1Ai

=(0.75+0.64+0.51+0.36+0.19)(0.1)

=2.45(0.1)

=0.245

In the above answer, we are finding the area of the "outer" rectangles. To find a better approximation, we could also find the area of the inner rectangles, and then average the 2 results. The graph for the inner rectangles is as follows:

And this is the sum of the areas for the inner rectangles (the 5th one has height 0, so area 0):

A=Σ5i=1Ai

=(0.64+0.51+0.36+0.19+0)(0.1)

=1.7(0.1)

=0.17

The average of the 2 areas is given by: 0.245+0.172=0.2075.

A third way of doing this problem would be to find the mid-point rectangles. The diagram for this would be:

This time our area is

(0.6975+0.5775+0.4375+0.2775+0.0975)×0.1 =0.20875

(The first one comes from f(0.55) = 1 − (0.55)2 = 0.6975).

This answer is slightly above the average of the outer and inner rectangles, and less work!

Here we can see a demonstration of the above answer using LiveMath with a graph animation:

(b) Find the area under the curve given in part (a), but this time use n =

10, using the sum of areas of rectangles method.

Answer

Since n = 10,

h=Δx=1−0.510=0.05

We take the outer rectangles and find the areas (10 of them) as follows:

A=Σ10i=1Ai

=(0.6975+0.64+0.19+...+0.9755+0)(0.05)

=3.7875(0.05)

=0.189375

You can play with this concept further in the Reimann Sums section.

Finding Areas using Definite Integration

There must be a better way than finding areas of rectangles! Integration was developed by Newton and Leibniz to save all this "adding areas of rectangles" work.

General Case

[NOTE: The curve is completely ABOVE the x-axis].

By a Δ-process, we can show that in general, the exact area under a

curve y = f(x) from x=a to x=bis given by the definite integral:

Area=∫baf(x)dx

How do we evaluate this expression?

If F(x) is the integral of f(x), then

∫baf(x)dx=[F(x)]ba=F(b)−F(a)

This means:

To evaluate a definite integral, follow these steps:

integrate the given function (do not include the K)

substitute the upper limit (b) into the integral

substitute the lower limit (a) into the integral

subtract the second value from the first value

the answer will be a number

This forms part of The Fundamental Theorem of Calculus.

Mini-Lecture

Example 2: Evaluation of Definite Integral

Evaluate: ∫1013x2dx

Answer

∫1013x2dx

=[3×x33]101 (We integrate)

=[x3]101

=103−13 (Substitute upper and lower values and substract)

=1000−1

=999

Example 3: Arches problem

Returning to our arches problem above...

If the arch is 2 m wide at the bottom and is 3 m high, (i) find the equation of the parabola (ii) find the area under each arch using integration.

Answer

Mini-Lecture

See the mini-lecture on the area under a curve.

Example 4

Find the (exact) area under the curve y = x2 + 1 between x = 0 and x =

4 and the x-axis.

Answer

This is the area we need to find:

The area is given by:

∫40(x2+1)dx

=[x33+x]40

=(433+4)−(033+0)

=763 units2

≈25.3 units2

4. The Definite Integral

by M. Bourne

In the last section, we used the expression

∫baf(x)dx=[F(x)]ba=F(b)−F(a)

to find the area under a curve.

F(x) is the integral of f(x);

F(b) is the value of the integral at the upper limit, x=b; and

F(a) is the value of the integral at the lower limit, x=a.

This expression is called a definite integral. Note that it does not involve a constant of integration and it gives us a definite value (a number) at the end of the calculation.

We will use definite integrals to solve many practical problems. First, we see how to calculate definite integrals.

Example 1

Evaluate ∫51(3x2+4x+1)dx

Answer

This question requires us to:

1) Find the integral and then write the upper and lower limits with square brackets, as follows:

[x3+2x2+x]51

The upper and lower limits are written like this to mean they will be substituted into the expression in brackets.

2) Next, substitute 5 (the upper limit) into the integral:

[(5)3+2(5)2+5]=125+50+5=180

3) Then substitute 1 into the integral:

[(1)3+2(1)2+1]=1+2+1=4

4) Subtract the result of (3) from the result of (2) for our final answer:

180−4=176

Normally, we would write our complete solution as follows:

∫51(3x2+4x+1)dx

=[x3+2x2+x]51

=[(5)3+2(5)2+5]−[(1)3+2(1)2+1]

=180−4

=176

Note that our final answer is a number and does not involve "+ K". We are now dealing with definite integrals.

Example 2

Evaluate ∫94(2x+3x√)dx

Answer

This requires:

(1) Find the integral

(2) Substitute 9 into the integral

(3) Substitute 4 into the integral

(4) Subtract the result of Step (3) from the result of Step (2).

∫94(2x+3x√)dx

=[x2+2x3/2]94

=[92+2(9)3/2]−[42+2(4)3/2]

=135−32

=103

Explanation: The integral of x√ is as follows (I have not included the

constant):

∫x√dx=∫x1/2dx

=x3/23/2

=2x3/23

Example 3

Evaluate ∫1.61.2(5+6x4)dx

Answer

This requires:

1) Find the integral

2) Substitute 1.6 into the integral

3) Substitute 1.2 into the integral

4) Subtract.

∫1.61.2(5+6x4)dx

=∫1.61.2(5+6x−4)dx

=[5x−2x−3]1.61.2

=7.5117−4.8426

=2.6691

Definite Integrals and Substitution

Recall the substitution formula for integration:

∫undu=un+1n+1+K (if n≠−1)

When we substitute, we are changing the variable, so we cannot use the same upper and lower limits. We can either:

Do the problem as an indefinite integral first, then use upper and lower limits later

Do the problem throughout using the new variable and the new upper and lower limits

Show the correct variable for the upper and lower limit during the substitution phase.

We will be using the third of these possibilities.

Example 4

Find

∫0−1x3(1−2x4)3dx

using a substitution.

Answer

Put u=1−2x4

Then du=−8x3dx

The question has x3dx so we write

−du8=x3dx

So we have:

∫0−1x3(1−2x4)3dx

=−18∫x=0x=−1u3du

=−18×[u44]x=0x=−1

=−132[u4]x=0x=−1

=−132[(1−2x4)4]0−1

=−132[(1−0)4−(1−2)4]

=−132[(1)−(1)]

=0

NOTE: If we take the second approach, i.e. leaving the variable changed, we have:

u=1−2x4

As x=−1→0 (this means "as x takes values from −1 to 0"),

then u=−1→1

So the problem becomes:

∫0−1x3(1−2x4)3dx

=−18∫1−1u3du

=−132[u4]1−1

=−132[14−(−1)4]

=0 as before.

This second approach is quite useful later when the substitutions become more involved (e.g.trigonometric substitution).

Application: Work

Einstein riding his bicycle.

In physics, work is done when a force acting upon an object causes a

displacement. (For example, riding a bicycle.)

If the force is not constant, we must use integration to find the work done.

We use

W=∫baF(x)dx

where F(x) is the variable force.

Example 5

Find the work done if a force F(x)=2x−1−−−−−√ is acting on an object and

moves it from x = 1 to x = 5.

Answer

To solve this, we need to evaluate

∫512x−1−−−−−√ dx

Put u=2x−1 then du=2dx

So du2=dx

So we have:

∫512x−1−−−−−√dx

=12∫x=5x=1u1/2du

=12×[23u3/2]x=5x=1

=13[(2x−1)3/2]51

=13[93/2−13/2]

=13[27−1]

=263

So the work done is 8.67 units..

Application: Average Value

The average value of a function f(x) in the region x = a to x = b is given

by:

Average=∫baf(x)dxb−a

Example 6

Find the average value of x(3x2 - 1)3 from 0 to 1.

Here is the graph of the situation:

Answer

Application: Displacement

If we know the expression, v, for velocity in terms of t, the time, we can find

the displacement (written s) of a moving object from time t = a to

time t = b by integration, as follows:

s=∫bav dt

Example 7

Find the displacement of an object from t = 2 to t = 3, if the velocity of the

object at time t is given by

v=t2+1(t3+3t)2

Answer

To find the displacement, we need to evaluate:

∫32t2+1(t3+3t)2dt

Put u=t3+3t, then du=(3t2+3)dt=3(t2+1)dt

So du3=(t2+1)dt

So we have:

∫32t2+1(t3+3t)2

=13∫t=3t=21u2du

=13∫t=3t=2u−2du

=−13[1u]t=3t=2

=−13[1t3+3t]32

=−13[133+3(3)−123+3(2)]

=−13[136−114]

=0.014550

So the displacement of the object from time t=2 to t=3 is 0.015 units.

See more on: displacement, velocity and acceleration as applications of integration.

NOTE 1: As you can see from the above applications of work, average value and displacement, the definite integral can be used to find more than just areas under curves.

NOTE 2: The definite integral only gives us an area when the whole of the

curve is above the x-axis in the region from x = a to x = b. If this is not

the case, we have to break it up into individual sections. See more atArea Under a Curve.

We now examine a definite integral that we cannot solve using substitution.

Not every integral can be integrated using substitution...

Consider this question.

Find: ∫10x2+1−−−−−√ dx

Attempted solution:

We try to put u=x2+1.

Then we find the differential:

du=2x dx

But the question does not have "2x dx" (it only has "dx") so we cannot

replace anything in the question with "du" properly. This means we can't

solve it using any of the integration methods used above. (Note: This question can be done using Trigonometric Substitution, however, but we don't meet trigonometric substitution until later.)

Explanation: Let's think about if the question said this instead:

∫10x2+1−−−−−√(2x)dx

This time, there is a "2x dx" involved in the question and so we would be

able to let:

u=x2+1

Then we would find the differential:

du=2x dx

Then we could proceed to find the integral like we did in the examples

above, by replacing 2x dx with duand the square root part with u√.

However, the problem ∫10x2+1−−−−−√ dx does not have a "2x" outside of

the square root so I cannot use the "u" substitution.

For now, we need to use numerical approaches to evaluate the integral

∫10x2+1−−−−−√ dx

(Note: Historically, all definite integrals were approximated using numerical methods before Newton and Leibniz developed the integration methods we have learned so far in this chapter.)

We can use two different numerical methods for evaluating an integral:

Trapezoidal rule Simpson's Rule

We meet these methods in the next 2 sections.

5. The Trapezoidal Rule

by M. Bourne

Problem: Find

∫10x2+1−−−−−√ dx

We put u=x2+1 then du=2x dx.

But the question does not contain an x dx term so we cannot solve it

using any of the integration methods we have met so far.

We need to use numerical approaches. (This is usually how software like Mathcad or graphics calculators perform definite integrals).

We can use one of two methods:

Trapezoidal rule Simpson's Rule (in the next section: 6. Simpson's Rule)

The Trapezoidal Rule

We saw the basic idea in our first attempt at solving the area under the arches problem earlier.

Instead of using rectangles as we did in the arches problem, we'll use trapezoids (trapeziums) and we'll find that it gives a better approximation to the area.

Recall that we write "Δx" to mean "a small change in x".

Let's see this in LiveMath. Note that our approximation is much better than using rectangles.

Now, the area of a trapezoid (trapezium) is given by:

Area=h2(p+q)

So the approximate area under the curve is found by adding the area of the trapezoids. (Our trapezoids are rotated 90° so that their new base is

actually the height. So h = Δx.)

Area≈12(y0+y1)Δx+12(y1+y2)Δx+12(y2+y3)Δx+...

We can simplify this to give us the Trapezoidal Rule, for n trapezoids:

Area≈Δx(y02+y1+y2+y3+...+yn2)

To find Δx for the area from x=a to x=b, we use:

Δx=b−an

and we also need

y0=f(a)

y1=f(a+Δx)

y2=f(a+2Δx)

…

yn=f(b)

Note

We get a better approximation if we take more trapezoids [up to a limit!].

The more trapezoids we take, Δx will tend to 0, that is, Δx→0.

We can write (if the curve is above the x-axis only

between x=a and x=b):

Area=∫baf(x)dx≈Δx(y02+y1+...+yn2)

Exercise

Using n=5, approximate the integral:

∫10x2+1−−−−−√ dx

Answer

Here, a=0 and b=1.

Δx=b−an=1−05=0.2

y0=f(a)= f(0)=02+1−−−−−√=1

y1=f(a+Δx)= f(0.2)=0.22+1−−−−−−√=1.0198039

y2=f(a+2Δx)= f(0.4)=0.42+1−−−−−−√=1.0770330

y3=f(a+3Δx)= f(0.6)=0.62+1−−−−−−√=1.1661904

y4=f(a+4Δx)= f(0.8)=0.82+1−−−−−−√=1.2806248

y5=f(a+5Δx)= f(1.0)=12+1−−−−−√=1.4142136

So we have:

Integral≈

0.2(12×1+1.0198039 +1.0770330+1.1661904 +1.2806248+12×1.4142136)

=1.150

So ∫10x2+1−−−−−√ dx≈1.150

6. Simpson's Rule

by M. Bourne

The Trapezoidal Rule is an improvement over using rectangles because we have much less "missing" from our calculations. We used straight lines to model the curve in trapezoidal Rule.

We seek an even better approximation. In Simpson's Rule, we use parabolas to approximate each part of the curve. This proves to be very efficient. (See more about Parabolas.)

We can show (by integrating the area under each parabola and adding these areas) that the approximate area is given the following. This is called Simpson's Rule:

Area=∫baf(x)dx

≈Δx3(y0+4y1+2y2+4y3+2y4...+4yn−1+yn)

Note: In Simpson's Rule, n must be EVEN.

Example

Approximate ∫32dxx+1using Simpson's Rule with n=4.

Answer

Δx=b−an=3−24=0.25

y0=f(a)=f(2)=12+1=0.3333333

y1=f(a+Δx)=f(2.25)=12.25+1=0.3076923

y2=f(a+2Δx)=f(2.5)=12.5+1=0.2857142

y3=f(a+3Δx)=f(2.75)=12.75+1=0.2666667

y4=f(b)=f(3)=13+1=0.25

So

Area=∫baf(x)dx

≈0.253(0.333333+4(0.3076923) +2(0.2857142)+4(0.2666667)+0.25)

=0.2876831

[The actual answer to this problem is 0.287682 so our Simpson's Rule

approximation has an error of only0.00036%.]