chapter three: applications of the derivative hare krsna hare krsna krsna krsna hare hare hare rama...

CHAPTER THREE:CHAPTER THREE:APPLICATIONS OF APPLICATIONS OF THE DERIVATIVETHE DERIVATIVE

Hare Krsna Hare Krsna Krsna Krsna Hare Hare

Hare Rama Hare Rama Rama Rama Hare Hare

Jaya Sri Sri Radha Vijnanasevara (Lord Krsna, the King of Math and Science)

KRSNA CALCULUS™ PRESENTS:

Edited on October 7, 2003

HARE KRISHNA!

Welcome back, dear students! I pray that you are all well.

In order to understand this chapter, you will have to know the following:

Definition of the derivative (at least the concept of it difference quotient.), derivatives of polynomials, trig functions, implicit differentiation, chain rule, and the rectilinear motion relationships (position, velcoity, and acceleration.)

If you do not understand this, please review the last chapter (Chapter 2) and keep practicing that chapter until you understand those topics above.

WHERE WE HAVE GONE?

As one mathematician said, “Calculus is a journey. A good tourist will think about the past here and use it to him or her in the future.”

Before we move onto the next chapter, let’s take a step back and see what we have covered so far, just to help us remember important things.

SUMMARY OF PREVIOUSLY STUDIED MATERIAL In chapter 1, we had studied limits. An informal and

most commonly used definition of the limit is simply the question, “When I make x closer and closer and closer to some number c, what f(x) am I approaching to?”

Asymptotes were explored using limits. Limits help determine end-behavior.

Continuity was briefly mentioned. Continuity is present at x=c is f(c) exists, if the limit as x approaches c from both sides exist and are the same value, and if the limit as x approaches c and f(c) are equal.

SUMMARY

With the help of limits, we were able to calculate instantaneous rates, and the slope of the line tangent to the curve at a certain point. By letting the change in x be infinitesimally small in the slope formula (difference quotient), we are able to get derivatives.

A differentiable function is continuous function, but a continuous function is not always differentiable.

DERIVATIVES.

Practice of derivative rules and formulas must be emphasized. Derivatives and Integrals (next chapter) will be the backbone to calculus.

Definitely study the differentiation rules from the last chapter!!!!!!

APPLICATIONS OF THE DERIVATIVE This chapter will underscore the uses of

derivatives in the following cases. How to graph functions by knowledge of

critical points and derivatives. How to differentiate inverse, exponential and

logarithmic functions. Differentials Related rates and optimization problems

CURVE SKETCHING

You can easily graph any function by knowing three things.

1) ZEROS AND UNDEFINED SPOTS 2) MAXIMUM AND MINIMUM POINTS 3) CONCAVITY AND INFLECTION POINTS.

CRITICAL POINTS

Let’s find the points where the function is zero or undefined. We want to graph f(x)=x3+2x2+x. Obviously, there are no undefined spots, due to the fact that this

is a polynomial function. All polynomial functions are continuous, thus differentiable.

To find the zeros, set the entire function to 0. 0= x3+2x2+x Original function 0=x(x2+2x+1) Factored 0=x 0=(x+1)(x+1) Factored the trinomial and set both factors

equal to zero. x=0, -1 x is solved for. Thus x=0 and x=-1

MAXIMUM OR MINIMUM POINTS Here is the graph of the

f(x).

MAXIMUM AND MINIMUMNotice that if you draw tangent lines at the maximum and the minimum, you will see that you will get horizontal lines. That means the slope is zero. Thus, the derivative is zero. In other words, at a maximum or minimum, the derivative is zero.

MAXIMUM AND MINIMUM POINTS Since the f’(x)=0, we can solve for the maximum and minimum

points. First, find the derivative of f(x)=x3+2x2+x. f’(x)= 3x2+4x+1 Differentiation using power rule and

sum/difference rule. 0 = 3x2+4x+1 Max/Min always have f’(x)=0. 0=(3x+1)(x+1)Factoring to solve for x. 0=3x+1 0=x+1 Setting factors equal to 0. x=-1/3 and x=-1 Solve for two answers of x. We know that x= -1/3 and x=-1 are the possible points, however,

without a graph, we cannot determine which is the max and which one is the min.

SIGN ANALYSIS TEST FOR MAX/MIN. You can do a sign chart and determine the sign between possible

zeros. Look how it is set up. What the sign of the two factors to the right of x=-1/3? What is the sign of the two factors between -1 and -1/3? What is the sign of the two factors to the left of -1? Since they are factors, you multiply them. Account for overall sign.

DERIVATIVE OF SIGNS

If the derivative goes from negative slope to zero to positive slope, then x is where the minimum is.

If the derivative goes from positive slope to zero to minimum slope, then x is where the maximum is.

Negative slope

zero

Pos

itive

Slo

pe

Negative slope

zero

Pos

itive

Slo

pe

WHAT WE KNOW

We know that this is a cubic function, with x=-1 and x=0 as the two zeros. No undefined points

We know that the maximum is at x = -1 and the minimum is at x= -1/3.

HORIZONTAL TANGENTS

Be careful when solving for x when f’(x) is 0. Perhaps you will have such function where the slope is zero at a certain point, but that point is neither maximum nor minimum. y(x)=x3 for example. Once you differentiate that, you get y’(x)=3x2. y’(x)=0. Therefore, you equate 0 and 3x2. You will see that x = 0. However, by doing a sign analysis chart, you will see that before and after 0, there is no sign change. The slope continues to be positive.

CONCAVITY

Concavity is very difficult to define. A good way to define concavity is using a

spoon. When you hold a spoon the right way (where the milk is in the spoon), the spoon is said to be concave up.

When the spoon is down (when the milk is not in the spoon… the spoon is not in the right way), then the spoon is said to be concave down.

CONCAVITY

An upward parabola is concave up. A downward parabola is concave down.

The f(x) is above the tangent line when it is concave up. The f(x) graph is under the tangent line when it is concave down.

The concavity is the rate of slope change. Hint hint!! RATE OF CHANGE!! The rate of change of the slope, means the derivative of the slope… The slope is the derivative of the function. So the concavity is the derivative of the derivative. The first derivative of slope and the second derivative of the function.

CONCAVITY GRAPHS

Both function have positive slopes. f’(x)>0.However, the red function is concave up, since it is above the tangent line. The blue function is concave down, since it is below its tangent lines.

Both function have negative slopes. f’(x)<0. The blue function is concave up, since it is above the tangent line. The red function is concave down, since it is below its respective tangent line.

REDf’(x)>0f’’(x)>0

BLUEf’(x)>0f’’(x)<0

BLUEf’(x)>0f’’(x)>0

REDf’(x)>0f’’(x)<0

ZERO CONCAVITY

Remember when the function, f(x), was at a maximum or minimum, the derivative, the slope, f’(x) was 0.

When the slope is at its maximum or minimum, the derivative of the slope is 0.

In order words, when f’(x) is at its maximum or minimum, then the derivative, f’’(x), the concavity will be 0.

Concavity is the slope of the slope. !!

Side note: Vertical tangents.

Let’s examine for a moment, the vertical line graph.

There is no such function that when graphed has a vertical line.

If you look at the slope of a line (y/x). You will see that the difference in y is some number. However, since there is no change in x, that will be equal to zero. n/0 is infinitely big that is undefined.

However, there are functions with vertical tangents. The cube root of x at x=0 has a vertical tangent. Therefore, the derivative of the cube root function will be defined everywhere except x=0.

Just be aware of that!

CONCAVITY AND INFLECTION POINTS The point where the change in derivative sign

(positive vs. negative slope) is called extrema (i.e. max/min points) f(max) and f(min), their derivatives at those points will both be 0.

The point where the change in concavity occurs is called the inflection point.

Since this is where the slope, f’(x), is the maximum, the steepest or the flattest, the derivative of slope, f’(x), will be 0.

After solving for x and doing a sign analysis, you will find the inflection points.

FINDING INFLECTION POINTS The original function was x3+2x2+x. Since we want to find the inflection point, we must obtain the second

derivative. f(x)= x3+2x2+x. GIVEN f’(x)= 3x2+4x+1. FIRST DERIVATIVE f’’(x)=6x+4 SECOND DERIVATIVE Since the inflection points are found by solving for x when f’’(x) = 0. 0=6x+4 f’’(x) =0 x= -2/3 SOLVED FOR X. Now we must do sign analysis. Everything right of -2/3 for the factor 6x+4 is positive. Everything left

of it is negative. Thus, x=-2/3 is an inflection point for x3+2x2+x. Everything right of x=-2/3 is concave up. Everything the to left of x=-

2/3 is concave down.

When sketching this curve…

Make sure your zeros are at the x values. Make sure your max, mis, and horizontal

tangents are drawn at the proper spots. Make sure your inflection point is accurately

drawn. A good way of thinking about inflection points

is when the tangent lines switch sides with respect to the graph.

YOUR GRAPH

INFLECTION POINT

MAXIMUMZERO

MINIMUM

ZERO

f’(x)>0 f’’(x)<0

f(x)=0f’(x)=0f’’(x)<0

f’(x)<0f’’(x)<0

f’(x)<0f’’(x)=0

f’(x)<0f’’(x)>0

f’(x)=0f’’(x)>0

f(x)=0

f’(x)>0 f’’(x)>0

ROLLE’S THEOREM

Some people hate calculus a lot. For example, they might like to find instantaneous velocity by saying distance/time. But we know that is not true.

Rolle was a mathematician who tried to disprove calculus. He disliked it very much. He wanted to a lot of geometry and algebra. Well, you know, if you hate something a lot, it is just as much as loving it a lot. Similarly, Rolle hates calculus so much, that he had a theorem that is very helpful IN calculus.

This theorem will not be tested very frequently in exams, but its best to know it anyway.

ROLLE’S THEOREM (MEAN VALUE THEOREM) There will always be an

average rate line that will be parallel to the tangent line of a certain point. This means they have the same slope.

You can see that the average rate line between x=-2 and x=2 is parallel to the tangent line of x=0. Thus, you can conclude, that they have the same slope.

)(')()(

cfab

afbf

DIFFERENTIATION OF INVERSE FUNCTIONS An inverse of a function is found by reflecting the

graph over the line y=x. The function is found simply by switching the y with

the x and vice verse.. For example: If y = 3x+2 with coordinates (0,2) as one possible point, then its inverse is x=3y+2 with coordinates (2,0). I just switched the x and y around. That’s all.

Sometimes, a function may have an inverse that is not a function. y=x2 is function and x=y2 is not. Remember, a function is a function if for any x, there is one and only one y value. The vertical line test determines that. (Course III Info for N.Y.S students).

DIFFERENTIATION OF INVERSE FUNCTIONS From pre-calculus, you know that if f(x) has its inverse

g(x), then the composition f(g(x)) is x. If f(x) has g(x) as its inverse function, then the formula

for finding the derivative of a function is….

))(('

1

xgfdx

dy

DERIVATIVE OF THE INVERSE OF y=x2. f(x) g(x) Composition of f and g dy/dx of f(x) Composition of f’ and g.

Inv. Deriv. Formula

Substitution.

xdx

dy

xgfdx

dy

xxgf

xxf

xxgf

xxg

xxf

2

1

))(('

1

2))(('

2)('

))((

)(

)(

2

2

CHECK You can check it also using the power rule.

xx

dx

dy

xxy

2

1

2

1 2

1

2

1

DERIVATIVE OF EXPONENTIAL FUNCTIONS The natural exponential function, f(x)=ex, has the base e.

e is a transcendental number (just like 16,108, , ...). It is named after its founder, Leonard Euler. e is derived and found many ways. e is a very special number in calculus. You will see why in a moment. e is approximately 2.718… If you define it using limits, e = …

x

x xe

11lim

DERIVATIVE OF THE EXPONENTIAL FUNCTION Derivative definition f(x)=ex

Law of exponents, ex+h=(ex)(eh) Factor out ex.

Previously proven limit

Product Rule of limits Derivative

x

x

h

h

hx

h

xhx

h

xhx

h

h

e

e

h

e

h

ee

h

eee

h

ee

h

xfhxf

)1(

11

lim

)1(lim

lim

lim

lim

0

0

0

0

0

THE DERIVATIVE OF THE EXPONENTIAL FUNCTION This is one of the two functions whose derivative is

its own function! d(ex)=ex! Amazing how transcendental functions work! Jaya! Jaya!

xx eedx

d

THE NATURAL LOGARITHM FUNCTION y=ln x The natural logarithm is NOT taking a piece of log

naturally from a tree, play rhythms on it! The natural logarithm is the inverse of the natural

exponential function. It has e as its base. There are many ways y=ln x is derived by.

y=ln x is defined for all x>0. We can find the derivative using the inverse

differentiation rule since ln x is the inverse of ex.

DIFFERENTIATING y=ln x. f(x) g(x) inverse of f(x) Composition of f(x) and g(x) dy/dx of f(x) Composition of f’(x) and g(x) If f’(g(x)) = f(g(x)), and if f(g(x)) = x,

then f’(g(x))=x. Inverse Differentiation Rule

Replacing f’(g(x)) with x.

xdx

dy

xgfdx

dy

xexgf

exf

exgf

xxg

exf

x

x

x

x

1

))(('

1

))(('

)('

))((

ln)(

)(

ln

ln

THEREFORE…

I find it pretty cool that an algebraic function like y = 1/x is the derivative of transcendental function y = ln x. Transcendental functions are functions that cannot be derived simply by algebra.

x

xdx

d 1ln

INVERSE TRIGONOMETRIC FUNCTIONS The inverse of the function y = sin x is y= arcsin x. NOTE: arcsin x is NOT a function. For example, arcsin (½) = /3, 5/6,

-7/6, -11/6, so on… since one x value produced an infinite number of y values, you know that this is not a function

However, Arcsin x IS a function (notice capital A, compared to previous lower-case a). Arcsin x is restricted from [-/2, /2], just as Arccos x is restricted from [0,] and Arctan x is restricted from (-/2, /2). Note that the () parenthesis represent the open interval excluding endpoints. Remember than tan (/2) = ∞.

If you look at the inverse cosine function, you will notice that it is just the inverse sine function reflected over the x axis. Therefore, the slope, the derivative, of the inverse cosine function will be negative the slope, derivative, of the inverse sine function. Same thing will occur with tangent and cotangent, secant and cosecant functions.

d(arcsin x)=-d(arccos x) d(arctan x)=-d(arccot x) d(arcsec x)=-d(arccsc x)

DIFFERENTIATION OF THE ARCSINE FUNCTION f(x) g(x) inverse function Composition of f and g. dy/dx of f Composition of f’ and g. Derivative formula for

finding inverses of functions

Applying f’(g(x)). What is cos(arcsin x)?

)cos(arcsin

1

))(('

1

)cos(arcsin))(('

cos)('

)sin(arcsin))((

arcsin)(

sin)(

xdx

dy

xgfdx

dy

xxgf

xxf

xxgf

xxg

xxf

The denominator

To simplify the denominator to more practical terms, we know that y=arcsin x is the same as sin(arcsin x)= sin y = x. Then to solve for cos y.

2

22

22

1cos

1cos

1cossin

xy

yx

yy

Pythagorean Trig Identity

sin y = x

Solved for cos y.

AFTER APPLYING cos y IN THE DENOMINATOR… We see that…

2

2

1

1arccos1

1arcsin

xdx

xdxdx

xd

USING THE SAME PROCESS… If you use the derivative of an inverse rule for the

arctangent and the arcsecant functions, you will see that

2

2

1

1cot

1

1arctan

xxarc

dx

dx

xdx

d

1

1csc

1

1sec

2

2

xxxarc

dx

d

xxxarc

dx

d

MISCELLANEOUS THINGS TO KNOW

Some textbooks will make great emphasis on this. Other textbooks won’t talk about it too much. Since the AP and many college calculus I courses do not test this topic greatly, I won’t discuss it in great depth.

You can treat dy/dx as a fraction of the infinitesimally small change in y over the infinitesimally small change in x.

You can consider y = x2, dy = 2x dx since dy/dx = 2x. dy is said, “a differential amount of y.” Differential means an infinitesimal change in a specific direction.

EXAMPLE DIFFERENTIAL PROBLEM If you take a cube with each side having

length 8, what is the differential volume if you have a cube that has each side having length 7.99?

PROBLEM WORKED OUT

The volume of a cube is given

dV/dx is the differential amount of volume over the differential amount of change in length. The change in length is simply 8-7.99=.01=dx. Multiply both sides by it so you isolate dV. Then plug in 7.99 for x, and plug in 0.01 for dx and calculate the differential amount of volume.

92.1

)01(.)99.7(3

32

2

3

dV

dV

dxxdV

xV

APPLICATION

At the Sri Sri Gaura Nitai temple, they want to make the rectangular temple room such that the perimeter is 1000 feet. Find the maximum length and width so that everyone can have the fortune to the Sri Sri Gaura Nitai. In addition, find the area.

OPTIMIZATION

Optima- what??? Optimization problems is a fancy way of saying

maximum minimum problems. The best way to start out is to draw a picture to help

yourself visualize the problem.

THEN..

Since we have two variables, ℓ and w, we have to solve everything in terms of one variable.

We know that A= ℓ w and Prect = 2 ℓ +2w. So we can solve for ℓ in either equation. Most

preferably use the P equation.

2

2rect wP

And finally…

Substite the expression for l into the area formula.

2rect

rect

rect

2

1

2

2

2

2

wwPA

wwP

A

wA

wP

According to the problem…

P = 1000 ft so…

But to find maximum w, the derivative will have to equal 0.

Then solve for w.

2500 wwA

w

w

wdw

dA

wwA

ft 250

2500

02500

500 2

Then…

Plug w in for the expression for length.

Coincidentially, the width and length are the same. Therefore, the temple room will be square so that everyone could see the Lord Sri Sri Gaura Nitai Jaya!

ft 250

250500

5002

21000

w

w

FIND THE AREA…

Simple.. The length is 250 ft. The width is 250 ft. Multiply them together..

A= 625,000 sq ft.

Nice big temple ain’t it?? HARI HARI BOL!!!!

RELATED RATES

Remember how I constantly emphasized, RATE = DERIVATIVE

Rate specifically deals with time. Here is an example: The diameter and height of a paper cup in the

shape of a cone are both 4 in. and water is leaking out at the rate of 0.5 in3/ sec. Find the rate at which the water level is dropping when the diameter of the surface is 2 in.

VARIABLES SOUND NICE.. (laughs) V is volume in cubic

inches h is height. t is time Volume formula

(known) 2r = h, since

diameter=h.

12

π

2π

3

1

π3

1

3

2

2

hV

hh

V

hrV

Differentiate with respect to time

Remember now… we are differentiating with time!!! So, we must differentiate implicitly!! Remember chapter two??

Since we know that the rate of leak is ½ cu in/ sec, dV/dt is – ½ since we are decreasing.

Solve for the rate of change for the height.

Since there is no variable for time, the rate of change for the height is constant for any value of t.

Replace the h with 2 in and the rate of the water level is dropping with the height and time is 1/2 in/sec. 2π

1

2π

4

2

1

π

4

2

12

1

4

π

12

π3

12

π

2

2

22

3

dt

dh

hdt

dhdt

dV

dt

dhh

dt

dhh

dt

dV

hV

SUMMARY

We covered a good amount of material in this chapter. Let’s go over what we learned.

You can find the maximum or minimum of f(x) by setting f’(x) = 0. Solve for x and do a sign analysis.

You can find concavity of f(x) by setting f’’(x)=0. Solve for x and do a sign analysis.

DERIVATIVES

We learned how to differentiate more functions using the inverse rule.

COMMIT THESE TO MEMORY AS WELL AS THE FUNCTIONS IN CHAPTER TWO!!! IF YOU DO NOT REMEMBER ANY FUNCTION’S DERIVATIVE, DO NOT GO ON TO CHAPTER FOUR!! COMMIT EVERYTHING TO MEMORY!!!!

2

2

2

2

2

2

1

1cot

1

1csc

1

1sec

1

1arctan

1

1arccos1

1arcsin

1ln

xxarc

dx

d

xxxarc

dx

d

xxxarc

dx

dx

xdx

dxdx

xdxdx

xdx

xdx

d

eedx

d xx

OPTIMIZATION/RELATED RATES For optimization problems, relate everything to one

variable. For example, area and perimeter with sides a and b. Solve for a, for example, and plug the a expression in for the area formula so you have an area function A(b). Then differentiate that and set it equal to 0. DON’T FORGET SIGN ANALYSIS!!!

For related rate problems, gather all of your variables together and use any geometry formulas you need. Differentiate with respect to TIME. You most likely will need to use implicit differentiation. It

It helps in both scenarios to draw a picture of the problem to get a better understanding.

TEST

Please visit the following website to take a practice exam and please look over the answers.

If you have any difficulty, please e-mail me at kksongs_1@hotmail.com.

Please read help statement prior to e-mailing me.

Good luck! Hare Krsna!