Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 5 pptx

The distance between the wall of the house and a point on the ladder is a function of the height, h, of the point on the ladder.. The number of pounds of sugar cane it takes to support a

iii The volume of liquid in this beaker, V , is 9π h We want to calculate height given volume, so we solve for h in terms of V :

9π.

The height is proportional to the volume We can see this from the function formula,

h = Cb(V ) =9π1 · V , or from the physical situation itself The walls of the beaker are perpendicular to the base; therefore all cross-sections are equal in area and the height

In the examples that follow we will work on expressing one variable as a function

of another These examples are chosen to highlight relationships that will arise repeatedly throughout our studies They will also serve as a review of some geometry including similar triangles and the Pythagorean Theorem (For a summary of some useful geometric formulas, refer to Appendix B.)

and paper A solution will have more impact, and stay with you longer, if you have spent a bit of time tackling the problem yourself Read the problems that follow and try each one on your own before reading the solutions The problem-solving strategies highlighted below should help you out Think of them as a way of coaching yourself through

a problem

Portable Strategies for Problem Solving

Draw a picture whenever possible

Label known quantities and unknown quantities so you can refer to them Make your labeling clear and explicit

Take the problem apart into a series of simpler questions Plan a strategic approach to the problem

Make the problem more concrete (or simpler) in order to get started Either solve the problem with the concrete numbers and try to generalize, or spot check your answer

by making sure it works in a concrete example or two

EXAMPLE 1.7 Functioning around the house. An 8-foot ladder is leaning against the wall of a house If

the foot of the ladder is x feet from the wall, express the height of the top of the ladder as

a function of x

SOLUTION We begin with a picture, labeling all known and unknown quantities Let y be the height of

the top of the ladder

ladder against wall

wall

8 ft.

y ft.

x ft.

Figure 1.8

The Pythagorean Theorem tells us how x and y are related

x2+ y2= 82

y2= 64 − x2

y = ±64 − x2, but height is never negative, so

y =64 − x2 Therefore, y = f (x) =√64 − x2, where x ∈ [0, 8]

EXAMPLE 1.8 Functioning around the house. A 13-foot ladder is leaning against the wall of a house The

foot of the ladder is 5 feet from the house and the top has a height of 12 feet The distance between the wall of the house and a point on the ladder is a function of the height, h, of the point on the ladder Write a formula for this function

SOLUTION Let d be the distance between the point on the ladder and the wall

ladder against wall

wall

13 ft.

12 ft.

5 ft.

13

12 – h

h

5 h

d

Figure 1.9

We must relate d and h We can do this by using similar triangles We’ll then solve for d in terms of h

d

5 12

12· (12 − h)

12 Therefore, d = f (h) = 5 −5h12

Thinking ahead can save energy.There were several different algebraic options when using similar triangles to relate d and h Since our goal is to solve for d, our work is simplified

by using a relationship that is written so that d is in the numerator of the ratio

EXERCISE 1.2 Functioning in the evening Late in the evening an elegant 512-foot-tall woman is standing

by a 14-foot-high street lamp on a cobbled road The length of the shadow she casts is a function of her distance from the lamppost Write a formula for this function, where x is the distance between the woman and the lamppost

elegant woman under streetlamp

14 ft

5.5 ft

x ft

woman

shadow length =

Figure 1.10 Answer

The length of her shadow is f (x) =1117x

EXAMPLE 1.9 Functioning in the morning.We’re making a pot of coffee using a conical coffee filter

The coffee filter holder is a right circular cone with a radius of 6 centimeters at the top and

a height of 12 centimeters Express the amount of liquid in the filter as a function of the height of the liquid

h

12 cm

6 cm

12-h

h r 6

12

coffee in a coffee filter

Figure 1.11

SOLUTION We can use the volume of a cone to relate the amount of liquid, V , and the height, h, of the

liquid

3π r

2h

Here r is the radius of the top of the cone of liquid, not of the conical filter itself, so r varies with the amount of liquid But now V is expressed as a function of two variables, r and h.

We would like the volume expressed as a function of h only Since r depends on h, we must express r in terms of h We can do this using similar triangles

h 12

2· h Therefore,

3π

 h 2

2 h

3π

h2

12π h

EXAMPLE 1.10 Functioning in the kitchen.Chocolate pudding is being served in hemispherical bowls

with a radius of 2 inches The top skin of the pudding forms a disk whose radius, r, depends upon the height, h, of the pudding Express r as a function of h

SOLUTION First, be sure you understand the question The radius of the pudding skin and the radius of

the bowl are generally different They are the same only if the bowl is filled all the way to

the top

2-h

h

2 2"

2"

2

r h

2-h h

h

r

r

2 bowl of chocolate pudding

searching for a way

to relate h and r

success! relate r and 2-h using

a triangle with hypotenuse 2

Figure 1.12

We can relate r and h by looking at a cross-sectional slice and using a right triangle

While it may be tempting to draw a triangle whose legs are h and r, this is not useful, since

the hypotenuse of the triangle is unknown Instead, we must draw a triangle that involves

the radius of the bowl itself This radius must emanate from the center of the sphere.

(2 − h) + r = 2

r2= 4 − (2 − h)2

r2= 4 − (4 − 4h + h2)

r2= 4h − h2

r = ±4h − h2 but r ≥ 0, so

r =4h − h2

EXAMPLE 1.11 Functioning with friends.Javier goes to a pizza shop intending to order a small pizza and

eat it When he enters the shop he sees some of his friends and they decide to split a large pizza If the radius of a large pizza is twice the radius of a small pizza, what fraction of the large pizza should be allocated to Javier to give him the amount of food he originally intended to eat?

small pizza

large pizza

Figure 1.13 SOLUTION We’ll use functional notation in solving the problem

Area of small pizza

2 π(2r)2= π r

2

4 Allocate a quarter of the large pizza to Javier

Notice that we did not need to know the actual radius of either the small or the large pizza We were able to determine the answer simply by calling the radius of the small pizza r

EXAMPLE 1.12 Functioning on a diet of sugar cane. Giant pandas eat sugar cane S pounds of sugar cane

provide N calories On average, each day a panda needs to take in C calories for every pound of the panda’s weight

i The number of pounds of sugar cane it takes to support a panda for a day is a function

of the weight, x, of the panda Express this function with a formula The formula will involve the constants S, N , and C

ii The number of pounds of sugar cane it takes to support a pair of pandas, one weighing

P pounds and the other weighing Q pounds, for a period of W weeks is a function of

The formula will involve the constants S, N , C, P , and Q.

SOLUTION i Let’s use the strategy of taking the problem apart into a series of simpler questions

We’ll use unit analysis to help us

We’re looking for the number of pounds of sugar cane it takes to support a panda weighing x pounds for one day In order to use unit analysis constructively we’ll need to

be very precise; we don’t want to confuse pounds of sugar cane with pounds of panda

Question:How many calories does the panda need for one day?

lbs of panda· x lbs of panda = Cx calories

Question:How many pounds of sugar cane will provide Cx calories?

If we know the number of pounds of sugar cane per calorie, then multiplying by

Cx, the number of calories needed, will give the answer S pounds of sugar cane provide

Ncalories We can express this as

Slbs of sugar cane

S N

lbs of sugar cane

Therefore, to support the panda for one day we need

S N

lbs of sugar cane

N xlbs of sugar cane

Number of pounds of sugar cane = f (x) =SCN x

ii Again, let’s break this down into a series of simpler questions

How many pounds of sugar cane are needed to support a pair of pandas, one weighing

Amount to support the pair for a day = amount to support one panda

+ amount to support the other

N

Amount to support the pair for a day =SCPN +SCQN

Whatever it takes to feed the pandas for a day, it takes 7 times that amount to feed them for a week and 7W times that amount to feed them for W weeks To feed the pandas for W weeks it takes

f (W ) = 7W · SCN · (P + Q)

Working with Rates Consider the statement “S pounds of sugar cane provide

Ncalories.” From this we can determine the amount of sugar cane equivalent to one calorie (pounds of sugar cane per calorie):

Slbs of sugar cane

S N

lbs of sugar cane

Similarly, we can determine the number of calories per pound of sugar cane:

N calories

N S

calories lbs of sugar cane

In other situations we may be given analogous sorts of rate information For instance,

if a typist types W words in H hours then he types at a rate of

W H

words hour .

If we know the number of hours the typist worked, we can estimate the number of words

he typed Unit analysis gives

words

Multiplying rate by time gives the amount done

Alternatively, we can calculate the time per word

H W

hours word.

If we know the number of words the typist must type, we can estimate the time it will take him

hours

Presenting a Function Graphically

While you may be accustomed to seeing functions described by formulas, frequently information about a function is obtained directly from its graph The function itself may be presented graphically, without a formula; think about a seismograph recording the size of tremors in the earth, an electrocardiogram machine measuring electric activity in the heart,

or the output of many standard measuring instruments of meteorologists These machines give us pictures, not formulas Alternatively, consider the function that takes time as input and gives as output the temperature at the top of the Prudential Center in downtown Boston The function can be quite easily presented in the form of a graph, while arriving at a formula

to fit past data is a much more complicated task (Even if such a formula is found, it will require frequent alteration as data comes in that does not fit the existing formula.) The function that gives the number of applicants to Harvard College as a function the calendar year can also be presented much more simply by a graph than by trying to fit a formula to past data

1990 1991 199219931994199519961997 19981999 13000

14000 15000 16000 17000

year

18000

Applicants to Harvard College temperature

time

Figure 1.14

When graphing functions we use the convention that the independent variable (input)

is plotted along the horizontal axis and the dependent variable (output) is plotted along the vertical axis.13The coordinates of points on the graph are of the form (input, corresponding output) Given the function f , for every x in the domain of f the point with coordinates (x, output of f corresponding to x) is a point on the graph of f Recall that “the output of f corresponding to x” can be written using the shorthand f (x) Thus the points on the graph are of the form

(input, corresponding output) (x, output of f corresponding to x) (x, value of f at x)

(x, f (x))

All four expressions carry the same meaning; we usually use the last one

The height of a graph at a point is the value of the function at that point.

EXAMPLE 1.13 CEis the calibrating function for a particular evaporating flask The graph of CEis drawn

below Use the graph to identify the domain and range of CE

13 When constructing a function to model a situation, you must decide which variable you consider input and which you

consider output For instance, when calibrating bottles, we consider the input (or independent) variable to be the volume and the output (dependent) variable to be the height When using a calibrated bottle for measuring, we consider the input variable to be

(11, 2)

Volume Volume → Height C E

Figure 1.15

SOLUTION By looking at the graph of CE we can tell that the domain of the function, the set of all

possible volumes, is [0, 11] and the range, the set of all possible heights for these volumes,

It is useful to be able to obtain information about a function from its graph, as we will see below

If x is in the domain of f then the point (x, f (x)) is on the graph of f

If (3, 16) lies on the graph of some function named g, then 3 is an acceptable input of the function, and 16 is the associated output, so g(3) = 16 If an input of 8 produces

an output of 3, then the point (8, 3) will be on the graph More generally, if (x1, y1)is

on the graph of f , then y1= f (x1) The converse14is also true If y1= f (x1), then the point (x1, y1)is on the graph of f

If a function is given by an equation y = f (x), then a point (x, y) lies on the graph of the function if and only if it satisfies the equation y = f (x)

The “if and only if” construction comes up frequently in mathematical discussions, so we’ll pause for a moment to clarify the meaning of an “if and only if” statement

Language and Logic: An Interlude.

“A if and only if B” means “A and B are equivalent statements.” Using symbols we write

A ⇔ B Specifically, “P is a square if and only if P is a rectangle with sides of equal length” says that the two statements “P is a square” and “P is a rectangle with sides of equal length” are equivalent They carry the same information

A ⇔ B means A ⇒ B and B ⇒ A

Aif and only if B means A implies B and Bimplies A

14 The word “converse” has a precise meaning in math Suppose that if A is true, then B is also true The converse reverses the two pieces, saying if B is true, A is also true The converse of a true statement is often not true For example, the statement “if