Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 58 ppsx

When the bug is at the point 3, 4 its y-coordinate is decreasing at a rate of 6 units per second the minus sign indicating a decrease.. So at this moment the top of the ladder is falling

In this section we will demonstrate the power and versatility of implicit differentiation and the Chain Rule in context The example that follows makes the transition from Section 17.3

to the applied problems of this section

words, the point (x, y) is moving on the circle of radius 5 centered at the origin Think of

a bug crawling around the circle; the coordinates of the bug at time t are given by (x, y) or (x(t ), y(t))

Suppose thatdydt = −6 units/second, when x = 3 and y > 0 What isdxdt at this moment?

y

x

5

5 (3, 4)

Figure 17.4

Let’s think about the question in terms of the bug We know that when x = 3, y = 4 (because

x2+ y2= 25 and y > 0) When the bug is at the point (3, 4) its y-coordinate is decreasing

at a rate of 6 units per second (the minus sign indicating a decrease) This tells us that the bug is traveling in a clockwise direction Knowing that the bug stays on the circle, we want

to find the rate of change of its x-coordinate (By thinking about the bug’s path, you can see that at (3, 4),dxdt must be positive Thinking harder can lead you to realize that dxdt >6.)

SOLUTION Let’s make the fact that x and y vary with t more explicit by writing

[x(t)]2+ [y(t)]2= 25

What We Know: dydt = −6

What We Want: dxdt, when x = 3 and y > 0

Using principle (i) we differentiate both sides of this equation with respect to t

d dt

[x(t)]2+ [y(t)]2 =dtd [25]

2x(t)dx

dt + 2y(t)dy

dt = 0

xdx

dt + ydy

dt = 0

25 − 9 = 4.

3dx

dt + 4(−6) = 0 dx

dt = 24/3 = 8

REMARKAt the moment we are interested in we know that x = 3 and y = 4, but because

x isn’t always 3 and y isn’t always 4, we can’t substitute x = 3 and y = 4 until after

differentiating (Similarly, if we are interested in f(2) we can’t evaluate at 2 until after differentiating If we were to evaluate f at 2 before differentiating, we would think that

f(2) is always zero, regardless of f )

is being pulled away from the wall at the constant rate of 0.5 feet per second

(a) How fast is the top of the ladder falling when the bottom of the ladder is 3 feet from the wall?

(b) Does the top of the ladder fall at a constant rate or does the rate depend on how far the bottom is from the wall? If the rate is not constant, how is it changing?

SOLUTION We need to draw a picture of the situation and label variables and constants in order to

express mathematically what we know and what we are trying to find (It’s crucial to determine what we know and what we’re looking for; otherwise it is impossible to figure out a strategy for proceeding.)

Below are two pictures: Figure 17.5(a) represents the situation at time t and Figure 17.5(b) represents a snapshot at the moment in question

Let x = the distance from the foot of the ladder to the wall

Let y = the height of the top of the ladder

The Pythagorean Theorem tells us that x and y are related by x2+ y2= 25 Relating the variables x and y allows us to relate their rates of change Because x and y vary with time,

we can write [x(t)]2+ [y(t)]2= 25

(a) What We Know: dx

dt = 0.5 feet per second

What We Want:dydt, when x = 3

Differentiate x2+ y2= 25 with respect to t to get

2xdx

dt + 2ydydt = 0

Equivalently,

xdx

5ft y

x

at time t

5

3 4

snapshot

Figure 17.5

When x = 3 we know that y = 4, because x and y are related by x2+ y2= 25 Evaluating (17.1) at x = 3, y = 4, and dxdt = 0.5 gives

3(0.5) + 4dy

dt = 0

4dy

dt = −3 2 dy

dt = −3

8.

dy

dt is negative because y is decreasing; the ladder is falling

So at this moment the top of the ladder is falling at a rate of 3/8 feet per second Notice that the mathematics involved in Example 17.10(a) is just like that in Example 17.9 even though the situation is very different

(b) We can start with the equation relatingdxdt anddydt

xdx

dt + ydydt = 0 The bottom of the ladder is being pulled away at a constant rate of 0.5 feet per second,

so we know thatdxdt = 0.5 Therefore,

0.5x + ydydt = 0 dy

dt =−0.5xy = − x

2√

25 − x2

As x gets larger x

2√ 25−x 2 gets larger; the numerator increases and the denominator decreases, both contributing to the growth of the fraction Therefore, the ladder starts falling slowly and speeds up as the bottom moves away from the wall

At the same moment a ferry located 60 miles directly west of the dock is traveling toward the dock at 25 mph Assume that the boats maintain their speeds for the next two hours How fast is the distance between the boats changing at 11:00 a.m.? Are the boats getting farther apart or closer together?

SOLUTION On the following page are two pictures of the situation The picture on the left is a general

picture; the one on the right is a snapshot taken at 11:00 a.m

y z

dx

dt = –25

dy

dt = 20

20

√1625

snapshot

at 11:00 am

Figure 17.6

Let x = the distance (in miles) between the ferry and the dock

Let y = the distance (in miles) between the fishing boat and the dock

Let z = the distance (in miles) between the two boats x, y, and z are functions of time

t, where t is measured in hours x = x(t), y = y(t), and z = z(t)

Approach 1

What We Know: dxdt = −25 mph (negative because x is decreasing with time)

dy

dt = 20 mph (positive because y is increasing with time) The Pythagorean Theorem tells us that x, y, and z are related by x2+ y2= z2

What We Want: dzdt, when t = 1

x2+ y2= z2

dt + 2ydydt = 2zdzdt

xdx

dt + ydydt = zdzdt When t = 1, x = 60 − 25 = 35, and y = 20 We use the relationship x2+ y2= z2to find that z =√1625

35(−25) + 20(20) =√1625dz

dt dz

dt =√−475

1625≈ −11.78

The distance between the boats is decreasing at a rate of about 11.78 mph.

Approach 2

Express x, y, and z explicitly in terms of t and then finddzdt at t = 1 Let t = 0 correspond

to 10:00 a.m

y(t ) = 20t (rate) · (time) = distance x(t ) = 60 − (distance traveled) = 60 − 25t z(t ) =(60 − 25t)2+ (20t)2

We must finddz and then evaluate it at t = 1 This is left as an exercise for the reader

P R O B L E M S F O R S E C T I O N 1 7 4

1 Suppose we toss a rock into a pond causing a circular ripple If the radius is increasing

at a rate of 3 feet per second when the diameter is 4, how fast is the area increasing?

2 The price and the demand for a certain item can be modeled by the equation 20p = −q + 200

(a) Express the rate of change of quantity demanded with respect to price in terms of

a derivative and evaluate it

(b) Suppose that price is determined by the world market, so that price and quantity can both be thought of as functions of time At a certain instant the price is $6 and

is increasing at a rate of $0.25 per week At what rate is the quantity demanded changing at this instant? Why is your answer negative?

3 A bug is walking around the circle x2+ y2= 169 At a certain instant the bug is at the point (−5, 12) and its y-coordinate is decreasing at a rate of 3 units per second (a) Is the bug traveling the circle in a clockwise direction or a counterclockwise direction?

(b) How fast is its x-coordinate changing at this instant?

4 A spherical balloon is losing air at a steady rate of 0.5 cm3/hour

(a) How fast is the radius decreasing when the diameter of the balloon is 30 cm? (b) How fast is the radius decreasing when the diameter of the balloon is 15 cm? (c) How fast is the diameter decreasing when the diameter of the balloon is 15 cm? (d) How fast is the radius decreasing when the circumference of the balloon is 12 π centimeters?

5 A cylindrical barrel with radius 3 feet is filled with oil The barrel stands upright and oil comes out a faucet at the base of the tank

(a) How are the volume of oil in the tank and the height of oil in the tank related? (b) What is the rate of change of volume with respect to height? Is it constant, or does

it depend upon the height? Does your answer make sense to you?

(c) As oil leaves the tank, both the volume and the height of oil in the tank change with respect to time

i When the height is decreasing at a rate of 0.5 feet per hour, how fast is the volume of oil in the tank changing?

ii When oil is leaving the tank at a rate of 3 cubic feet per hour, how fast is the height of oil in the tank changing?

6 At 7:00 a.m a truck is 100 miles due north of a car The truck is traveling south at a constant speed of 40 mph, while the car is traveling east at 60 mph How fast is the distance between the car and the truck changing at 7:30 a.m.?

7 As shown on the following page, a wheelbarrow is being wheeled down a perfectly straight 13-foot ramp The top of the ramp is 5 feet high and the ramp covers a horizontal distance of 12 feet When the wheelbarrow has moved a horizontal distance of 6 feet

it is moving horizontally at a rate of 0.5 feet per second At what rate is it moving vertically downward?

5 ft

6 ft

12 ft

8 Valentina is sipping from a straw she has stuck in a conical cup of lemonade The cup

is an inverted circular cone of height 12 cm and radius 6 cm When the height of the lemonade in the cup is 10 cm, Valentina is sipping lemonade at a rate of 2 cm3/second

At this moment, how fast is the height of the lemonade in the cup decreasing?

6 cm

12 cm

9 Two vehicles are on parallel roads that are 0.5 miles apart At noon the vehicles are half a mile apart and traveling in the same direction One is traveling at 30 mph and the other at 40 mph

(a) After 2 hours, how fast is the distance between the cars increasing?

(b) When the cars are 40 miles apart, how fast is the distance between them increasing? (c) What would be the answer to part (a) if the cars were traveling in opposite directions?

10 In Barcelona there is a beautiful Spanish castle set 1/4 of a mile back from a straight road A bicyclist rides by the castle at a velocity of 15 mph Assuming that the biker maintains this speed, how fast is the distance between the biker and the castle increasing

20 minutes later?

11 A conical container is used to hold oil It is positioned upright with the tip of the cone

at the bottom Oil comes out a faucet at the base of the container We know that the volume of a cone is V = 13π r2h

(a) As oil leaves the cone the height, radius, and volume of oil in the container change with time FinddV

dt in terms of r, h,dr

dt, and dh

dt (b) Suppose the container has a height of 24 inches and a radius of 12 inches Express the volume of oil in the container as a function of the height of oil in the container

(Hint: Use similar triangles to express the radius of the oil in terms of the height

of the oil.) (c) Suppose oil is leaking out of the container at a rate of 5 cubic inches per hour How fast is the height of the oil in the container decreasing when the height is 10 inches? When the height is 4 inches?

Do the relative sizes of your answers make sense to you intuitively?

12 An airplane is flying at a speed of 600 mph at a constant altitude of 1 mile It passes directly over an air traffic control tower How fast is the distance between the control tower and the airplane increasing when the plane is 10 miles away from the tower?

13 Sand is being dumped into a conical pile whose height is 1/2 the radius of its base Suppose sand is being dumped at a rate of 5 cubic meters per minute

(a) How fast is the height of the pile increasing when it is 9 meters high?

(b) How fast is the area of the base increasing at this moment?

(c) How fast is the circumference of the base increasing at this moment?

(d) Will the height be increasing more slowly, more rapidly, or at a steady pace as time goes on?

14 It is evening and a woman 5 and a half feet tall is standing by a 14-foot-high street lamp

on a cobbled road, waiting for her ride home from work When she sees her husband pull his car up to the curb she begins to walk away from the light at a rate of 4 feet per second How fast is the length of her shadow changing?

woman under streetlamp

14 ft

5.5 ft

x ft

woman

shadow length =

VI An Excursion into Geometric Series

18

Geometric Sums, Geometric Series

An Introductory Example

Consider the modeling we’ve done with exponential functions: When modeling invest-ments, we’ve restricted ourselves to the condition that no deposits and no withdrawals are made; when modeling the elimination of a drug from the body, we’ve restricted our model

to an investigation of one dose of the drug, and one dose only When you stop to think about this, we’ve been rather unrealistic about our restrictions Why is this?

The reason for the restrictions we’ve put on the situations we have modeled is that

we have been restricting ourselves to differentiable functions, functions whose graphs are continuous and smooth If you withdraw $1000 from a bank account, the amount in the account plummets by $1000 The graph of M(t), the amount of money in the account, will not be continuous and therefore will not be differentiable Similarly, when a person on medication swallows a pill, the amount of medication in his body leaps up In this chapter

we will tackle these situations, situations in which a lump sum of some quantity is either introduced or removed at discrete intervals This gives the chapter a different flavor from the work we have been doing so far

559

proportional to the amount of digitalis in the body.1Suppose 30% of the drug present in the body at any moment is eliminated within 24 hours If 0.04 mg of digitalis is taken at the same time every morning, how many milligrams are in the body one week later, immediately after the person has taken his eighth dose? How many milligrams will be in his body several years later, assuming he continues to take the drug regularly? (There are fluctuations, since there are 0.04 more mg of digitalis in the body immediately after taking a pill than there were right before taking the pill Figure out the medication level immediately after the morning pill.)

SOLUTION Immediately after the first pill of digitalis the person has 0.04 mg in his body By the next

morning 30% has been eliminated, so 70% of the 0.04 mg remains Right after the second pill, the number of milligrams rises to

.04 + 7(.04)

The term 0.04 represents the medication from the second pill; the term 0.7(0.04) represents what remains of the first pill By the following morning only 70% of what was previously

in the body remains

.7[.04 + 7(.04)], or equivalently, (.7)(.04) + (.7)2(.04) remains

Immediately after the third pill is taken the person has

.04 + (.7)(.04) + (.7)2(.04) mg of digitalis, (3rd pill) +what is leftof 2nd pill

 +what is left

of 1st pill

 Day Amount of digitalis in the person’s system right after taking a pill

2 04 + (.7)(.04)

3 04 + (.7)(.04) + (.7)2(.04)

4 04 + (.7)(.04) + (.7)2(.04) + (.7)3(.04) Let’s focus for a minute on an individual dosage With each passing day the amount of digitalis remaining in the body from this individual pill decreases by 30%, or equivalently,

is multiplied by 0.7

After one week, medication from the first pill has been in the body for seven days, the second pill for six days, the third pill for five days, and so on We are looking for the amount

of medication in the body immediately after the eighth pill is taken, so we write a sum with eight terms, each term representing what is left of a pill that has been taken

.04 + (.7)(.04) + (.7)2(.04) + (.7)3(.04) + (.7)4(.04) + (.7)5(.04) + (.7)6(.04) + (.7)7(.04)

8th

pill



+what is left

of 7th pill

 +what is left

of 6th pill

 +what is left

of 5th pill

 +what is left

of 4th pill

 +what is left

of 3rd pill

 +what is left

of 2nd pill

 +what is left

of 1st pill



1 The half-life of digitalis in a patient with good renal function if 1.5–2 days The figures given are consistent with that.