We’ll use the rate at which water is decreasing today to approximate the change in water level over the next two days.. The tangent line to√ xat x = 16 is the best linear approximation o
Trang 1We’ll use the rate at which water is decreasing today to approximate the change in water level over the next two days
dW
dt = lim
t →0
W
t , so dW
dt ≈ W t for t small
W ≈dWdt t
If we let t = 2 and use dWdt
t =0= −115, we have
W ≈ −115 · 2 = −230
This makes sense; the water is decreasing at a rate of 115 gallons/day so after two days we’d expect it to decrease by about 230 gallons
W (2) ≈ G0− 230 Because W (t) is concave up, we expect the actual water level to be a bit higher than this
EXAMPLE 8.2 Approximate√
16.8 Use a first derivative to get a good approximation
SOLUTION Let’s begin by sketching√
xand getting an off-the-cuff approximation of√
16.8 This will help us see how a tangent line can be of use We see that√
16.8 is a bit larger than 4
y
x
16 (16, 4) y = √x
Figure 8.3
Question: How do we know this?
Answer:16.8 is close to 16 and√
16 = 4
Question: How do we know√
16.8 is a tad more than 4?
Answer:We know that√
xis increasing between 16 and 16.8
Question: How can we estimate how much to add to 4 to get a good approximation of
√ 16.8?
Trang 2Answer: This depends on the rate at which√
xis increasing near x = 16 This is where the derivative comes into the picture The derivative of√x
at x = 16 gives the rate of increase d
dx
√ x
x=16
2√ x
x=16
2√
16=1
8. Here we can adopt one of two equivalent viewpoints
Viewpoint (i): The Derivative as a Tool for Adjustment. We begin with an off-the-cuff approximation of√
16.8 grounded in a nearby value of√x
that we know with certainty Knowing the rate of increase of√
x at x = 16 enables us to judge how to adjust our approximation (in this case, 4) to fit a nearby value of x dydx≈ x y for x small (because dy
dx = lim x→0 y x) In this example x = 16.8 − 16 = 0.8
∆ x
∆y
16 16.8
slope = dy
dx
∆y
∆ x
≈
Figure 8.4
dy
dx =1
8≈ y 0.8⇒ y ≈0.8
10
So y ≈ 0.1 and we have the approximation√16.8 ≈ 4 + 0.1 = 4.1
In other words, the “new” y-value is approximately the “old” y-value, 4, plus y, where we approximate y bydxdy· x
Viewpoint (ii): Tangent Line Approximation. Again we begin with an off-the-cuff ap-proximation grounded in a value of√
x we know with certainty We choose x = 16 The tangent line to√
xat x = 16 is the best linear approximation of√xaround x = 16, so we use that tangent line to approximate√
16.8
16.8 16
y = √x
tangent line at x = 16
slope = 1 8
Figure 8.5
Trang 3The y-value corresponding to x = 16.8 on the tangent line is very close to the corre-sponding y-value on the curve
The tangent line has slope18and passes through the point (16, 4) Its equation is of the form y − y1= m(x − x1), fromy−y1
x−x 1 = m where m =18and (x1, y1) = (16, 4)
y − 4 = 1
8· (x − 16)
y = 4 +1
8 · (x − 16) Notice that this just says that the “new” y-value is approximately the “old” y-value, 4, plus the adjustment term—as in the first approach
The y-value on the tangent line when x = 16.8 is
y = 4 +1
8 · (16.8 − 16) = 4.1
Is the approximation√
16.8 ≈ 4.1 an overestimate or an underestimate? We know that d
dx
√
x =2√1
x; thus, as x increases the derivative decreases Therefore√x
is concave down, the tangent line lies above the graph of the function, and this approximation is a bit too big Comparing our approximation, √
16.8 ≈ 4.1, with the calculator approximation,
√ 16.8 ≈ 4.09878, we see that the approximation is good The error is about 0.00122;
we are off by about 0.03% REMARK If we wanted to further refine our approximation, we could use information supplied by the second derivative to make the required adjustment.1 Further successive
refinements involve higher-order derivatives and give us what is called a Taylor polynomial
approximationto the function at x = 16
EXERCISE 8.1 Approximate √
101 using a tangent line approximation Compare your answer to that obtained using a calculator
EXERCISE 8.2 Suppose you use a tangent line approximation to estimate√
4.8 Do you expect the dif-ference between your approximation and the actual answer to be greater or less than the analogous difference we found in Example 8.2? Explain, and then check your reasoning by computation and using a calculator
EXERCISE 8.3 We argued that f (x) =√xis concave down by reasoning that fis decreasing Show that
f is concave down by showing that f(x) <0 Use the limit definition of derivative to compute f(x) If you have trouble, look back at Section 5.1 to see how the derivative of
f was computed
Answers to Exercises 8.1, 8.2, and 8.3 are supplied at the end of this section.
Generalizing Our Method
To approximate f ($) using a tangent line approximation, choose an x near $ for which the value of f is known Let’s call this x-value a Find the tangent line to f at x = a The slope 1
The larger the value of the second derivative at x = 16 the larger the rate of change of the slope near x = 16.
Trang 4is f(a)and a point on the line is (a, f (a)) Therefore, the equation of the tangent line is
y − f (a) = f(a) · (x − a) or y = f (a) + f(a) · (x − a)
Evaluating at x = $ gives the corresponding y-value on the tangent line
Answers to Selected Exercises
Answer to Exercise 8.1
√
101 ≈ 10 + 1
2√
100 · (101 − 100)
y ≈ 10 + 1
20· (1) = 10.5
Answer to Exercise 8.2
The slope of√
x is changing more rapidly at x = 4 than at x = 16 In both cases we are using the tangent line to approximate the value of the function at a distance 0.8 from the known value Therefore we expect the error to be greater for√
4.8 than for√
16.8
√ 4.8 ≈ 2 + 1
2√
4 · (4.8 − 4)
≈ 2 +14 · (0.8) = 2.2 The difference between the approximation, 2.2, and the actual value, 2.19089 , is approximately 0.009
Answer to Exercise 8.3
f(x) = −1
4(√ x)3
P R O B L E M S F O R S E C T I O N 8 1
1 Use the tangent line approximation (linear approximation) of f (x) =√xat x = 25 to approximate the following Use the graph of√
xand its tangent line at x = 25 to predict the relative accuracy of your approximations Check the accuracy using a computer or calculator
(a) √
24.9 (d) √
27
2 Suppose we want to use a tangent line approximation of f (x) =√x at x = a to approximate a particular square root numerically Which values of a should we choose
to approximate each of the following?
(a) √
115.5
3 Use a tangent line approximation to f (x) = 1x at x = 2 to approximate1.91
Trang 54 Approximate √
98 using the appropriate first derivative to help you Explain your reasoning
5 Use the fact that dxdx1/3=13x−2/3to approximate√3
30 Do you expect your an-swer to be an over-approximation or an under-approximation? Explain Compare your answer to the approximation supplied by your calculator
6 A steamboat is traveling down the Mississippi River It is traveling south, making its way from point St Paul, Minnesota, to Dubuque, Iowa At noon it departs St Paul traveling at 10 mph and is accelerating It continues to accelerate over the next 10 minutes Between noon and 12:10 p.m it has covered 5 miles Let s(t) be the distance the steamboat has traveled from point St Paul, where t is measured in minutes We’ll use noon as benchmark time of t = 0
(a) Determine the sign of s(0), s(0), and s(0) Which of these expressions are you given enough information to specify numerically?
(b) Sketch s(t) over the time interval [0, 10]
(c) Find good upper and lower bounds for the distance the boat has traveled between noon and 12:05 p.m Use a sketch to illustrate your reasoning graphically
7 Consider the solid right cylinder with a fixed height of 10 inches and a variable radius Let V (r) be the volume of the cylinder as a function of r, the radius, given in inches Interpret dV /dr geometrically Explain why your answer makes sense by looking at
V geometrically
Trang 6Exploratory Problems for Chapter 8
Circles and Spheres
You know how to express the area and the circumference of a circle in terms of its radius; you also know how to express the volume and surface area of a sphere in terms of its radius Having completed the Exploratory Problems for Chapter 7, you know how to take the derivative of both the area function for the circle and the volume function for the sphere Have you observed the following?
A(r) = πr2is the area of a circle of radius r dA
dr = 2πr is its circumference
V (r) =43π r3is the volume of a sphere of radius r dV
dr = 4πr2is its surface area These relationships are not coincidental The problems that follow ask
you to make sense out of this observation (Notice that the area of a square with sides of length s is given by A(s) = s2and its perimeter
is 4s, not 2s )
r
∆r
circle of radius r and radius r + ∆r
sphere of radius r and radius r + ∆r
∆ s
∆ s
square of side s and side s + ∆ s
s
s
s + ∆ s
Figure (a) Figure (b) Figure (c)
1.Let A(r) give the area of a circle as a function of its radius
dA
dr = lim r→0
A
r where A = A(r + r) − A(r) Geometrically, A corresponds to the shaded area in Figure (a) above
(a) Imagine wrapping a piece of yarn around the outside of a circle
of radius r, thereby increasing the radius of the circle by the thickness of the yarn The thickness of the yarn corresponds
to r Let r be small in comparison to r Approximate the additional area by clipping the yarn and laying it out straight
Explain why, for r very small, A r ≈ 2πr = circumfer-ence of circle
(b) Now approach the problem more algebraically Instead of clipping a piece of yarn, look at the algebraic expression for
Trang 7Aand show that it is 2π r r + π( r)2 Argue that for r
very small in comparison to r, we can approximate A by
2π r r and that as r tends toward zero this approximation
gets better and better
2.Investigate the relationship between the derivative of the volume
of a sphere and the surface area of a sphere by approximating
V, when r is very small in comparison to r As an aid in
visualization imagine a grapefruit without the skin as the sphere
of radius r and the same grapefruit with the skin as the sphere of
radius r + r V corresponds to the volume of the skin of the
grapefruit
3.Let A(s) be the area of a square with sides of length s The
derivative of the area of a square is not its perimeter; it is only half
of the perimeter Explain this by looking at A geometrically and
approximating it (Refer to Figure (c).)
Trang 88.2 THE FIRST AND SECOND DERIVATIVES IN CONTEXT:
MODELING USING DERIVATIVES
We’ve been interpreting the first derivative in context since its introduction but we have not yet used the second derivative extensively in context Yet, according to a
tongue-in-cheek article from The Economist, second-order derivatives were the rage in the 1990s.
Below is an excerpt from the article “The Tyranny of Differential Calculus” published in
The Economiston April 6, 1991
“The pace of change slows,” said a headline on the Financial Times’s survey of world
paints and coatings last week Growth has been slowing in various countries—slowing quite quickly in some cases Employers were invited recently to a conference on
“techniques for improving performance enhancement.” It’s not enough to enhance your performance, Jones, you must improve your enhancement
Suddenly, everywhere, it is not the rate of change of things that matters, it is the rate
of change of the rate of change Nobody cares much about inflation; only whether it is going up or down Or rather, whether it is going up fast or down fast No respectable budget director has talked about the national debt for decades; all talk sternly about the need to reduce the budget deficit, which is, after all, roughly the rate at which the national debt is increasing Indeed, in recent years it is not the absolute size of the deficit that has mattered so much as the trend: Is the rate of change of the rate of change of the national debt positive or negative?
(© 1991 The Economist Newspaper Group, Inc Reprinted with permission Further reproduction prohibited www.economist.com)
Consider the comment about the paint industry that reads “Growth has been slowing
in various countries.” If we let P (t) be the size of the paint industry at time t, the graph of
P (t )is roughly given below The graph is increasing because the industry is growing; it is concave down to reflect that the growth is slowing
P (t)
t
Figure 8.6
dP
dt >0 because P is increasing with time; d2P
dt 2 =dtd dPdt<0 because the growth is slowing
EXERCISE 8.4 Suppose that in countries A and B the growth of the paint industry is slowing, and at time
t∗growth is slowing more quickly in country A than in country B Let PA(t )and PB(t )be the sizes of the paint industries in countries A and B, respectively, at time t
Trang 9(a) What are the signs of PA(t∗)and PB(t∗)?
Can you determine which is larger?
(b) What are the signs of P
A(t∗)and P
B(t∗)?
Can you determine which is larger?
(c) What are the signs of PA(t∗)and PB(t∗)?
Can you determine which is larger?
Answers are provided at the end of this section.
EXAMPLE 8.3 When politicians talk about the budget “deficit” they are referring to the difference between
the government’s revenues and expenditures for a given year When they speak of “the
national debt” they mean the cumulative deficit (or surplus) over all the years of the
government’s recorded transactions Let x(t) be the deficit for year t and let y(t) be the national debt incurred up to and including year t In our model x(t) and y(t) are both continuous and differentiable x(t) will be positive if there is a deficit, negative if there is a surplus
(a) There has been considerable talk about a balanced budget What would a balanced budget mean in terms of x(t)? In terms of y(t)?
(b) Politicians also often speak of deficit reduction plans What would such a reduction mean in terms of x(t) and y(t) and their derivatives?
(c) The article from The Economist tells us that the budget deficit is roughly the rate at
which the national debt is increasing What, then, is the calculus-based relationship between x(t) and y(t)?
SOLUTION (a) A balanced budget would mean that x(t) = 0 and y(t) is constant
(b) Deficit reduction would mean x(t) is positive and decreasing
x(t ) >0 and x(t ) <0
Deficit reduction would mean y(t) is increasing at a decreasing rate,
y(t ) >0 and y(t ) <0
(c) y(t ) = x(t)
This makes sense both practically speaking and in terms of the answers to parts (a) and (b)
Answers to Selected Exercises
Answers to Exercise 8.4
(a) PA(t∗)and PB(t∗)are both positive because the size of the paint industry won’t be negative We cannot determine which will be larger
(b) P
A(t∗)and P
B(t∗)are both positive because both industries are growing We cannot determine which rate of growth is larger
(c) P
A(t∗)and P
B(t∗)are both negative, because growth is slowing
Determining which is larger is semantically tricky Since growth is slowing more rapidly in country A than in country B, P
A(t∗)is more negative than P
B(t∗) More negative means smaller, so P(t∗)is smaller than P(t∗)
Trang 10P R O B L E M S F O R S E C T I O N 8 2
1 Water is being poured into a bucket at a steady rate h(t) gives the height of water at time t Let t∗be the time when the bucket is half full What can you say about the signs
of h(t∗)and h(t∗) Explain your reasoning precisely in plain English
2 Suppose that the revenue, R, brought in each month by the after-eight shows at a movie theater is a function of the price p of a ticket Suppose that R is measured in thousands
of dollars and that p is measured in dollars
Interpret the following statements in words
(a) R(3.5) = 50 (b) R(7.50) = −15
3 A company is making industrial-size rolls of paper towels A machine is wrapping paper around a roll at a steady rate By this we mean that the same number of sheets
of paper towels are added to the roll every minute Let D(t) be the diameter of the roll
of paper towels at time t Determine the sign of the following Explain your answers using plain English
(a) D(t) (b) D(t ) (c) D(t )
4 Let Y (t) be the number of Japanese yen exchangeable for one U.S dollar, where t is the number of days after January 1, 1996
(a) What is the practical significance of the values of t for which Y(t )is positive? (b) What is the practical significance of the values of t for which Y(t )is negative and
Y(t )is negative?
(c) What is the meaning of the statement Y(5) = 0.8?
(d) Interpret the quantity Y (5)−Y (3)2
AND POWER FUNCTIONS
In this section we shift our focus from interpreting and using derivatives to actually comput-ing them Fruits of our labors with limits will be general procedures for differentiatcomput-ing sums, products, and quotients, and the ability to easily compute the derivative of any function of the form xn, where n is an integer
Sum Rule and Constant Multiple Rule
Section 7.4 dealt with principles for working with limits We restate two of these principles here for reference