4100 AWL/Thomas_ch03p147-243 8/19/04 11:17 AM Page 227
3.8
EXAMPLE 5
Linearization and Differentials
227
Finding Differentials of Functions
(a) dstan 2xd = sec2s2xd ds2xd = 2 sec2 2x dx
sx + 1d dx - x dsx + 1d
x
x dx + dx - x dx
dx
(b) d a
b =
=
=
2
2
x + 1
sx + 1d
sx + 1d
sx + 1d2
Estimating with Differentials
Suppose we know the value of a differentiable function ƒ(x) at a point a and want to predict how much this value will change if we move to a nearby point a + dx. If dx is small,
then we can see from Figure 3.51 that ¢y is approximately equal to the differential dy.
Since
ƒsa + dxd = ƒsad + ¢y,
the differential approximation gives
ƒsa + dxd L ƒsad + dy
where dx = ¢x. Thus the approximation ¢y L dy can be used to calculate ƒsa + dxd
when ƒ(a) is known and dx is small.
EXAMPLE 6
dr 0.1
Estimating with Differentials
The radius r of a circle increases from a = 10 m to 10.1 m (Figure 3.52). Use dA to estimate the increase in the circle’s area A. Estimate the area of the enlarged circle and compare your estimate to the true area.
a 10
Solution
dA = A¿sad dr = 2pa dr = 2ps10ds0.1d = 2p m2 .
A ≈ dA 2 a dr
FIGURE 3.52 When dr is
small compared with a, as it is
when dr = 0.1 and a = 10 , the
differential dA = 2pa dr gives
a way to estimate the area of the
circle with radius r = a + dr
(Example 6).
Since A = pr 2 , the estimated increase is
Thus,
As10 + 0.1d L As10d + 2p
= ps10d2 + 2p = 102p.
The area of a circle of radius 10.1 m is approximately 102p m2 .
The true area is
As10.1d = ps10.1d2
= 102.01p m2 .
The error in our estimate is 0.01p m2 , which is the difference ¢A - dA.
Error in Differential Approximation
Let ƒ(x) be differentiable at x = a and suppose that dx = ¢x is an increment of x. We
have two ways to describe the change in ƒ as x changes from a to a + ¢x:
The true change:
The differential estimate:
¢ƒ = ƒsa + ¢xd - ƒsad
dƒ = ƒ¿sad ¢x.
How well does dƒ approximate ¢ƒ?
Copyright © 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley
4100 AWL/Thomas_ch03p147-243 8/19/04 11:17 AM Page 232
232
Chapter 3: Differentiation
Linearizing Trigonometric Functions
y
y f (x)
In Exercises 11–14, find the linearization of ƒ at x = a . Then graph
the linearization and ƒ together.
11. ƒsxd = sin x
at
sad x = 0,
sbd x = p
f f(x 0 dx) f(x 0)
12. ƒsxd = cos x
at
sad x = 0,
sbd x = -p>2
13. ƒsxd = sec x
at
sad x = 0,
sbd x = -p>3
14. ƒsxd = tan x
at
sad x = 0,
(x 0, f (x 0 ))
dx
Tangent
sbd x = p>4
0
The Approximation s1 xdk « 1 kx
15. Show that the linearization of ƒsxd = s1 + xdk at x = 0 is
Lsxd = 1 + kx .
16. Use the linear approximation s1 + xdk L 1 + kx to find an approximation for the function ƒ(x) for values of x near zero.
a. ƒsxd = s1 - xd6
c. ƒsxd =
b. ƒsxd =
1
e. ƒsxd = s4 + 3xd1>3
f. ƒsxd =
3
a1 -
1
b
2 + x
Derivatives in Differential Form
In Exercises 19–30, find dy.
21. y =
20. y = x21 - x
2x
1 + x2
22. y =
2
21x
3s1 + 1xd
23. 2y 3>2 + xy - x = 0
24. xy 2 - 4x 3>2 - y = 0
25. y = sin s51xd
26. y = cos sx 2 d
27. y = 4 tan sx 3>3d
29. y = 3 csc s1 - 2 1xd
3
33. ƒsxd = x - x,
34. ƒsxd = x 4,
x0 = 1,
x0 = 1,
dx = 0.1
x0 = -1,
dx = 0.1
dx = 0.1
dx = 0.1
x0 = 0.5,
dx = 0.1
x0 = 2,
dx = 0.1
Differential Estimates of Change
2
18. Find the linearization of ƒsxd = 2x + 1 + sin x at x = 0 . How
is it related to the individual linearizations of 2x + 1 and sin x
at x = 0 ?
19. y = x - 31x
x0 = 1,
32. ƒsxd = 2x 2 + 4x - 3,
36. ƒsxd = x 3 - 2x + 3,
B
17. Faster than a calculator Use the approximation s1 + xdk L
1 + kx to estimate the following.
3
a. s1.0002d50
b. 2
1.009
3
31. ƒsxd = x 2 + 2x,
x
x 0 dx
x0
35. ƒsxd = x -1,
2
1 - x
d. ƒsxd = 22 + x 2
21 + x
df f '(x 0 ) dx
In Exercises 37–42, write a differential formula that estimates the
given change in volume or surface area.
37. The change in the volume V = s4>3dpr 3 of a sphere when the radius changes from r0 to r0 + dr
38. The change in the volume V = x 3 of a cube when the edge
lengths change from x0 to x0 + dx
39. The change in the surface area S = 6x 2 of a cube when the edge
lengths change from x0 to x0 + dx
40. The change in the lateral surface area S = pr2r 2 + h 2 of a
right circular cone when the radius changes from r0 to r0 + dr
and the height does not change
41. The change in the volume V = pr 2h of a right circular cylinder
when the radius changes from r0 to r0 + dr and the height does
not change
42. The change in the lateral surface area S = 2prh of a right circular
cylinder when the height changes from h0 to h0 + dh and the radius does not change
28. y = sec sx 2 - 1d
Applications
30. y = 2 cot a
43. The radius of a circle is increased from 2.00 to 2.02 m.
1
b
1x
Approximation Error
In Exercises 31–36, each function ƒ(x) changes value when x changes
from x0 to x0 + dx . Find
a. the change ¢ƒ = ƒsx0 + dxd - ƒsx0 d ;
b. the value of the estimate df = ƒ¿sx0 d dx ; and
c. the approximation error ƒ ¢ƒ - dƒ ƒ .
a. Estimate the resulting change in area.
b. Express the estimate as a percentage of the circle’s original
area.
44. The diameter of a tree was 10 in. During the following year, the
circumference increased 2 in. About how much did the tree’s diameter increase? The tree’s cross-section area?
45. Estimating volume Estimate the volume of material in a cylindrical shell with height 30 in., radius 6 in., and shell thickness
0.5 in.
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