Section 3.10 Linear Approximations and Differentials
48. When blood flows along a blood vessel, the flux F (the volume of blood per unit time that flows past a given point) is proportional to the fourth power of the radius R of the blood vessel:
F = kR4
(This is known as Poiseuilles Law; we will show why it is true in Section 8.4.) A partially clogged artery can be expanded by an operation called angioplasty, in which a balloon-tipped catheter is inflated inside the artery in order to widen it and restore the normal blood flow.
Show that the relative change in F is about four times the relative change in R. How will a 5% increase in the radius affect the flow of blood?
Solution:
SECTION 3.10 LINEAR APPROXIMATIONS AND DIFFERENTIALS ¤ 261 38. (a) sin = 20
⇒ = 20 csc ⇒
= 20(− csc cot ) = −20 csc 30◦cot 30◦(±1◦)
= −20(2)√
3
± 180
= ±2√ 3 9 So the maximum error is about ±29
√3 ≈ ±121 cm.
(b) The relative error is ∆
≈
=±29
√3 20(2) = ±
√3
180 ≈ ±003, so the percentage error is approximately ±3%.
39. = ⇒ =
⇒ = −
2 . The relative error in calculating is∆
≈
= −( 2)
= −
. Hence, the relative error in calculating is approximately the same (in magnitude) as the relative error in .
40. = 4 ⇒ = 43 ⇒
=43
4 = 4
. Thus, the relative change in is about 4 times the relative change in . So a 5% increase in the radius corresponds to a 20% increase in blood flow.
41. (a) =
= 0 = 0 (b) () =
() =
=
(c) ( + ) =
( + ) =
+
=
+
= +
(d) () =
() =
+
=
+
= +
(e)
=
=
−
2 =
−
2 = −
2 (f ) () =
() = −1
42. (a) () = sin ⇒ 0() = cos , so (0) = 0 and 0(0) = 1. Thus, () ≈ (0) + 0(0)( − 0) = 0 + 1( − 0) = .
(b)
[continued]
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
50. In physics textbooks, the period T of a pendulum of length L is often given as T ≈ 2πpL/g, provided that the pendulum swings through a relatively small arc. In the course of deriving the formula, the equation aT = −g sin θ for the tangential acceleration of the bob of the pendulum is obtained, and then sin θ is replaced by θ with the remark that for small angles, θ (in radians) is very close to sin θ.
(a) Verify the linear approximation at 0 for the sine function:
sin θ ≈ θ
(b) If θ = π/18(equivalent to 10◦) and we approximate sin θ by θ, what is the percentage error?
(c) Use a graph to determine the values of θ for which sin θ and θ differ by less than 2%. What are the values in degrees?
Solution: SECTION 3.10 LINEAR APPROXIMATIONS AND DIFFERENTIALS ¤ 283 50. (a) () = sin ⇒ 0() = cos , so (0) = 0 and 0(0) = 1.
Thus, () ≈ (0) + 0(0)( − 0) = 0 + 1( − 0) = .
(b) The relative error in approximating sin by for = 18 is18 − sin(18)
sin(18) ≈ 00051, so the percentage error is 051%.
(c)
We want to know the values of for which = approximates = sin with less than a 2% difference; that is, the values of for which
− sin sin
002 ⇔ −002 − sin
sin 002 ⇔
−002 sin − sin 002 sin if sin 0
−002 sin − sin 002 sin if sin 0 ⇔
098 sin 102 sin if sin 0 102 sin 098 sin if sin 0 In the first figure, we see that the graphs are very close to each other near = 0. Changing the viewing rectangle and using an intersect feature (see the second figure) we find that = intersects = 102 sin at ≈ 0344.
By symmetry, they also intersect at ≈ −0344 (see the third figure).
Thus, with measured in radians, sin and differ by less than 2% for −0344 0344. Converting 0344radians to degrees, we get 0344(180◦) ≈ 197◦, so the corresponding interval in degrees is approximately
−197◦ 197◦.
51. (a) The graph shows that 0(1) = 2, so () = (1) + 0(1)( − 1) = 5 + 2( − 1) = 2 + 3.
(09) ≈ (09) = 48 and (11) ≈ (11) = 52.
° 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
1
SECTION 3.10 LINEAR APPROXIMATIONS AND DIFFERENTIALS ¤ 283
50. (a) () = sin ⇒ 0() = cos , so (0) = 0 and 0(0) = 1.
Thus, () ≈ (0) + 0(0)( − 0) = 0 + 1( − 0) = .
(b) The relative error in approximating sin by for = 18 is18 − sin(18)
sin(18) ≈ 00051, so the percentage error is 051%.
(c)
We want to know the values of for which = approximates = sin with less than a 2% difference; that is, the values of for which
− sin sin
002 ⇔ −002 − sin
sin 002 ⇔
−002 sin − sin 002 sin if sin 0
−002 sin − sin 002 sin if sin 0 ⇔
098 sin 102 sin if sin 0 102 sin 098 sin if sin 0 In the first figure, we see that the graphs are very close to each other near = 0. Changing the viewing rectangle and using an intersect feature (see the second figure) we find that = intersects = 102 sin at ≈ 0344.
By symmetry, they also intersect at ≈ −0344 (see the third figure).
Thus, with measured in radians, sin and differ by less than 2% for −0344 0344. Converting 0344radians to degrees, we get 0344(180◦) ≈ 197◦, so the corresponding interval in degrees is approximately
−197◦ 197◦.
51. (a) The graph shows that 0(1) = 2, so () = (1) + 0(1)( − 1) = 5 + 2( − 1) = 2 + 3.
(09) ≈ (09) = 48 and (11) ≈ (11) = 52.
° 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
52. Suppose that we dont have a formula for g(x) but we know that g(2) = −4 and g0(x) =√
x2+ 5 for all x.
(a) Use a linear approximation to estimate g(1.95) and g(2.05).
(b) Are your estimates in part (a) too large or too small? Explain.
Solution:
262 ¤ CHAPTER 3 DIFFERENTIATION RULES
We want to know the values of for which = approximates = sin with less than a 2% difference; that is, the values of for which
− sin sin
002 ⇔ −002 − sin
sin 002 ⇔
−002 sin − sin 002 sin if sin 0
−002 sin − sin 002 sin if sin 0 ⇔
098 sin 102 sin if sin 0 102 sin 098 sin if sin 0 In the first figure, we see that the graphs are very close to each other near = 0. Changing the viewing rectangle and using an intersect feature (see the second figure) we find that = intersects = 102 sin at ≈ 0344.
By symmetry, they also intersect at ≈ −0344 (see the third figure). Converting 0344 radians to degrees, we get
0344
180◦
≈ 197◦≈ 20◦, which verifies the statement.
43. (a) The graph shows that 0(1) = 2, so () = (1) + 0(1)( − 1) = 5 + 2( − 1) = 2 + 3.
(09) ≈ (09) = 48 and (11) ≈ (11) = 52.
(b) From the graph, we see that 0()is positive and decreasing. This means that the slopes of the tangent lines are positive, but the tangents are becoming less steep. So the tangent lines lie above the curve. Thus, the estimates in part (a) are too large.
44. (a) 0() =√
2+ 5 ⇒ 0(2) =√
9 = 3. (195) ≈ (2) + 0(2)(195 − 2) = −4 + 3(−005) = −415.
(205) ≈ (2) + 0(2)(205 − 2) = −4 + 3(005) = −385.
(b) The formula 0() =√
2+ 5shows that 0()is positive and increasing. This means that the slopes of the tangent lines are positive and the tangents are getting steeper. So the tangent lines lie below the graph of . Hence, the estimates in part (a) are too small.
LABORATORY PROJECT Taylor Polynomials
1.We first write the functions described in conditions (i), (ii), and (iii):
() = + + 2 () = cos
0() = + 2 0() = − sin
00() = 2 00() = − cos
So, taking = 0, our three conditions become
(0) = (0): = cos 0 = 1
0(0) = 0(0): = − sin 0 = 0
00(0) = 00(0): 2 = − cos 0 = −1 ⇒ = −12
The desired quadratic function is () = 1 −122, so the quadratic approximation is cos ≈ 1 −122.
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
2
