AP Calculus AB / Mr. Hansen 2/1/2005

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Answers to Chapter 6 Review Questions

	Note: Odd-numbered answers are already in the back of your book. However, all the answers are listed here for your convenience. In most cases, the work is missing, and only the answer is given. However, I have provided a full solution for #37 in §6.1.
	§6.1
34.	y = ln |x| – 3
37.	y ′′′ = t^–3 Antidifferentiate to get y ′′ = t^–2/(–2) + C1 Using initial condition y ′′(1) = 2, we plug in to get 2 = 1^–2/(–2) + C1 2 = –½ + C1 5/2 = C1 Since by algebra, C1 = 5/2, our earlier equation that we obtained by antidifferentiating becomes y ′′ = t^–2/(–2) + 5/2, or to write it in a somewhat simpler way, y ′′ = –½t^–2 + 5/2 Now continue antidifferentiating: y ′ = –½t^–1/(–1) + 5t/2 + C2 y ′ = ½t^–1 + 5t/2 + C2 Using initial condition y ′(1) = 3, we plug in to get 3 = ½(1^–1) + 5(1)/2 + C2 By algebra, C2 = 0. Therefore, y ′ = ½t^–1 + 5t/2, with no additional constant required. Antidifferentiate one last time: y = ½ ln |t| + 5t^2/4 + C3 Using initial condition y ′(1) = 3, we plug in to get 1 = ½ ln |1| + 5(1^2)/4 + C3 1 = ½(0) + 5/4 + C3 –¼ = C3 Final particular solution: y = ½ ln |t| + 5t^2/4 – ¼. You should always check your answer before proceeding. First, take the derivatives to verify that y ′′′ = 1/t^3 as claimed. y = ½ ln |t| + 5t^2/4 – ¼ y ′ = ½ (1/t) + 10t/4 – 0 = ½t^–1 + 5t/2 y ′′ = ½(–t^–2) + 5/2 = –½t^–2 + 5/2 y ′′′ = t^–3 + 0 ü So far, so good. Now make sure that all 3 initial conditions are satisified: y(1) = ½ ln |1| + 5(1)^2/4 – ¼ = 0 + 5/4 – ¼ = 1ü y ′(1) = ½(1)^–1 + 5(1)/2 = ½ + 5/2 = 3ü y ′′(2) = –½(1)^–2 + 5/2 = –½ + 5/2 = 2ü
42.	s(t) = –cos t – t + 2
51.	c(x) = x^3 – 6x^2 + 15x + 400
52.(a)	x^2e^x + C
(b)	x sin x + C
(c)	–x^2e^x + C
(d)	–x sin x + C
(e)	x^2e^x + x sin x + C
(f)	x^2e^x – x sin x + C
(g)	x^2/2 + x^2e^x + C
(h)	x sin x – 4x + C
	§6.2
1.	–⅓ cos 3x + C
2.	¼ sin 2x^2 + C
3.	½ sec 2x + C
4.	(7x – 2)^4 + C
5.	⅓ tan^–1(x/3) + C
6.
7.	⅔ [1 – cos (t/2)]^3 + C
8.	⅔ (y^4 + 4y^2 + 1)^3 + C
9.	1/(1 – x) + C
10.	tan (x + 2) + C
11.	⅔ (tan x)^3/2 + C
12.	sec (q + p/2) + C
13.	ln (ln 6)
14.	⅔
15.	⅓ sin (3z + 4) + C
16.	–⅔ (cot x)^3/2 + C
17.	1/7 (ln x)^7 + C
18.	¼ tan^8(x/2) + C
19.	¾ sin (s^4/3 – 8) + C
20.	–⅓ cot 3x + C
21.	½ sec (2t + 1) + C
22.	–6/(2 + sin t) + C
23.	0
24.	ln 9 – ln 2
25.	½ ln 5
26.	2p
27.	–⅓ ln |cos 3x| + C = ⅓ ln |sec 3x| + C
28.
31.	14/3
32.	⅓
33.	–½
34.	0
35.	10/3
36.	0
37.	2Ö3
38.	¾
39.	Book’s answer: It is also correct to say The advantage of the first form is that it defines a family of functions. The second form is somewhat easier to find, but it describes a family of relations. The two collections are identical, however.
40.	y = [tan^–1(x^2/4) + C]^2
41.	y = –ln (C – e^sin^x)
42.	y = ln (e^x + C)
43.	y = (x^2 + 3)^–1
44.	y = (ln x)^4; it is not necessary to say ln |x| here, since the presence of ln x in the original diffeq. allows us to infer that x > 0
	§6.3
11.	e^x(x^2 – 7x + 7) + C
	§6.4
19.	Diffeq. dT/dt = –k (T – Tambient) has solution T = Tambient + (T0 – Tambient)e^–kt. Therefore, as in the practice test and solutions, we can use the conditions of the problem to compute k = –0.1 ln (4/7) » 0.0559615788, after which the remaining questions are straightforward.
(a)	By calc., intersection of function T and the constant function 35 occurs at t » 27.527 minutes, which is 17.527 minutes longer.
(b)	We can use the same k, since k is determined by the insulating properties of the soup’s container and other factors that do not change. Now, however, we plug everything into a solution that uses a different value (namely, –15) for Tambient. Using our calculator’s intersection finder or root finder, we get t » 13.258 minutes. This is reasonable, since the cooling should occur roughly twice as fast in a freezer.
	§6.5
20.	First, we must double the number of bees from 5,000 to 10,000. By the yellow box on p. 332 (a.k.a. Rule of 69), that will take (ln 2)/k = (ln 2)/(¼) » 2.7725 . . . years. Second, we must follow a new exponential growth curve (less steep) to grow the population from 10,000 to 25,000. By inspection, the second curve has equation y = 10,000e^t^/12, where t = years since the switchover to the second growth curve occurred, and y = bee population. Use the intersection finder or root finder to solve for t when y = 25,000. Answer: 10.9954 . . . additional years. Add the two answers together: 13.768 years.
	§6.6
15.	Exact particular solution: y = –(x^2 – 2x + 2)^–1 Exact value for y(3) = –1/5 EULER2 estimates for y(3):      –0.1850562975 using Dx = 0.2 (absolute error 0.0149, relative error 7.5%)      –0.1928547867 using Dx = 0.1 (absolute error 0.007145, relative error 3.6%)      –0.1965026239 using Dx = 0.05 (absolute error 0.003497, relative error 1.7%)
16.	Exact particular solution: y = 2e^x + 1 Exact value for y(1) = 2e + 1 » 6.43656 EULER2 estimates for y(1):      5.97664 using Dx = 0.2 (absolute error –0.4599, relative error –7.1%)      6.18748 using Dx = 0.1 (absolute error –0.24908, relative error –3.9%)      6.306595 using Dx = 0.05 (absolute error –0.12997, relative error –2.0%)