AP Calculus AB / Mr. Hansen 10/14/2003

Name: ________________________

HW Answer Key for §§3-5 through 3-9

§3.5	#Q1-Q10 all, 1, 2, 3, 5, 6, 9, 11.
Q1.	none (discriminant is < 0)
Q2.	dy/dx = 10x
Q3.	y¢ = –51x^–4
Q4.	f ¢(x) = 1.7x^0.7
Q5.	d/dx (3x + 5) = 3
Q6.	f (3) = 45
Q7.	f ¢(3) = 30
Q8.	limx®3 5x^2 = 45
Q9.	e
Q10.	definite integral
1.	See p. 682.
2.	v(t) = –1.2t^–5 – 5 a(t) = 6t^–6
3.	See p. 682.
5abc.	See p. 682.
d.	v(t) = 0 Þ –3t^2 + 26t – 35 = 0 This latter equation has two roots on [0, 9], namely t = 7 or t = 5/3. (Use root finder or factoring. Factoring the equation as (–3t + 5)(t – 7) = 0 is a bit easier, because then it is obvious that the product on the left, which represents v(t), changes sign only at 5/3 and at 7.) For t < 5/3, v(t) < 0. For 5/3 < t < 7, v(t) > 0. For t > 7, v(t) < 0 once again. A relative max. occurs when v = 0 and the derivative is in the process of changing from pos. to neg. Clearly only t = 7 satisfies these conditions. Later in the text, we will learn a “second derivative test” that sometimes works. If you look at the second derivative when t = 7, you notice that a(7) = –6(7) + 26 < 0 Þ negative concavity Þ t = 7 must give a relative max. The problem with the second derivative test is that students often forget that if the second derivative is 0, the test provides no information.
e.	Use result of part (d) to see a relative min. at t = 5/3. Note that x is still positive there, and x does not drop below 0 until after t > 9. Conclusion: No, x is never negative for t Î [0, 9].
6a.	v(t) = 4t^3 – 33t^2 + 76t – 48 a(t) = 12t^2 – 66t + 76
b.	v(t) = 0 when t » 1.048, 2.371, or 4.832
c.	Make the graph and transcribe it accurately onto your paper. (That is a skill that is assumed for the AP exam.) Notice that the curve for x(t) has a relative max. or min. whenever the curve for v(t) has a root.
d.	Overlay the a(t) curve onto your sketch. Notice two things: (1) the x(t) curve has a point of inflection at each place where a(t) crosses the t-axis, and (2) the v(t) curve has a relative max. or min. at each of these places.
9.	See p. 682.
11.	See p. 682.
§3-6	#1, 2, 3, 7.
1.	Make graph. (A neat sketch on non-graph paper is acceptable.) See p. 682.
2.	Since f (2) » .909 and f ¢(2) » –.416, you would sketch a line passing through the point (2, .909) having slope m = –.416. This line should appear to be tangent to the sine curve. The equation of this line, using point-slope form, is y – .909 = –.416(x – 2) or, in calculator syntax, Y2 = –.416(x – 2) + .909 When you overlay this line on your sine graph, you should see that the line and the curve have the same slope when x = 2, which means that they have the same instantaneous rate of change. The derivative, of course, equals instantaneous rate of change. The graph does not prove that d/dx (sin x) = cos x, but it at least provides some corroborating evidence, since the rates of change appear to match when x = 2.
3.	See p. 682.
7a.	Inside: 3x	Outside: sin u
b.	Inside: sin x	Outside: u^3 (the cubing function)
c.	Inside: x^3	Outside: sin u
d.	Inside: cos x	Outside: 2^u (exponential function with base 2)
e.	Inside: tan x	Outside: 1/u (reciprocal function)
f.	Inside: sec x	Outside: log u (logarithmic function)
§3-7	#1, 2, 3-21 mo3.
1.	See p. 683.
2a.	Let u = x^2 – 1. Since d/dx u^3 = 3u^2 du/dx (by power rule and chain rule), we have f ¢(x) = d/dx u^3 = 3(x^2 – 1)^2(2x). On the AP exam, you would show only the following amount of work: f ¢(x) = d/dx [(x^2 – 1)^3] = 3(x^2 – 1)^2(2x).
b.	By the binomial theorem, (x^2 – 1)^3 = x^6 – 3x^4 + 3x^2 – 1. (Please review Algebra II if necessary.) This polynomial is easy to differentiate: f ¢(x) = 6x^5 – 12x^3 + 6x.
c.	Answer from (a) is 3(x^2 – 1)^2(2x) = 6x(x^2 – 1)^2 = 6x(x^4 – 2x^2 + 1) = 6x^5 – 12x^3 + 6x (Q.E.D.)
3.	f ¢(x) = –3 sin 3x
6.	h ¢(x) = (cos x^5)(5x^4), or you could write 5x^4 cos x^5 Note: You cannot write cos x^5(5x^4), because that makes the argument of the cosine function appear to be everything that follows.
9.	y ¢ = (6 sin^5 x)(cos x), or you could write 6 sin^5 x · cos x or 6(sin^5 x)cos x Note: You should not write 6 sin^5 x cos x (even though that is what your book gives on p. 683), because that makes x cos x appear to be the argument of the sine function.
12.	f ¢(x) = 20 sin(–5x)
15.	f ¢(x) = 160(sin^2/3 4x)(cos 4x) Note: Again, as in #9, note that your book’s answer is not in the best format.
18.	f ¢(x) = 9(x^2 + 8)^8(2x) = 18x(x^2 + 8)^8 Note: Simplification of this type is not required on the AP exam. You may leave your answer in the unsimplified form if you wish, namely f ¢(x) = 9(x^2 + 8)^8(2x).
21.	y ¢ = –200x cos^99(x^2 + 3) sin(x^2 + 3)
§3-8	#Q1-Q10 all, 1-4 all.
Q1.	f ¢(x) = 9x^8
Q2.	dy/dx = –3 sin x
Q3.	y ¢ = 2.4(5x^6 + 11)^1.4(30x^5) = 72x^5(5x^6 + 11)^1.4, but simplification is optional
Q4.	s ¢(x) = 0
Q5.	The limit is 12 because the discontinuity at x = 7 is removable.
Q6.	The limit is 2^0 = 1 since the function is continuous.
Q7.	Yes; the only point of discontinuity is at x = 3.
Q8.	f (x) = –cos x^2 + C
Q9.	cos^2 x + sin^2 x = 1 for all values of x, both real and complex (This is called the “Pythagorean identity.”)
Q10.	Your derivative function must pass through the point (1, 0) since the function has a relative extreme point when x = 1. Because the curve appears to be a parabola (degree 2), a linear derivative function is appropriate. The instantaneous slope of the book’s curve appears to be –1 when x = 0 and +1 when x = 2, which suggests that the derivative is probably the line y = x – 1 (slope 1, intercept –1).
1ab.	See p. 683.
c.	See p. 683 for the first two parts. The third part of #1(c) was omitted in your book’s answer key. By graphing y ¢(t), you can see that the peaks and valleys (i.e., most rapid change of y) occur at t = 8, 18, 28, 38, 48, etc. These all correspond to a height of 25, which is the axle height.
d.	See p. 683. The book added an additional answer, which addresses the third question in part (c), but provided insufficient explanation.
2a.	d(t) = 80 + 30 cos(p/3(t – 1.3))
b.	d ¢(t) = –10p sin(p/3(t – 1.3))
c.	d ¢(5) = d ¢(11) » 21.021 cm/sec (moving away from the wall) The answers are the same because the motion is periodic with period 6 seconds, and these two times are exactly one period apart.
d.	d ¢(20) » –21.021 cm/sec (moving toward the wall) Reason: d(t), which is distance from wall, is decreasing if d ¢(t) < 0.
e.	The greatest absolute value of d ¢(t) occurs when t = 2.8, 5.8, 8.8, 11.8, etc., since these t values make the sine expression in part (b) evaluate to ±1. The resulting speeds (i.e., absolute value of d ¢(t)) are thus 10p » 31.416 cm/sec, and these occur when d(t) = 80 cm.
f.	The first root of d ¢(t) occurs when t = 1.3 seconds. By using the givens of the problem, or by plugging in to d(t), we see that t = 1.3 gives the pendulum’s maximum distance from the wall, 110 cm.
3.	See p. 683.
4a.	Let L(d) = length (in minutes) of daylight on day d. L(d) = 729 + 114 cos(2p/365(d – 172)) August 7 is 47 days after June 21, i.e., day number 172 + 47 = 219. Plugging in, we have L(219) » 807.680 minutes, or about 13 hr. 28 min.
b.	L¢(d) = –228p/365 sin(2p/365(d – 172)) L¢(219) » –1.420 minutes per day
c.	As in #2(e), the greatest absolute value of L¢(d) occurs when the sine expression is ±1, and this rate would be 228p/365 » 1.962 min./day. These extremes occur (by calc.) on days approx. 80.746 and 263.247, which round to March 22 and September 20.
§3-9	#2-16 even, 17-24 all.
2.	f (x) = x^10 + C
4.	f (x) = x^5/5 + C
6.	f (x) = x^–1065/(–1065) + C
8.	f (x) = –cos x + C
10.	f (x) = 77x^7/3/(7/3) + C = 3/7 · 77x^7/3 + C = 33x^7/3 + C [simplification is optional]
12.	f (x) = (sin 4x)/4 + C
14.	f (x) = (8x + 3)^6/48 + C Explanation: We divide by 6 to account for the exponent, but we must also divide by 8 to account for the chain rule. Remember, you can always check your work by taking the derivative and seeing if you get the given result for f ¢(x).
16.	f (x) = x^3/3 – 10x^2/2 + 7x + C You may wish to simplify the middle term as –5x^2, but that is not required.
17.	See p. 684.
18.	f (x) = x^8/8 + 99.875
19.	See p. 684.
20.	f (x) = –cos x + 7
21.	See p. 684.
22.	f (x) = x^3/3 + 6x^2 – 7x – 52
23.	See p. 684.
24a.	v(t) = –6 cos 3t + C
b.	Since v(0) = –6 = –6 cos(3 · 0), apparently C = 0. Thus the particular equation is simply v(t) = –6 cos 3t Negative velocity, as in #2 of §3-8, is to be interpreted as swinging backwards (i.e., toward the pusher, if there is one).
c.	Calculator keystrokes: Store –6cos(3x) into Y1 and then perform the following: 2nd TBLSET TblStart=0 DTbl=.2 Auto Auto 2nd TABLE
d.	Maximum velocity (6 ft/sec) occurs at t = p/3 seconds and at infinitely many other times. Acceleration at these times is 0. Perhaps this is not surprising, since you know that the maximum “zoom” you experience on a swing occurs precisely at the low point, where gravity switches over from giving you positive acceleration to giving you negative acceleration (or vice versa, if you are on a backswing). For that instant, acceleration is zero.