Calculus AB / Mr. Hansen / 12/10/2003 Chapter 5 Test (Version 1, Corrected)

Name: ________KEY_____________ Key for Part I: EDBA EAAD

Part I.	Multiple Choice (8 questions, 6 points each). NO CALCULATOR ALLOWED.
	Wrong answers are –1½ points each to discourage guessing. Omitted problems count as 0.
1. ___	Give the equation of the line that is tangent to the curve y = –x^2 + 2x + 4 at the point (–1, 1).
	(A) y = x + 1 (B) y = 3x/5 + 52/5 (C) y = x + 2	(D) y = 3x/5 – 2 (E) y = 4x + 5
2. ___	If y = (x + 5)(3x), what is dy?
	(A) (3x + 15) dx (B) 3 dx (C) 4 dx	(D) (6x + 15) dx (E) (3 + 3x) dx
3. ___	equals . . .
		(A) 1/4 sin (x/2) (B) 1/2 sin (x/2) (C) sin (x/2) (D) 2 sin (x/2) (E) 0
4. ___	dx equals . . .
	(A) (1/9)(3x^2 + 4)^3/2 + C (B) (2/3) (3x^2 + 4)^3/2 + C (C) 2(3x^2 + 4)^1/2 + C	(D) 6x(3x^2 + 4) + C (E) (3/5)x^2 + (4/3)x^3 + C
5. ___	On what domain is the function e^x/(x + 1) decreasing?
	(A) (0, ¥) (B) (–1, 0) (C) (0, 1)	(D) (–¥, –1) (E) (–¥, –1) È (–1, 0)
6. ___	Where does the function have a vertical tangent line?
	(A) x = 1 (B) x = 0 (C) x = –1	(D) x = 1/3 (E) Nowhere
7. ___	What is the slope of the tangent line to the curve when x = 3?
	(A) –3/8 (B) 3/2 (C) 1/2 (D) –1/4 (E) –3/16
8. ___	If h is a small positive constant, then is an approximation for . . .
	(A) f (h) (B) (C)	(D) (E)

Part II.	CALCULATOR PERMITTED.
Show your work. Be sure that all final answers include units and are accurate to at least 3 decimal places.
9.	(20 points)
(a, b, c)	Given the function y = 2x^3 – x^2/3 on the interval [2, 6], estimate the definite integral three ways: (a) using the midpoint rule with 4 subintervals, M4, (b) using the trapezoid rule with 4 subintervals, T4, and (c) using Simpson’s rule with 8 subintervals, S8.
	i	xi	yi	wt. (mdpt.)	wt. · yi (mdpt.)	wt. (trap.)	wt. · yi (trap.)	wt. (Simpson)	wt. · yi (Simpson)
	0	2.0	2(2^3) – 2^2/3 = 14.412 …			1	14.412…	1	14.412…
	1	2.5	29.407 …	1	29.407…			4	117.631…
	2	3.0	51.919 …			2	103.839…	2	103.839…
	3	3.5	83.444 …	1	83.444…			4	333.779…
	4	4.0	125.480 …			2	250.960…	2	250.960…
	5	4.5	179.524 …	1	179.524…			4	718.097…
	6	5.0	247.075 …			2	494.151…	2	494.151…
	7	5.5	329.634 …	1	329.634…			4	1318.536…
	8	6.0	428.698 …			1	428.698…	1	428.698…
				Subtotals	622.0112454		1292.062784		3780.107766
answer (a)	» M4 = Dx(y1 + y3 + y5 + y7) » 1(622.0112454) » 622.011
answer (b)	» T4 = Dx/2 (y0 + 2y2 + 2y4 + 2y6 + y8) » 1/2 (1292.062784) » 646.031
answer (c)	» S8 = Dx/3 (y0+4y1+2y2+4y3+2y4+4y5+2y6+4y7+y8) » 0.5/3 (3780.107766) » 630.018
	[Note that Dx = 0.5 for Simpson’s Rule, since there were twice as many subintervals.]
(d)	Check your work by comparing (2M4 + T4)/3 to S8. If you have done everything correctly, the two results should match. If they do not match, say so and move on. (Or, if you have time, you may try to correct your work.)
answer (d)	(2 · 622.011 + 646.031) = 630.018 to 3 decimal places, as claimed ü [Notice how accurate Simpson’s Rule is. The result agrees to 3 places with the TI-83’s fnInt.]
10.	(24 points) Radio station WSTA operates 24 hours a day with a power consumption of 5 kilowatts. If electricity were priced at a flat 10 cents per kilowatt hour, a day’s worth of power would cost 24 · 5 · $.10 = $12.00. However, Pepco has instituted a “demand pricing” function, a smooth sinusoid. The cost per kilowatt hour, in dollars, is C(t) = .07 + .05 sin(pt/12), where t is time in hours since the start of the day. Compute the cost of operating WSTA for 24 hours, and explain why your method is necessary and appropriate.
answer 10.	The cost of operating WSTA for a brief sliver of time equals C(t) times 5 times the length of time used, or 5C(t) Dt. Adding these costs produces a Riemann sum, and the limit as Dt ® 0 is $8.40 by calc. The foregoing explains why our method is appropriate. The reason we must use a definite integral is that the cost function C(t) is not constant.
11.	(8 points) Essay problems were selected randomly from those listed on the study guide.