AP Calculus AB / Mr. Hansen 2/18/2005

Name: _________________________

Solutions to Chapter 7 Review Problems (p. 413)

1.	This problem is easy with a calculator: ò0^5 |v(t)| dt = ò0^5 |t^2 – .2t^3| dt » 10.417 ft. Remember that the absolute value function (abs) is found under 2nd CATALOG ENTER on your calculator, since abs is the first function listed alphabetically. However, make sure you can also do this problem without a calculator. First, you must rewrite v(t) as t^2(1 – .2t) so that you can see that v(t) is a product of nonnegative factors on [0, 5]. Why is this so important? You need to make sure that there are no pesky sign changes to worry about. Now apply FTC1 to find the answer. This is somewhat tedious, but it is like eating your vegetables: you feel better afterward. Since v(t) ³ 0 on [0, 5], ò0^5 |v(t)| dt = ò0^5 v(t) dt. We evaluate as follows: ò0^5 (t^2 – .2t^3) dt = (t^3/3 – .2t^4/4) |0^5 = 5^3/3 – .2(5^4)/4 – (0 – 0) = 5^3/3 – (5^3)/4 = 5^3(⅓ – ¼) = 5^3(1/12) = 125/12 ft.
4.	This is a standard “variable-factor” problem that you can anticipate seeing on your test. The real world is crawling with such problems, which is one of the main reasons for studying the calculus in the first place. If density were uniform (as in Form II math and science), we could simply multiply density times length to find the total mass of the rod. But, of course, density is varying here. The density function is not p but rather r (the Greek letter rho). We must assemble a Riemann sum and then use our standard limiting trick to convert the Riemann sum to a definite integral. In most problems, density equals mass divided by volume. However, in this problem, the rod has a certain mass per unit length (not per unit volume), and it is this quantity that we use for density. Since density = M/x if both M and x are fixed, we know that density » DM/Dx even when both are changing, provided Dx is small. Therefore r(x) » DM/Dx. At this point, some of you are saying, “It looks as if the density function, r(x), is actually nothing more than M ¢, namely the derivative of M.” Well, you’re right! By algebra, we can manipulate the equation r(x) » DM/Dx to get DM » rDx This is a very useful equation, and it has a real-world explanation: A little chunk of mass (DM) approximately equals density times a little bit of length. We say “approximately,” since if density is always changing as a function of x, there is no way we can make Dx small enough to avoid some fluctuation of density. This is unavoidable, and we accept the small error that occurs. We need to add up all the rDx values (Riemann-sum style) in order to approximate the total value of M. We do this as follows: M » S (DM) = S (rDx) Now, we apply the limiting process, the magical thing known as the definite integral, and the “approximately equal” sign becomes a true equal sign: M = ò□^□ (dM) = ò0^2 (rdx). Why did I use boxes as placeholders in the first definite integral? Answer: because the limits for M are not the same as the limits for x. I don’t really care what the limits for M are, but I have to show something there to remind myself that I have a definite integral, not an indefinite integral. This is a fine point; don’t worry too much about it. Also, note that the density function r satisfies r(x) ³ 0 on [0, 2]. (It had better! After all, how could we have negative density?) For people who understand the previous discussion, it is OK to omit all of the earlier stuff and start showing work only at this point: M = ò0^2 (rdx) = ò0^2 (11 – 4x) dx = (11x – 2x^2) |0^2 = 11(2) – 2(2^2) – (0 – 0) = 22 – 8 = 14 g.
5.	This problem also involves a variable-factor product, and again, you should be able to work the entire problem without a calculator. I will shorten the explanation below. If you have any uncertainty at all, please reread #4 above and make sure that you understand every single word, symbol, and punctuation mark in #4. By inspection, E(t) ³ 300 on [0, 24]. Reason: Since the cosine of anything cannot exceed 1, the quantity (2 – cos(blah blah blah)) must be greater than or equal to 1. Therefore, E(t) = 300(2 – cos(blah blah blah)) = (300 times something greater than or equal to 1). Therefore, E(t) ³ 0 on [0, 24], meaning that there are no sign changes to worry about. Total power consumption = ò0^24 E(t) dt = ò0^24 300(2 – cos(pt/12)) dt = 300 ò0^24 (2 – cos(pt/12)) dt = 300 (2t – (12/p) sin(pt/12)) |0^24 = 300 (2·24 – (12/p) sin 2p – (0 – (12/p) sin 0)) = 300 (48 – 0 – (0 – 0)) = 300·48 = 14,400 kwh
10.	We did this one in class. Let there be two functions of y, namely x1(y) = (y^2 – 4)/4 x2(y) = (y + 16)/4 You must plot these to see what is going on. Note that x1 is a parabola opening to the right, with vertex at (–1, 0). If you have forgotten how to get that, please review your Precalculus class notes or contact me. There is insufficient time to reteach basic material like that in our class. Note that x2 is a straight line with slope 4, x-intercept at (4, 0), and y-intercept at (0, –16). The two curves intersect where they both have the same values for x and y. The easiest way to find the points of intersection is to set 4x = 4x and therefore y^2 – 4 = y + 16. By Algebra I (steps omitted), y = –4 or y = 5. Since x2 is always to the right of x1 when y Î [–4, 5], our Riemann sum to find area must have as its variable factor the difference x2 – x1. (Think of a number line. Since x2 always lies to the right of x1 for the interval of interest, x2 – x1 is a positive quantity that tells us the width of the Riemann sum rectangles. The height is always Dy, which becomes dy in the integral.) Area = ò–4^5 (x2 – x1) dy = 30.375 by calc. You should also be able to compute the answer by FTC1, as we did in class on Thursday, 2/17/2005. (Please refer to your class notes.)
16.	Although this problem does not require a calculator, it is certainly much easier if you use one. Let y1 = 1 + cos x y2 = 2 – cos x and observe that for one of the larger “beads,” y2 is always the upper function. Consider the first bead in Quadrant I, and use root finder or intersection finder to determine the x-values of the left and right bounds of the bead, namely a = 1.047... and b = 5.235… You should also be able to find the exact values by algebra: a = p/3, b = 5p/3. Note: I strongly recommend that you use the calculator feature X®A ENTER after finding the first root and X®B ENTER after finding the second one. That way, you need not write down all the digits, and you will be guaranteed to minimize the roundoff errors. Click here for a 2-point bonus. Finally, area = òa^b (y2 – y1) dx » 7.653 by calc. You should also practice solving the problem by FTC1, and you should prove that the exact answer equals 4p/3 + 2Ö3. For additional practice, do #15 and make sure you can get the answer given on p. 702.
18.	Since we did this one in class, you should already have it in your notes. A calculator is required for this problem. Let the functions be y1 and y2, respectively, where y1 is the upper function. By root finder or intersection finder, the x values of intersection are a = –1.893... and b = 1.893... (symmetric about y-axis since both functions are even). Area = òa^b (y1 – y2) dx » 5.731 by calc.