MATH141: Calculus II Exam #1 review 6/8/2017 Page 1

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1 MATH: Calculus II Eam # review /8/7 Page No review sheet can cover everything that is potentially fair game for an eam, but I tried to hit on all of the topics with these questions, as well as show you some of the different kinds of things that I could ask.. I ll use this first item to list some of the things you should be able to do. You should know how to calculate the volume obtained by revolving the graph of a function around either the or the y ais. Know when you use the shell method and when you use the washer method. You should know the integrals that calculate arc length, both in rectangular and in parametric equations. You should know Hooke s law and how to use it. You should be comfortable with pumping water out of tanks, especially cones and hemispheres. You should know the formulas for the coordinates of the center of gravity of an object. You should be able to parametrize line segments and circles. You should be able to apply L Hôpital s Rule.. The base of a solid is the triangle with vertices, ),, ), and, ), and the crosssections of the solid perpendicular to the -ais are semicircles. Find the volume of the solid. Solution. Here s a picture of the triangle. y The cross-sections are semicircles, and the diameter of each of those semicircles is the distance between the two slanted lines. The top line is y = and the bottom line is y =, so the distance between the two lines is their difference, namely =. If that s the diameter, then the radius is, and so the area of the semicircle is πr =

2 MATH: Calculus II Eam # review /8/7 Page ). Integrating the contribution of each from to yields π V = π ) π = 8 d = π 8 = π 8 8 = π.. Take the portion of the parabola y = that s in the first quadrant and revolve it around the line y =. What s the resulting volume? [Hint. Try to figure out what the cross-sections look like. Volume is still the integral of cross-sectional area.] V = Solution. A representative value will contribute the area of a washer to the overall volume. This washer has inner radius and outer radius ) +, since it s to reach the horizontal ais and you have to go one more unit to reach the ais of rotation. Since the area of a washer is πr πr, we integrate this formula to find the overall volume: π5 ) π) ) d = π 5 + ) d = π + ) d. Now it s a slog. On an eam I m happy if you leave the answer uncombined, like ) 5 π Find the arc length of the portion of the graph of y = between = and =. Solution. The formula for arc length is L = + f )) d. The derivative of is, which is when squared. The integral + d can be tackled with a u-substitution, by letting u = +, so du = d. This will make a factor of come out of the integral. u = + d = u du = = + ) = 7). 5 5 =

3 MATH: Calculus II Eam # review /8/7 Page. A spring is naturally five feet long. It takes twenty pounds of force to stretch it to seven feet. Which takes more work: stretching the spring from its natural length to being eight feet long, or stretching it from eight feet long to nine feet long? Solution. Stretching it two feet past equilibrium requires pounds of force, so the spring constant comes from the equation = k. So k =, and the force required to stretch the spring feet from dismplacement is. Now for two work integrals: W = W = d = 5 = 5 ft-lbs, and d = 5 = 5 ft-lbs. So stretching it from rest to three feet takes a little bit more work. 5. This question appeared on the Spring final eam.) A tank is in the shape of the curve y = 8 for revolved around the y ais, with the and y aes labeled in feet. Suppose the tank is filled with water weighing.5 pounds per cubic foot. Draw a picture of the situation. Then write down the integral for the work required to bring the water to the top of the tank until there is a depth of foot of water in the tank. Do not evaluate the integral. Solution. I don t know how to make a picture revolve around the y-ais in this graphics program, but I can at least draw y = 8 : 8 y Since we re going to pump the water vertically, and that s represented in the picture by the y ais, we want our integral to be in terms of y. We re going to move the water that s between height and height 8, so the integral will go between and 8. We lift to height 8, so an 8 y) term will account for this. So far we ve got y) Ay) dy,

4 MATH: Calculus II Eam # review /8/7 Page where Ay) represents the area of the cross-section at height y. Those cross-sections are circles, and the radius of the circle is the distance from the y-ais to the graph of the function. That s. The problem is that the integral is in terms of y, so we need to get in terms of y as well. Writing y = 8 and solving for gives = y/8. This is our radius, and Ay) = πr, so the work is given by the integral 8 ) y.5 8 y) π dy. 8. Find the center of gravity of the trapezoid with vertices, ),, ),, ), and, ). You can do this with point masses, but this is also a chance to practice the integrals.) Solution. As with the previous one, I ll start with a picture. y In order to use the formulas we re going to need the equations for the boundary lines. The bottom horizontal line is y =, the top horizontal line is y =, and the slanty line is y = +5 which I m going to write as 5 ). We ll have to break the integrals in two since the top function changes, but as you ll see they re not very strenuous. M = ) d + = + + ) d = ) ) d In that last step I incorporated the into the antiderivative. On an eam, stop there. Since I want to do a sanity check at the end, this latter integral is, so M = + =.

5 MATH: Calculus II Eam # review /8/7 Page 5 On to M y! M y = ) d + 5 ) ) d = + ) d = + = + = 8. Finally, area. Let s use geometry for that: the figure is a square with side length two area: ) together with half a square of side length two area: ). So the total area is. Thus we get My, y) = A, M ) 8 = A 8, ).5,.78).. 8 The answer seems reasonable to me unlike the first draft of this solution where the center of gravity wasn t inside the figure). 7. Give a parametrization for the circle with radius, centered at, ), traversed counterclockwise, which takes π time to complete one rotation. Solution. The equations should be like { = r cosat + B) + h y = r sinat + B) + k for appropriate values of the constants. The radius is r, so that letter is. Because the center is at, ) we want h = and k =. The problem doesn t say where the circle starts, so I ll just make B =. It also doesn t specify how many rotatios to do, or anything about the domain, so I won t specify anything either. If for eample it said to make two full rotations, we d want t π.) The last constant we need is A, which governs the period. A circle completes one rotation in π time. If this is equal A to π, then the equation π = π simplifies to A =. Therefore one possibility is A { = cost) y = sint) +.

6 MATH: Calculus II Eam # review /8/7 Page 8. Suppose a particle moves along the ellipse parametrized by = cost), y = sint). Determine the speed of the particle when t = π. ) + dy ). dt For this problem, t) = 8 sint) Solution. The speed formula is d dt and y t) = cost). Hence, s = sin t) + cos t). We lose the negative sign when we square.) When t = π the trig functions are evaluated at π, where the sine is and cosine is. We get ) ) s = + = + = Calculate the following limits. a) lim ln). e b) lim. c) lim ). Solution. a) This was a gotcha problem. Plugging in = gives ln) =, which is not an indeterminate form. That s just equal to zero. b) This time plugging in = gives, so we will need L Hôpital s rule. Applying it, e lim e = lim =. c) This one is indeterminate of the form. We ll use the ep-log trick to tame it. lim ) = e lim ln ) ) = e lim ln ). This is a typographical nightmare, so I m going to drop the e for now, and we ll just remember to eponentiate our answer at the end. The eponent is now ln) =, so we push the or the, but the is the problem) into the denominator with the one-over-one-over trick: lim ln ) = lim ln )

7 MATH: Calculus II Eam # review /8/7 Page 7 and now we finally have, so we can apply the rule. Taking derivatives, lim ln The answer is e. ) = lim /) = lim =.