MAC Calculus II Spring Homework #6 Some Solutions.

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1 MAC Calculus II Spring 23 Homework #6 Some Solutions. 1. Find the centroid of the region bounded by the curves y = 2x 2 and y = 1 2x 2. Solution. It is obvious, by inspection, that the centroid is at (,.5). 2. A cone of base radius r and height h can be obtained by rotating the region bounded by the curves (or lines) of equations y =, x =, and y = (h/r)x + h about the y-axis. Use this fact, and the Theorem of Pappus, to obtain the formula for the volume of a cone. Solution. The area of the triangle is A = 1 2rh. Next we find the coordinates of the centroid. We get x = 1 r x( ( h A r )x + h) dx = 1 r (hx h A r x2 ) dx = 1 ( 1 A 2 hx2 1 ) r 3r hx3 = 1 6A hr2. ȳ = 1 r ( ( ) ) 2 h r ( x + h dx = h2 1 x ) 2 h 2 r dx = 2A r 2A r 6A. Since we rotate about the x-axis, we only need the x value; it is the radius of the circle described by the centroid. Replacing A by its value, x = r 3. By Pappus Theorem, the volume of the cone is V = 2π xa = 1 3 πr2 h. 3. A sphere of radius r is inscribed in a right circular cylinder of radius r. Two planes perpendicular to the central axis of the cylinder and a distance h apart cut off a spherical zone of two bases on the sphere. Show that the surface area of the spherical zone equals the surface area of the region that the two planes cut off on the cylinder. (This is part (c) of problem 3 of the Section Problems Plus of Chapter 8 of the textbook (p. 578), where you can find a picture illustrating the problem.) Solution. Set up axes by taking for the (x, y)-plane any plane that goes through the center of the sphere and the north pole, place the origin at the center of the sphere and let the y-axis be the line containing the north

2 and south poles so that the north pole has coordinates (, r). Then the y-axis coincides with the axis of the circumscribed cylinder. Suppose the lower of the two planes perpendicular to this axis is at y = c; then the upper one is at y = c + h. We should have r c c + h r so that both planes also intersect the sphere. The so called spherical zone of two bases determined by these planes can be obtained by rotating the arc of equation x = r 2 y 2, c y c + h, about the y-axis. The area is thus given by Now hence ds = 1 + S = 2π c+h c x ds. ( ) 2 dx r dy = dy r2 y dy = r 2 x dy, S = 2π c+h c r dy = 2πrh. Turning to the cylinder, the portion between the two planes is a cylinder of base radius r, height h; its area is thus 2πrh; the same as the area of the spherical zone. 4. Since we are in the plus section, Problem 4 of Section Problems Plus of Chapter 8, p Solution (a) We set up (x, y)-coordinates exactly as in the previous Exercise (Chapter 8, Exercises plus, #3). The observer is then at the point (, r+h) of the (x, y)-plane. If from this point we draw all the tangents to the sphere, their points of contact with the sphere determine a circle bounding the part of the sphere visible to the observer. That circle lies in a plane perpendicular to the (x, y)-plane and intersects the (x, y)-plane at two points; if one of these points has coordinates (c, d), the other one is ( c, d). The observed zone can thus be described as the surface obtained by rotating the arc x = r 2 y 2, d y r, about the y-axis. As in exercise 3, ds = (r/x) dy so that the area of this surface is S = 2π r d x ds = 2π r d r dy = 2πr(r d). Pictures illustrating what follows were drawn in class. We consider the line through the points (c, d) and (, r + H). This line is tangent 2

3 to the sphere, hence also to the circle obtained by the intersection of the sphere with the (x, y)-plane. The point (c, d) is on the circle thus the radius of the circle to (c, d) has slope d/c; the line we are considering is perpendicular to this radius, hence has slope c/d. On the other hand, because it goes through the points (c, d), (, r + H), its slope is (r + H d)/( c). Equating the two expressions for the slope gives c d = r + H d c from which one gets c 2 + d 2 = d(r + H). Using that (c, d) is on the sphere, we get c 2 + d 2 = r 2, thus d = r2 r + H. Returning to the expression we had for S, we see that the observed region has surface area ) S = 2πr (r r2 = 2πr2 H r + H r + H as was desired. (b) For this part, we only need one axis of coordinates. We let the x- axis be the line joining the centers of the two spheres and (quite arbitrarily) we place the origin at the center of the sphere of radius r. The center of the sphere of radius R is then at (, d). We assume R r. Suppose we place a light source at the point x where (so we are between the spheres) r x d R. The point is then at distance x r from the first sphere (of radius r) and d x R from the second sphere. By part (a) of this exercise, the total area illuminated on the first sphere is 2πr 2 (x r) r + (x r) = 2πr2 (x r), x on the second sphere The total area is thus 2πR 2 (d x R) R + (d x R) = 2πR2 (d x R). d x A(x) = 2πr2 (x r) x + 2πR2 (d x R). d x We have to maximize this quantity, for r x d R. It is easy to get some horrible expressions when differentiating, which is why I am going to simplify a bit; namely, I ll use x r x = 1 x r, d x R d x 3 = 1 R d x,

4 to get Then Setting to results in A(x) = 2πr 2 ( 1 x r ) ( + 2πR 2 1 R ). d x ( ) r A 3 (x) = 2π x 2 R3 (d x) 2. We see that x = R 3/2 d R 3/2 + r 3/2. ( ) r A 3 (x) = 4π x 3 + R3 (d x) 3 < ; thus, IF THE POINT WE FOUND IS INSIDE THE INTERVAL, it will be both a single critical point and a local maximum, hence THE point at which the maximum occurs. But is it in the interval? Do we have r 3/2 d r < < d R? R 3/2 + r3/2 The second inequality always holds, if R r. In fact, it reduces to and since d > r + R, we do have dr 3/2 > Rr 3/2 + R 5/2, dr 3/2 > rr 3/2 + R 5/2 Rr 3/2 + R 5/2. The first inequality, on the other hand, does not always hold; it requires (is equivalent to) d > r + r 1/2 R 3/2. For example, if the small radius were equal to 4 inches, the large one equal to 16 inches, then the condition works out to d > (16)3/2 = 36; the spheres must be at least 36 inches apart (and not a mere r + R = 2 inches). If this condition is not valid, then A (x) = happens to the left of our admissible interval; A is always negative, so A is decreasing and hence negative in the admissible interval. The conclusions are: i. If r + R d r + r 1/2 R 3/2, the light source should be placed on the surface of the smaller sphere. 4

5 ii. If d > r + r 1/2 R 3/2, then the light source should be placed a distance r 3/2 d x = r 3/2 + R 3/2 away from the center of the smaller sphere. 5. Chapter 1, Section 1, Exercise 6, p Solution. (a) We set up a table of values to plot. t x y Plot upon request. It CLEARLY is not the graph of a function of the variable x. (b) Because the curve is not the graph of a function of the variable x, we have two choices in solving and writing things in terms of cartesian coordinates. Choice 1, two functions of x: Choice 2, x as a function of y: y = x 3/2, and t = x 3/2. x = y 2/3. 6. Chapter 1, Section 1, Exercise 1, p Solution. (a) y = sin 2 θ = 1 cos 2 θ = 1 x2 4. (b) The curve described by these equations is an arc of the parabola y = 1 (1/4)x 2, joining the points ( 2, ) and (2, ), gone through back and forth repeatedly. 7. Chapter 1, Section 1, Exercise 22, p Note: The exercise says not to use a graphing device and to give reasons. I have no way of checking that you do not use a graphing device, and giving reasons for every curve might be too time consuming. But I want to have GOOD reasons for the match, independent of a graphing device, for at least three (3) of the matches. Solution. 5

6 (a) When t = we get (x, y) = (, ); this curve must go through the origin. So if it is one of the shown curves, it must be III, IV or V. But we can exclude V; as t increases, both x and y increase; the curve in V stays bounded. Looking at the equations more carefully, x is a cubic in t, thus ranges from to. On the other hand y = t 2 t = (t 1 2 ) This forces us to select the curve in IV, since it is the one in which y remains bounded below. We could run a few more tests, but if we accept it has to be one of the given curves, then it must be IV. Thus (a)-iv. (b) This time we have a curve in which x ranges from to while y = 2 t 2 2. The only possibility is VI. (c) This is a bounded periodic motion through the origin ((x, y) = (, ) for t =, ±π, ±2π,...); only V foots the bill. (d) If t is large positive, or large negative, then the equation is almost x = t, y = t; which describes the line y = x. The added sine terms are just a little wiggle added to this line, most noticeable for small t. The curve matching this description is III. (e) These equations correspond to I. It has to be a periodic, bounded curve, so only I, II, and V are possibilities. The picture in I is pretty bad, it seems as if the curve has x equal zero in a whole interval. Similarly for y =. But that could be fuzziness. We can look at points where x =. For the current equations, x = gives sin(t + sin t) =, hence t + sin t = kπ for some integer k. The equation is certainly satisfied if t = kπ. By using a bit of Calculus 1, using the fact that d (t + sin t kπ) = 1 + cos t, dt one sees that there cannot be another solution. By periodicity, we can restrict ourselves to the interval t < π and we see that there are only two t-values for which x =, namely t =, π. Looking at the corresponding y values, we conclude that the only points on the curve with first coordinate equal to are the points (, cos(1) and (, cos(π 1)). Because these two points, while on opposite sides of on the y-axis, are not symmetric with respect to the origin, II, V are excluded. (f) The answer is II. 8. Chapter 1, Section 1, Exercise 38, p Solution. We give two solutions. For both we refer to the picture on page 647 and let θ be the angle between the segment OB and the x-axis. Solution 1. The line of which OB is a part (i.e., the line through O and B) is a line through the origin of slope tan θ. It intersects the circle of equation (x a) 2 + y 2 = a 2 at the point A. Setting for a moment 6

7 m = tan θ, after a few calculations the coordinates of the point A are seen to be 2a ( m 2 + 1, 2ma m ). The coordinates of B are, of course, (2a, 2am). From all this it follows that AB = 2am2 m The rest is very easy, since it is clear (?) that x = OP cos θ = AB cos θ = 2am2 cos θ m2 + 1, y = OP sin θ = AB sin θ = 2am2 sin θ m Finally, recalling that m = tan θ, we have that sqrt1 + m 2 = secθ = secθ (since in [, π/2), sec θ is positive). Thus AB simplifies all the way to 2a sin 2 θ)/ cos θ and the final expression for the parametric equations of the cissoid of Diocles are x = 2a sin 2 θ, y = 2a sin3 θ cos θ. Solution 2. One can get a simpler derivation if one uses the well known (?) geometric fact that angles inscribed in semicircles are right angles. Thus, denoting by C the intersection of the positive x-axis with the circle, the angle OAC is a right angle, which implies that OA = 2a cos θ. The angle OCB is also a right angle, from which we get that OB = 2a/ cos θ. We get at once that We get as before. AB = OB OA = x = 2a sin 2 θ, 2a cos θ 2a cos θ = 2a sin2 θ cos θ. y = 2a sin3 θ cos θ. 9. Chapter 1, Section 2, Exercise 14, p Solution. dy dx = 2t 3t 2 + 2t = 2 3t + 2, d 2 y dx 2 = d ( ) dy = d ( ) dy dt dx dx dt dx dx 7

8 = d dt = ( 2 ) dt 3t + 2 dx = 6 dt (3t 2) 2 dx 6 (3t 2) 2 dx dt = 6 t(3t + 2) Chapter 1, Section 2, Exercise 2, p Solution omitted. 11. Chapter 1, Section 2, Exercise 28, p Solution. We need to find two (or more) different values of t at which we get the same (x, y) values. Calling these values t 1, t 2, they must simultaneously solve the equations 1 2 cos 2 t 1 = 1 2 cos 2 t 2, (tan t 1 )(1 2 cos 2 t 1 ) = (tan t 2 )(1 2 cos 2 t 2 ). We should also consider that both tan and cos 2 are periodic of period π, so all has to happen, if it happens, in the interval π/2 < t < π/2; the domain of definition of the main branch of tan. From the first equation, we get cos 2 t 1 = cos 2 t 2. Using this in the second equation, we get tan t 1 = tan t 2, except if 1 2 cos 2 t 1 =. Since tan t 1 = tan t 2 implies t 1 = t 2, the only chance for a self-intersection is to have cos 2 t 1 = cos 2 t 2 = 1/2. Since cos t > in the interval ( π/2, π/2), we get cos t 1 = cos t 2 = 1/ 2, and the solutions are (if we take t 1 < t 2, as we may) t 1 = π/4, t 2 = π/4. For these values of t we get x =, y =, so (, ) is the point of selfintersection. The slope at a point of parameter t is given by dy dt dx dt = sec2 t(1 2 cos 2 t) + 4 tan t cos t sin t. 4 cos t sin t This could be simplified somewhat, but replacing t by ±π/4, we see that the two slopes are ±1. The equations of the two tangents are y = x andy = x. 12. Chapter 1, Section 2, Exercise 32, p Solution. The equation of a (non-vertical) line through the point (4,3) is y 3 = m(x 4). The equation of a line tangent to the curve in question is given by y y(t) = m(t)(x x(t)), wherem(t) = y (t) x (t). We have to find values of t so that the second equation of a line equals the first. Rewriting these equations in the form y = mx + 3 4m, y = m(t)x + y(t) m(t)x(t), 8

9 we see that we must have m(t) = m and y(t) m(t)x(t) = 3 4m; thus, y(t) m(t)x(t) = 3 4m(t). Now m(t) = (dy/dt)/(dx/dt) = t so that the last equation becomes 2t t(3t 2 + 1) = 3 4t, ort 3 3t + 2 =. This equation has one obvious solution, namely t = 1. Reducing by it, we see that 1 is a double solution and get the other solution t = 2. For t = 1 we get m = 1, x = 4, y = 3, so this tangent line is y = x 1. (Is there something to explain why t = 1 was a double solution?) With t = 2, the values are m = 2, x = 13, y = 15 and the equation of the tangent line is y = 2x Finally, we check for a vertical tangent line, which would have the equation x = 4. Is there a value of t for which x = 4 and dx/dt =. The answer is no, since dx/dt = happens only for t = and x = 1 for t =. 13. Chapter 1, Section 2, Exercise 36, p Solution. By symmetry, and from the formulas, the area is given by A = 4 θ= θ=π/2 y dx = 4 π/2 (a sin 3 θ)( 3a cos 2 θ sin θ) dθ. Why are the limits of integration given in that funny order? The answer is that if θ is allowed to go from to π/2, the curve is described negatively. Think about it. We have: A = 12a 2 sin 4 θ cos 2 θ dθ. Here is a somewhat sneaky way to compute the integral. If we change variables by θ = π/2 t, dθ = dt, we get A = 12a 2 sin 4 θ cos 2 θ dθ = 12a 2 π/2 Now sin(t π/2) = cos t, cos 2 (t π/2) = sin t, so that sin 4 (t π/2) cos 2 (t π/2) dt. A = 12a 2 cos 4 t sin 2 t dt = 12a 2 cos 4 θ sin 2 θ dθ, the last inequality being due to the fact that the variable inside the integral has no meaning outside of it. Adding the two expressions of the area and using sin 4 θ cos 2 θ + cos 4 θ sin 2 θ = cos 2 θ sin 2 θ(sin 2 θ + cos 2 θ) = cos 2 θ sin 2 θ, 9

10 we get A+A = 12a 2 ( sin 4 θ cos 2 θ + cos 4 θ sin 2 θ ) dθ = 12a 2 cos 2 θ sin 2 θ dθ; that is, Finally, using A = 6a 2 cos 2 θ sin 2 θ dθ. cos 2 θ sin 2 θ = (cos θ sin θ) 2 = ( 1 2 sin 2θ)2 = 1 4 sin2 2θ = 1 (1 + cos 4θ), 8 A = 3 4 a2 14. Chapter 1, Section 2, Exercise 28, p (1 + cos 4θ) dθ = 3πa2 8. This one was supposed to be #38. The solution is omitted. 15. Chapter 1, Section 2, Exercise 42, p Solution. A picture showing the area of grazing of the cow can be found at the URL I refer to that picture for what follows. In this picture, the grazing area is in (two shades of) green and the silo in black. The area A + B above the x-axis equals the area C + D below the x-axis. Thus, the grazing area is S = 2(A + B). The area A is easy to figure out; it is the area of a quarter circle of radius the length of the rope. Clearly, the length of the rope is πr, thus A = (1/4)π(πr) 2 = (1/4)π 3 r 2, so that S = πr B; we must figure out what B is. If we look at the picture, we see that if we want to describe B as area below a curve, we have to look at things from the point of view of the y-axis; think of y as the independent variable. In this case we can describe B as the area of the region bounded below by the line x = r, above by the curve of equation x = r(cos θ + θ sin θ), y = r(sin θ θ cos θ), for θ π, minus the area of half the silo the portion above the x-axis. The area of half of the silo is (1/2)πr 2, hence B = π (x ( r))dy 1 2 πr2 = π π = r 2 ( θ cos θ sin θ + θ 2 sin 2 θ + θ sin θ ) dθ 1 2 πr2. (r(cos θ + θ sin θ) + r)r(θ sin θ) dθ 1 2 πr2 The integrals are not too hard to compute by integration by parts, and it all works out to B = 1 6 π3 r 2. 1

11 We thus get for the grazing area. S = 5πr Chapter 1, Section 3, Exercise 6, p Solution. We have ds 2 = dx 2 + dy 2 = [(aθ cos θ) 2 + (aθ sin θ) 2 ] dθ 2 = a 2 θ 2 dθ 2. Thus ds = a θ dθ. Notice the absolute value. As it turns out, we won t need it because the parameter interval is in the positive θ region; still, one should remember that c 2 = c, not to c. Now L = π θ dθ = π 17. Chapter 1, Section 3, Exercise 18, p Solution omitted. 18. Chapter 1, Section 2, Exercise 24, p Solution omitted. θ dθ = aπ