MATH 241 - Spring 2014 Instructor: Patrick Collins Homework #5 Solutions §3.5 #20 Find the derivative of q = 2 r - r2 . Solution dq d BI2 r dr = dr 1 2 = - r2 M F 12 I2 r - r2 M -12 d A2 r dr 2-2 r = 2 2 r-r2 1-r = - r2 E . 2 r-r2 §3.5 #36 Find the derivative of r = sec Solution Θ tanI Θ M. 1 dr dΘ d Bsec dΘ = = sec sec 1 Θ tan Θ Θ Θ d B dΘ Θ tanJ N 1 Θ tan = 2 Θ F tanI Θ M + sec - Θ d 1 AtanI Θ ME dΘ Θ F tanI Θ M + sec 1 Θ sec2 J N 1 sec Θ Θ2 . Θ sec2 I Θ M 1 d 1 A E dΘ Θ §3.5 #40 Find dy dt if y = sec2 Πt. Solution dy = 2 sec Πt dt d @sec ΠtD dt = 2 sec Πt sec Πt tan Πt = 2 Π sec2 Πt tan Πt. d @ΠtD dt §3.6 #14 Use implicit differentiation to find dy dx if y2 cosH1 yL = 2 x + 2 y. Solution 2y 2y dy dx dy dx cosJ y N + 1 y2 sinJ N dy y 1 y2 cos J y N + sinJ y N 1 1 dx dy dx J2 y cosJ y N - 2 + sinJ y NN 1 1 dy dx dy dx dy = 2+2 = 2+2 = 2 = 2 y cosJ N-2+sinJ N dx dy dx 2 1 1 y y . 2 HW5_Solutions.nb y2 sinJ N dy 1 y cosJ N + dx y dx y2 1 2y 2y dy cos J y N + sinJ y N dy 1 dx 1 dy dx J2 y cosJ y N - 2 + sinJ y NN 1 1 dy dx dy dx dy = 2+2 = 2+2 = 2 = 2 y cosJ N-2+sinJ N dx dy dx 2 §3.6 #28 1 1 y y . Find the slope of the curve Ix2 + y2 M = Hx - yL2 at the points H1, 0L and H1, -1L. 2 Solution We have Thus, at H1, 0L: 2 Ix2 + y2 M J2 x + 2 y N = 2 Hx - yL J1 - 2 I12 + 02 M J2 H1L + 2 H0L 4=2-2 and at H1, -1L : dy dx dy dx dy dx dy dx N. dy dx N = -1, 2 I1 + H-1L2 M J2 H1L + 2 H-1L 8-8 N = 2 H1 - 0L J1 - dy dx dy dx dy dx =2-2 dy dx dy dx N = 2 H1 - H-1LL J1 - dy dx N = 1. §3.6 #40 Power Rule for Rational Exponents. Let p and q be integers with q > 0. If y = x pq , differentiate the equivalent equation yq = x p implicitly and show that, for y ¹ 0, p d Ax pq E = q dx Solution qyq-1 dy dx dy xHpqL-1 . px p-1 = px p-1 = dx d Ax pq E dx qyq-1 p x p-1 q Ix pq Mq-1 = p x p-1 q x pHq-1Lq = = p q x p-1- pHq-1L q . But p-1which establishes the result. pHq-1L q = Hp-1L q-pHq-1L q = pq-q-pq+p q = p-q q = p q - 1, HW5_Solutions.nb 3 §3.6 #44 The Folium of Descartes (a) Find the slope of the folium of Descartes x3 + y3 - 9 xy = 0 at the points H4, 2L and H2, 4L. (b) At what point other than the origin does the folium have a horizontal tangent? (c) Find the coordinates of the point A in Figure 3.23, where the folium has a vertical tangent. Solution (a) Implicit differentiation gives dy 3 x2 + 3 y2 dy x2 + y2 Thus at H4, 2L we have: dx H2L2 + H4L2 dx dy dx dy dx -9y-9x -3y-3x dx H4L2 + H2L2 and at H2, 4L we have dx dy dy dx dy = 0. dx - 3 H2L - 3 H4L dx - 3 H4L - 3 H2L dx = 5 4, dy = 4 5. (b) In order to have a horizontal tangent we must have = 0 dy dy dy =0 =0 = 0. This occurs if 3 y - x2 = 0, or if y = x2 3. Thus we have to determine which points on the parabola y = x2 3 lie on the folium. Plugging in x2 3 for y gives dx x3 + J 3 N - 9 xJ 3 N = 0 x2 3 x3 + 6 x 27 6 x6 27 x2 - 3 x3 = 0 - 2 x3 = 0 x - 54 x3 = 0 x3 Ix3 - 54M = 0. Thus the folium has a horizontal tangent at x = 0 (the origin) and at x = (c) In order to have a vertical tangent we must have in y2 3 for x gives dy dx 3 54 . undefined. This occurs if y2 - 3 x = 0, or if x = y2 3. Plugging J 3 N + y3 - 9 J 3 N y = 0, y2 3 y2 which is precisely the same equation as before but with x replaced with y. Thus the folium has a vertical tangent where y = 3 54 . §3.7 #8 Diagonals. If x, y, and z are lengths of the edges of a rectangular box, the common length of the box’s diagonals is s= x2 + y2 + z2 . (a) Assuming that x, y, abd z are differentiable functions of t, how is ds dt related to dx dt, dy dt, and dz dt? (b) How is ds dt related to dy dt and dz dt if x is constant? 4 HW5_Solutions.nb (c) How are dx dt, dy dt, and dz dt related if s is constant? Solution (a) s2 ds dt 2s ds dt (c) If s is constant then ds dt = 0, and so x = dx dt dx +2y dt 1 dx Jx dt + y s = 2x ds dt (b) If x is constant then dx dt = 0, and so x2 + y2 + z2 = = 1 s Jy dy dy +y +z dt +z dt dz dt dy dt dy dt +2z +z dz N. dt dz dt dz N. dt = 0. §3.7 #12 Changing Dimensions in a Rectangular Box. Suppose that the edge lengths x, y, and z of a closed rectangular box are changing at the following rates: dx dt = 1 m sec, dy dt = -2 m sec, dz dt = 1 m sec. Find the rates at which the box’s (a) volume, (b) surface area, and (c) diagonal length are changing at the instant when x = 4, y = 3, and z = 2. Solution (a) The volume is given by V = xyz. We have dV dt dx dt = yz + x dy z + xy dt dz , dt and so dV ¥ dt Hx,y,zL=H4,3,2L = H1L H3L H2L + H4L H-2L H2L + H4L H3L H1L = 6 - 16 + 12 = 2 m3 sec. (b) The surface area is given by S = 2 xy + 2 xz + 2 yz. We have dS dt =2 dx dt y+2x dy dt +2 dx dt z+2x dz dt +2 dy dt z+2y dz , dt and so dS ¥ dt Hx,y,zL=H4,3,2L (c) By #8 we have = 2@H1L H3L + H4L H-2L + H1L H2L + H4L H1L + H-2L H2L + H3L H1LD = 2 H3 - 8 + 2 + 4 - 4 + 3L = 0 m2 sec. ds dt = 1 s Jx dx dt +y dy dt +z dz N. dt So we must first determine what s is when x = 4, y = 3, and z = 2. s= H4L2 + H3L2 + H2L2 = §3.7 #22 16 + 9 + 4 = ds ¥ dt Hx,y,zL=H4,3,2L Since 29 , we have = 29 29 HH4L H1L + H3L H-2L + H2L H1LL = 0 m2 sec. Hauling in a dinghy. A dinghy is pulled toward a dock by a rope from the bow through a ring on the dock 6 ft above the bow. The rope is being hauled in at the rate of 2 ft/sec. (See the figure on page 162.) (a) How fast is the boat approaching the dock when 10 ft of rope are out? HW5_Solutions.nb 5 (b) At what rate is the angle Θ changing at this instant? Solution Let L be the length of rope out and D the distance from the bow at time t. We are given that dL dt = -2 ft sec. (a) We have L2 = D2 + 36. Differentiating gives When L = 10 we have D = 100 - 36 = dL dt dL dt dD dt dD . dt 2L = 2D L = D 64 = 8, and so H10L H-2L = H8L dD ¥ dt L=10 (b) We have sin Θ = D L. Differentiating gives When L = 10 we have cos Θ = 6 10. Thus, dD cos Θ 6 dΘ 10 dt §3.7 #30 = dD dt 20 -8 dΘ dt = dt = - 2 ft sec. 5 L-D L 2 dL dt . H-52L H10L-H8L H-2L 9 = - 100 H10L2 dΘ 9 10 9 3 = I- 100 M I 6 M = - 60 = - 20 . dt = A Moving Shadow. A man 6 ft tall walks at the rate of 5 ft/sec toward a streetlight that is 16 ft above ground. At what rate is the tip of his shadow moving? At what rate is the length of his shadow changing when he is 10 ft from the base of the light? Solution Let x and z be the distance between the man and the lightpost and the tip of the shadow and the lightpost, respectively, as in the diagram below. We are given that dx dt = -5. 16 6 x z-x Similar triangles gives z 16 = z-x 6 6 z = 16 z - 16 x 10 z = 16 x z Therefore dy dt = dz dt - dz 8 = 5 H-5L = -8 ft/sec. dt dx = -8 - H-5L = -3. dt = 8 5 x. Letting y = z - x be the length of the shadow, we have that 6 8 = 5 H-5L = -8 ft/sec. dt dx = -8 - H-5L = -3. dt HW5_Solutions.nb dz Therefore dy dt = dz dt - Letting y = z - x be the length of the shadow, we have that
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