Math 1102: Calculus I (Math/Sci majors) MWF 3pm, Fulton Hall 230 Homework 2 solutions


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1 Mth 1102: Clculus I (Mth/Sci mjors) MWF 3pm, Fulton Hll 230 Homework 2 solutions Plese write netly, nd show ll work. Cution: An nswer with no work is wrong! Do the following problems from Chpter III: 6, p. 100 (40,4,24); 9, p. 115 (5,8,11,23). 6, p. 100: 40. Let f be function such tht f (u) = 1 1+u 3. Let g(x) = f(x 2 ). Find g (x) nd g (2). Do not ttempt to evlute f(u). Solution. Let h(x) = x 2 so tht g(x) = (f h)(x). By the chin rule, g (x) = f (h(x)) h (x). By the power rule, h (x) = 2x. Moreover, the sttement of the problem instructs us tht f (u) = 1 1+u 3. Thus g (x) = 1 2x 2x = 1 + (h(x)) x, 6 nd g (2) = = , supplementl exercises, p. 100: 4. Find the derivtive of (7x 2) 81. Solution. By the power rule nd chin rule, d ( ) (7x 2) 81 = 81(7x 2) dx 24. Find the derivtive of 2x 3 x + 1.
2 Solution. By the power rule (which we know pplies to the squre root s well!) nd chin rule, we hve d ( 2x3 x + 1) = 1 1 dx 2 2x3 x + 1 6x 2 1 (6x2 1) = 2 2x 3 x , p. 115: 5. A cube is expnding in such wy tht its edge is chnging t rte of 5 m/sec. When its edge is 4 m long, find the rte of chnge of its volume. Solution. Let x indicte the length of the side of the cube, so tht its volume V is given by V = x 3. By the chin rule, dv = 3x2 dx. As the rte of chnge of the side length is 5 m/sec, we hve dx/ = 5 m/sec. When x = 4 m, we hve dv = 3(4 m)2 5 m/sec = 240 m 3 /sec. 8. A point moves long the grph of y = 1/(x 2 + 4) in wy such tht dx/ = 3 units per second. Wht is the rte of chnge of its ycoordinte when x = 2? Solution. Applying the chin rule to y = 1/(x 2 + 4), we find tht As dx = 3, when x = 2 we hve dy = (x2 + 4) 2 2x dx. dy = (22 + 4) 2 4 ( 3) = A prticle moves differentibly on the prbol y = x 2. At wht point on the curve re its x nd ycoordintes moving t the sme rte? (You my ssume dx/ nd dy/ 0 for ll t.)
3 Solution. Applying the chin rule to the eqution defining the prbol, we see tht dy = 2x dx. The x nd ycoordintes re moving t the sme rte precisely when dx/ = dy/. Since this common vlue is nonzero we my divide both sides by it, nd find 1 = 2x, so tht x = 1/2. As y = x 2, the desired point is (1/2, 1/4). 23. Wter runs into conicl reservoir t the constnt rte of 2 m 3 /min. The vertex is 18 m down nd the rdius of the top is 24 m. How fst is the wter level rising when it is 6 m deep? Solution. Let V denote the volume of the wter, let h denote the height, nd let r denote the rdius. Note tht V = πr 2 h/3, so tht by the chin rule nd product rule dv = π ( 2rh dr 3 + r2 dh ). Moreover, r nd h re relted for this conicl reservoir, s Figure?? mkes cler pir of similr tringles: = h r, so tht r = 4h/3. Differentiting the former, we find tht dr/ = 4/3 dh/. Plugging these reltionships into the originl formul, we find tht ( dv = π 3 ( 4h 3 ) ) 2 dh 2 4h 3 h 4 dh 3 + = π ( ) h 2 dh 3 = 16π 9 h2 dh Solving for dh/, dh = 9 dv 16πh 2. Becuse we know tht dv/ = 2 m 3 /sec, t h = 6 m we hve dh = 9 16π 36 m 2 2 m 3 /sec = 1 32π m/sec.
4 Figure 1. An ellipse inscribed in squre. Problems A, B, nd C ll concern the ellipse P defined by x y2 = 1. When you finish Problem C, you will hve shown tht every ellipse cn be inscribed in squre! (See Figure 1). Problem A. For ech vlue of s stisfying < s <, there re two points on P with xcoordinte s. Find the two ycoordintes of these points, nd use implicit differentition to find the slopes of the two tngent lines to P t these two points. Your nswer should involve s,, nd b. (See Figure 2). Solution A. Isolting y in the eqution for P, we find ( ) y 2 = 1 x2. 2 Tking squre roots, we find tht the ycoordintes on P when x = s re given by y = ±b 1 s2 = ± b 2 s 2 2. Differentiting the eqution tht defines P, we see tht 2x + 2y 2 y = 0.
5 Figure 2. The sitution in problem A. (s,?) m =? (s,?) m =? Solving for y, we see tht y = 2x 2 2y = b2 x 2 y. At the point ( s, b 2 s 2), we clculte tht y is given by y s = 2 b 2 s = b s 2 2 s, 2 while t the point ( s, b 2 s 2), we see tht y is given by y = b s 2 s 2. Problem B. Find the vlues of s for which these two lines re perpendiculr. (Recll tht the slopes of perpendiculr lines re negtive reciprocls of ech other). Wht re the slopes of the two tngent lines t these points? Solution B. Let m 1 indicte the slope of the tngent line t ( s, b 2 s 2), nd let m 2 indicte the slope of the tngent line t ( s, b 2 s 2). In order to hve these lines be perpendiculr, we need m 1 = 1 m 2. Using
6 our computtions from Problem A, this eqution mounts to b s 2 s = 1 = 2 s 2. 2 b s b s 2 s 2 Clering denomintors, this eqution simplifies to More simply, so tht Tking squre roots, s 2 = 2 ( 2 s 2 ) = 4 2 s 2. s 2 ( 2 + ) = 4, s 2 = s = ± 2 + b. 2 Ech of these two vlues for the xcoordinte yields pir of points on P whose tngent lines re perpendiculr. The corresponding y coordintes cn be computed using the formul from Problem A: y = ± b 2 s 2 = ± b = ± b = ± b 2 ( ± ) b = ± b 2 ( 2 + ) b = ± Using the positive vlue of s bove, nd the formul we derived in Problem A for m 1, we see tht m 1 = b = 1. Since, by construction, m 1 nd m 2 re slopes of perpendiculr lines, we hve tht m 2 = 1 m 1 = 1. Similrly, using the negtive vlue of s will yield tht m 1 = 1 nd m 2 = 1. Problem C. Consider the four tngent lines to P obtined from Problem B. These lines outline qudrilterl. You ve shown in Problem B
7 tht pir of the opposite ngles of this qudrilterl re right ngles. Show tht this qudrilterl is in fct squre. (It follows tht the ellipse is inscribed. See Figure 1). Solution C. The slopes of the four tngent lines re 1, 1, 1, nd 1. Since these split into equl pirs the qudrilterl hs two pirs of prllel sides, i.e. it is prllelogrm. Since pir of nonprllel sides hve slopes 1 nd 1, every pir of djcent sides meet t right ngles, i.e. the qudrilterl is rectngle. If we clculte the lengths of the sides nd find tht they re equl, then we will hve demonstrted tht the qudrilterl is squre. For ech of the four lines, we know the slope nd point on the line (nmely, the point of tngency with P ). We hve: Line L 1 through 1 ( 2, 1 ) with slope 1; line L through ( 2, ) with slope ; line L 3 through ( 2, ) with slope 1; nd line L through 1 ( 2, ) with slope 1. We thus hve the four equtions: 2 + L 1 : L 2 : y y 2 + = x = x L 3 : y + L 4 : y = x = x Let p 1 indicte the intersection of L 1 with L 2, p 2 tht of L 2 with L 3, p 3 tht of L 3 with L 4, nd p 4 tht of L 4 with L 1. We clculte the coordintes of p 1, p 2, p 3, nd p 4 directly: The intersection of L 1 nd L 2 cn be found by dding the two defining equtions together, which yields 2y Simplifying nd solving for y, we find 2b2 2 + b = b. 2 y = = 2 +,
8 8 while the difference of the two equtions shows tht x = 0. Thus p 1 = (0, 2 + ). Adding the defining equtions of L 2 nd L 3 yields simply y = 0, nd solving for x in the defining eqution for L 2 we see tht 2 + b = x b, 2 so tht x = = 2 +, nd p 2 = ( 2 +, 0). Adding the defining equtions of L 3 nd L 4 yields 2 2y b = b. 2 Simplifying nd solving for y, we find y = = 2 +, while the difference of the two equtions shows tht x = 0. Thus p 3 = (0, 2 + ). Adding the defining equtions of L 4 nd L 1 yields y = 0, nd solving for x in the defining eqution for L 1 we see tht 2 + b = x b, 2 so tht x = = 2 +, nd p 4 = ( 2 +, 0). The distnce between every pir of djcent points is 2( 2 + ), so tht the qudrilterl is rectngle with equl side lengths, hence squre. See Figure 3 for schemtic with lbels.
9 L 1 (0, 2 + ) 1 2 +( 2, ) 1 2 +(2, ) L 3 ( 2 +, 0) ( 2 +, 0) L ( 2, ) 1 2 +(2, ) (0, 2 + ) L 4 Figure 3. The solution to Problem C with lbels.
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