# 57 ft approximately a the function ln x x02 is

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Unformatted text preview: 18y ) = 0, 32x + 18y = 0, y = − x so dt dt dt dt dt 9 9 9 16 9 16 256 2 400 2 81 2 2 x = 144, x = 144, x = , x = ± . If x = , then y = − = − . Similarly, 16x + 9 81 9 25 5 5 95 5 16 9 16 9 16 9 . The points are ( , − ) and (− , ). if x = − , then y = 5 5 5 5 55 38. 32x 1 1 1 ds 1 dS ds 1 dS + = we get − 2 −2 = 0, so given that = −2. From dt s=10 dt s=10 s S 6 s dt S dt 2 S ds 1 1 1 dS 225 dS = − 2 . If s = 10, then + = which gives S = 15. So (−2) = 4.5 cm/s. =− dt s dt 10 S 6 dt s=10 100 The image is moving away from the lens. 39. Find 129 Chapter 4 40. Suppose that the reservoir has height H and that the radius at the top is R. At any instant of time let h and r be the corresponding dimensions of the cone of water (see ﬁgure). We want to show that dh dV is constant and independent of H and R, given that = −kA where V is the volume of water, dt dt A is the area of a circle of radius r, and k is a positive constant. The volume of a cone of radius r and 2 1 R R 1 R r height h is V = πr2 h. By similar triangles = , r = h thus V = π h3 so 3 h H H 3 H dV =π dt R H 2 h2 dh dt (1) R r 2 R dV = −kA or, because A = πr2 = π h2 , But it is given that dt H 2 R dV = −kπ h2 , which when substituted into equation (1) gives dt H 2 2 R R dh dh = −k . h2 = π h2 , −kπ H H dt dt 41. H h dr is a constant Let r be the radius, V the volume, and A the surface area of a sphere. Show that dt 4 dV = −kA, where k is a positive constant. Because V = πr3 , given that dt 3 dV dr = 4πr2 (1) dt dt dV dV = −kA or, because A = 4πr2 , = −4πr2 k which when substituted into dt dt dr dr = −k . equation (1) gives −4πr2 k = 4πr2 , dt dt But it is given that 42. Let x be the distance between the tips of the minute and hour hands, and α and β the angles shown in the ﬁgure. Because the minute hand makes one revolution in 60 minutes, 2π dα = = π/30 rad/min; the hour hand makes one revolution in 12 hours (720 minutes), thus dt 60 dα 2π dx dβ dβ given that = = π/360 rad/min. We want to ﬁnd = π/30 and = π/360. dt 720 dt α=2π, dt dt β =3π/2 Using the law of cosines on the triangle shown in the ﬁgure, x2 = 32 + 42 − 2(3)(4) cos(α − β ) = 25 − 24 cos(α − β ), so dα dβ dx = 0 + 24 sin(α − β ) − , 2x dt dt dt 12 dα dβ dx = − sin(α − β ). When α = 2π and β = 3π/2, dt x dt dt x2 = 25 − 24 cos(2π − 3π/2) = 25, x = 5; so 11π 12 dx (π/30 − π/360) sin(2π − 3π/2) = in/min. = dt α=2π, 5 150 4 x 3 b a β =3π/2 43. Extend sides of cup to complete the cone and let V0 be the volume 1 of the portion added, then (see ﬁgure) V = πr2 h − V0 where 3 2 4 1 1 1 1 r h = = so r = h and V = π πh3 − V0 , h − V0 = h 12 3 3 3 3 27 1 9 dV dh 9 20 dh dh dV = πh2 , = , cm/s. = (20) = 2 dt 2 dt 9 dt dt πh dt h=9 π (9) 9π 4 r 6 2 6 h Exercise Set 4.7 130 EXERCISE SET 4.7 1. x2 − 4 (x − 2)(x + 2) x+2 2 = lim = lim = + 2x − 8 x→2 (x + 4)(x − 2) x→2 x + 4 3 5 2 − lim 2 2x − 5 x→+∞ x lim = = 7 x→+∞ 3x + 7 3 3 + lim x→+∞ x (a) lim x→2 x2 (b) 2. sin x cos x sin x = sin x = cos x so lim =0 x→0 tan x tan x sin x x2 − 1 x2 − 1 (x − 1)(x + 1) x+1 2 = =2 so lim 3 = x→1 x − 1 x3 − 1 (x − 1)(x2 + x + 1) x +x+1 3 (a) (b) 3. lim 1/x =1 x→1 1 4. 5. lim ex =1 x→0 cos x 6. 7. sec2 θ =1 θ→0 1 8. 9. lim lim x→π + cos x = −1 1 10. lim 2 cos 2x = 2/5 5 cos 5x lim 1 = 1/5 6x − 13 lim tet + et = −1 −et lim cos x = +∞ 2x x→0 x→3 t→0 x→0+ 3e3x 9e3x = lim = +∞ x→+∞ 2x x→+∞ 2 11. 1/x =0 x→+∞ 1 13. lim+ − csc2 x −x −1 = lim+ = lim+ = −∞ 2 x→0 sin x x→0 2 sin x cos x 1/x lim+ −1/x x = lim+ 1/x = 0 2 )e1/x x→0 e (−1/x 14. 15. 16. lim x→0 x→0 20. 21. 22. 23. lim 100x99 (100)(99)x98 (100)(99)(98) · · · (1) = lim = · · · = lim =0 x x→+∞ x→+∞ x→+∞ e ex ex √ cos x/ sin x 2/ 1 − 4x2 = lim+ cos2 x = 1 lim+ 17. lim =2 x→0 sec2 x/ tan x x→0 x→0 1 lim 1− 18. 12. lim x→0 1 1 1 1 + x2 = lim = 2 x→0 3(1 + x2 ) 3x 3 lim (x − π ) tan(x/2) = lim x→π x→π 19. x 1 = lim x = 0 x x→+∞ e x→+∞ e lim xe−x = lim x→+∞ x−π 1 = lim = −2 cot(x/2) x→π −(1/2) csc2 (x/2) sin(π/x) (−π/x2 ) cos(π/x) = lim = lim π cos(π/x) = π x→+∞ x→+∞ x→+∞ 1/x −1/x2 lim x sin(π/x) = lim x→+∞ lim+ tan x ln x = lim+ x→0 lim x→(π/2)− x→0 ln x 1/x − sin2 x −2 sin x cos x = lim+ = lim+ = lim+ =0 2x x→0 x→0 cot x x→0 − csc x 1 sec 3x cos 5x = lim x→(π/2)− −5(+1) 5 cos 5x −5 sin 5x = lim − = =− cos 3x x→(π/2) −3 sin 3x (−3)(−1) 3 131 24. Chapter 4 x−π 1 = lim =1 x→π sec2 x tan x lim (x − π ) cot x = lim x→π x→π ln(1 − 3/x) −3 = lim = −3, lim y = e−3 x→+∞ x→+∞ 1 − 3/x x→+∞ 1/x 25. y = (1 − 3/x)x , lim ln y = lim 26. y = (1 + 2x)−3/x , lim ln y = lim − 27. y = (ex + x)1/x , lim ln y = lim 28. y = (1 + a/x)bx , lim ln y = lim 29. y = (2 − x)tan(πx/2) , lim ln y = lim 30. y = [cos(2/x)]x , lim ln y = lim x→+∞ x→0 32. ln(ex + x) ex + 1 = lim x = 2, lim y = e2 x→0 x→0 e + x x...
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## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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