57 ft approximately a the function ln x x02 is

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

Ask a homework question - tutors are online