Math 22B Solutions Homework 1 Spring 2008

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1 Mth 22B Solutions Homework 1 Spring 2008 Section A sphericl rindrop evportes t rte proportionl to its surfce re. Write differentil eqution for the volume of the rindrop s function of time. Solution Let V = volume, t = time, nd S = surfce re. Then dv = ks, for k > 0. Since the volume of the rindrop is given by V = 4 3 πr3, where r is its rdius, nd its surfce re is given by S = 4πr 2, we cn solve for r nd substitute to get S = 4π ( )2 3 3 V 3. 2 Thus, dv = cv 3, 2 for some 4π constnt c > The temperture of n object chnges t rte proportionl to the difference between the temperture of the object itself nd the temperture of its surroundings. The mbient temperture is 70 o F, the rte constnt is 0.05 (min) 1. Write differentil eqution for the temperture of the object t ny time. Solution Let T be the temperture of the object, nd t be time. Then we hve dt = c(t 70) = 0.05(T 70) degrees F/min. Notice tht the coefficient c is negtive becuse the object is cooling. Section Consider the differentil eqution dy = y + b, where both nd b re positive numbers. () Solve the differentil eqution Solution Since dy = y +b dy nd solve for y to get 1 y+b dy y + b = =. Now we integrte both sides ln y + b = t + c 1

2 (b) see figure 1 (c) ln y + b 1 = e t+c = c e t y + b = c e t y(t) = b ce t (i) As increses the equilibrium solution gets smller. The convergence rte increses s well. (ii) As b increses, the equilibrium solution gets lrger. The convergence rte stys the sme. (iii) As nd b both increse, the equilibrium solution stys the sme, but the convergence rte increses for ll solutions. 8. () Find the rte constnt r if the popultion doubles in 30 dys. Solution The generl solution is p(t) = p 0 e rt where time is mesured in months, so 30 dys is 1 month for t = 1. We plug in constnts to get 2p 0 = p 0 e r 1. If we solve for r, we get r = ln(2) per month. (b) Find r if the popultion doubles in N dys. Solution Agin, time is in months, so t = N months. We solve for r s 30 in prt () to get 2p 0 = p 0 e r N 30 r = ln(2) 30 per month. N 17. () R dq = V. In order to seprte vribles we need RdQ = V Q C. Integrting both sides we get + Q C dq = V Q R C dq V Q C = R c ln V Q C = t R + C ln V Q C = t + c V Q C = ce t 2

3 See figure 2 for more reference. Q = CV Cce t Q(0) = 0 = CV Cce t Q(t) = CV (1 e t Section 1.3 Determine the order nd stte whether it is liner or nonliner. 1. t 2 d2 y + t dy + 2y = sint 2 Solution Since the highest derivtive is second order, the eqution is second order. Moreover, the eqution is liner since it hs the form n (t)y (n) (t)y = g(t) (1) 3. d4 y 4 + d3 y 3 + d2 y 2 + dy + y = 1 Solution The highest derivtive is fourth order, so the eqution is fourth order. This eqution lso hs the form (1), so it is lso liner. 4. dy + ty2 = 0 Solution There is only one derivtive, nd it is of order 1, so the eqution is first order. The eqution is non-liner becuse it hs the term y d2 y 2 + sin(t + y) = sint Solution The eqution is second order becuse the highest order derivtive is of order two. Since sin(t + y) is term, the eqution is not liner. Verify tht ech is solution of the differentil eqution 9. ty y = t 2, y = 3t + t 2 Solution We differentite y to obtin y = 3 + 2t. Substituting in to the differentil eqution gives ty y = t(3t + 2t) (3t + t 2 ) = 3t + 2t 2 3t t 2 = t 2, s desired. Determine the vlues of r for which the given differentil eqution hs solutions of the form y = e rt. 3

4 15. y + 2y = 0 Solution We substitute y = e rt nd y = re rt into the eqution to get re rt + 2e rt = 0. This implies tht re rt = 2e rt, which gives us tht r = y y = 0 Solution We substitute y = e rt, y = re rt, nd y = r 2 e rt into the eqution to get r 2 e rt e rt = 0. Then we hve (r 2 1)e rt = 0, which implies tht r 2 1 = 0 since e rt is never zero. Thus, r = ± See textbook pge 25 for question. Solution () see figure (3) (b) We hve tht F = m, but for the tngentil direction. Let F θ = m t het. The only force cting in the tngentil direction is the weight, so F θ = mgsinθ. The ccelertion t het is the liner ccelertion long the pth t het = L d2 θ, s given in the problem. So mgsinθ = ml d2 θ. See 2 2 figure 4 for more reference. (c) From prt (b), we hve mgsinθ = ml d2 θ, so L d2 θ + gsinθ = 0, 2 2 which implies tht d2 θ + g sinθ = 0. 2 L 30. See book pgs for problem. () The kinetic energy of mss m is given by T = 1 2 mv2, where v is the velocity. A prticle in motion on circle of rdius L hs speed L dθ, where θ is ngulr position, dθ is ngulr speed. So T = 1 2 m(ldθ )2. (b) Potentil energy is given by mgh, where h is the height nd g is the grvity. If we tke v = 0, the lowest point of the swing is h = L. If the pendulum is not t the lowest point of the swing, h = L(1 cosθ). So V = mgl(1 cosθ) See figure 5 for more reference. (c) E = T + V E = 1 2 ml2 ( dθ )2 + mgl(1 cosθ). So then de = ml2dθ d 2 θ + mglsinθdθ 2 by the Chin Rule. Then we set dt = 0 to get 0 = ml 2dθ d 2 θ + mglsinθdθ 2 4

5 Dividing both sides by ml dθ we get 0 = L d2 θ 2 + gsinθ, or 0 = d2 θ L gsinθ 5