HOMEWORK SOLUTIONS MATH 1910 Sections 7.9, 8.1 Fall 2016


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1 HOMEWORK SOLUTIONS MATH 9 Sections 7.9, 8. Fll 6 Problem Show tht for ny constnts M,, nd, the function yt) = )) t ) M + tnh stisfies the logistic eqution: y SOLUTION. Let Then nd Finlly, y = y M yt) = ). ) M + tnht )). yt) M = ) tnht )), yt) yt) M ) = 4 M tnh y t) = 4 M t ) )) = 4 M t ) ). t ) ) = yt) yt) M ) d Problem Solve the integrl 4 SOLUTION. d 4 = d/) = cosh ) ) + C. Problem ) Show tht y = tnh t stisfies the differentil eqution dy/dt = y with initil condition y) =. b) Show tht for rbitrry A, B, the function y = A tnhbt) stisfies dy dt = AB B A y, y) =
2 c) Let vt) be the velocity of flling object of mss m. For lrge velocities, ir resistnce is proportionl to the squre velocity vt). If we choose coordintes so tht vt) > for flling object, then by Newton s Lw of Motion, there is constnt > such tht dv dt = g m v Solve for vt) by pplying the result of b) with A = gm/ nd B = g/m. d) Clculte the terminl velocity lim t vt). e) Find if m = 5lb nd the terminl velocity is mph. SOLUTION. ) First, note tht if we divide the identity cosh t sinh t = by cosh t, we obtin the identity tnh t = sech t. Now, let y = tnh t. Then, Furthermore, y) = tnh =. b) Let y = A tnhbt). Then dy dt = sech t = tnh t = y. ) dy dt = AB sech Bt) = AB tnh Bt)) = AB y A = AB By A Furthermore, y) = A tnh =. c) Mtching the differentil eqution with the templte from prt b) yields dv dt = g m v dv dt = AB B A v AB = g nd B A = m. Solving for A nd B gives A = mg nd B = g b. Thus vt) = A tnhbt) = mg tnh ) g b t. d) lim t vt) = mg lim t tnh ) g mg b t =
3 e) Substitute m = 5lb nd g = 3ft/sec = miles/hr into the eqution for the terminl velocity obtined in prt d) nd then solve for. This gives = ) = 78.8lb/mile Problem 8..6 Solve tn d using integrtion by prts, with u = tn nd dv = d. SOLUTION. Using u = tn nd v = gives us u = tn, v = = u = +, v =. Integrtion by Prts gives us tn d = tn + )d. For the integrl on the right we ll use the substitution w = +, dw = d. Then we hve tn d = tn + )d = tn dw w = = tn ln w + C = tn ln + + C. Problem 8..3 Solve the integrl sin d SOLUTION. Let u = nd v = sin. Then we hve u = v = cos u = v = sin Using Integrtion by Prts, we get sin d = cos ) cos )d = cos + cos d. We must pply Integrtion by Prts gin to evlute cos d. Ting u = nd v = cos, we get cos d = sin sin d = sin + cos + C. Plugging this into the originl eqution gives us sin d = cos + sin + cos ) + C = cos + sin + cos + C. ln Problem 8.. Solve the integrl d
4 SOLUTION. Let u = ln nd v =. Then we hve u = ln v = u = v = Using Integrtion by Prts, we get ln d = ln )d = ln + d = ln + C = ln + ) + C. lnln )d Problem Solve the integrl. 8.. SOLUTION. Let u = lnln ) nd dv = d/. This gives u = Applying integrtion by prts formul, we hve: lnln ) d = ln lnln ) ln Problem Solve the integrl ln d. ln ) ln = ln, nd v = ln. d = ln lnln ) ln + c ln SOLUTION. Let u = ln nd dv = d. This gives, u = /d nd v = /. Using integrtion by prts formul we hve: ln d = ln ) d = ln ) 4 ) = ln 4 ) = ln 3 4 Problem Derive the reduction formul n e d = n e n n e d. SOLUTION. Let u = n nd dv = e d. Then du = n d, v = e, nd n e d = n e n n e d. Problem Find f), ssuming tht f)e d = f)e e d 4
5 SOLUTION. We see tht Integrtion by Prts ws pplied to f)e d with u = f) nd dv = e d, nd therefore f ) = u =. Thus, f) = ln + C for ny constnt C. Problem Find the re enclosed by y = ln nd y = ln ). SOLUTION. The two grphs intersect t = nd = e, nd ln is bove ln ), so the re is e ln ln ) ) d = e ln d e ln ) d Using integrtion by prts for the second integrl, let u = ln ), dv = d; then du = ln nd v =, so tht e ln ) d = ln ) ) e Substituting this bc into the originl eqution gives e e ln d = e ln ln ) ) e d = 3 ln d e e ln d We use integrtion by prts to evlute the remining integrl, with u = ln nd dv = d; then du = d nd v =, so tht e ln d = ln e int e d = e e ) = nd thus, substituting bc in, the vlue of the originl integrl is e ln ln ) ) e d = 3 ln d e = 3 e Problem Define P n ) by n e d = P n )e + C Use the reduction formul in Problem 6 to prove tht P n ) = n np n ). Use this recursion reltion to find P n ) for n =,, 3, 4. Note tht P ) =. SOLUTION. From 8..6 we hve n e d = n e n n e d = P n )e + C ) 5
6 nd lso n e d = P n )e + D ) If we substitute the result of ) into ) nd compre the coefficients in front of e we get: P n )e + C = n e np n )e + D) = e n np n )) nd which gives P n ) = n np n ). P ) = P ) = P ) = P ) = ) = + P 3 ) = 3 3P ) = ) = P 4 ) = 4 4P 3 ) = ) = Problem 8..9 Set I, b) = ) b d, where, b re whole numbers. ) Use substitution to show tht I, b) = Ib, ). b) Show tht I, ) = I, ) = +. c) Prove tht for nd b, I, b) = I, b + ) b d) Use b) nd c) to clculte I, ) nd I3, ). e) Show tht I, b) =!b! + b + )!. SOLUTION. ) Let u = = du = d nd the bounds of u go from to. I, b) = ) b d = b) For b = from prt ) we get I, ) = Ib, ) = u) u b du) = d = + + = + u) u b du = Ib, ) c) Let us try to trnsform the RHS of wht we wnt to prove, by using integrtion by prts, where u = ) b+ nd dv = d. This gives du = b + ) b d nd v =. Then we hve: I, b+) = b + ) b+ + b + ) b + ) b d = ) b d = I, b) 6
7 d) ) 3 I, ) = I, + ) = I, ) = I, ) = I3, ) = I, 3) = + 3 I, 4) = 4 5 I, 5) = I5, ) = d = e) Apply result c) times nd lso pply result from prt b) once. I, b) = I, b+) = b + = = 3 = 6 5 d = 6 = 6 b + I, b+) = b + b + I 3, b+3) = = b + b + 3 ) ) I, + b) = b b + ) b + ) b + ) b + ) ) ) b b + ) b + ) b + ) b + ) b + + =! b! + b + )! Problem 8..9 Let I n = n cos )d nd J n = n sin )d. ) Find reduction formul tht epresses I n in terms of J n. n cos )) Hint: Write X n cos ) nd b) Use the result of ) to show tht I n cn be evluted eplicitly if n is odd. c) Evlute I 3. SOLUTION. ) Let u = n, dv = cos ) = du = n ) n d nd dv = dsin )) = v = sin ). Applying integrtion by prts on I n we get: I n = n cos ))d = n sin ) n n sin )d = n sin ) n J n b) If n is odd then n, n 4, n 6,..., 3, re ll odd numbers, nd recursively we cn relte using result from prt ), I n to J n, J n 4,..., J in the end. And we cn esily compute J, which mens tht we cn esily compute I n for every n odd. c) I 3 = sin )/ J = sin )/ sin )d = sin )/ + dcos ))/ = sin )/ + cos )/ + C
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