MAS113 CALCULUS II SPRING 2008, QUIZ 4 SOLUTIONS

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1 MAS113 CALCULUS II SPRING 8, QUIZ 4 SOLUTIONS Quiz 4a Solutions 1 Find the area of the surface obtained by rotating the curve y = x 3 /6 + 1/x, 1/ x 1 about the x-axis. We have y = x / 1/x. Therefore, x 4 = dx x = 4 Therefore, the surface area is x 3 π x x 1/ + 1 x x + 1 x = x + 1 x. x 6 dx = π 7 + x = 63 x 1/ 56 π. Solve the differential equation y = y + 5 with the initial condition y = y. T 1 = e t. e t y = 5e t = y = 5 + ce t. y = y gives that c = y 5. Consequently, the solution is yt = 5 + y 5e t. 3 Determine the values of r for which the differential equation y + y = has solutions of the form y = e rt. y = e rt gives that y = re rt. We must have re rt + e rt =, or r + =. r =. Quiz 4b Solutions 1 Find the length of the curve y = 4x, y. We have dx/ = y/. Therefore, the length of the curve is y. Making the change of variables u = y/1, we get that the above integral is u u du = u + 1 lnu + 1 u 1 = + 1 ln.

2 Solve the differential equation y = y + 5 with the initial condition y = y. T = e t. e t y = 5e t = y = 5 + ce t. y = y gives that c = y 5/. Consequently, the solution is yt = 5 + y 5 e t. 3 Determine the values of r for which the differential equation y y = has solutions of the form y = e rt. y = e rt gives that y = re rt and y = r e rt. We must have r e rt e rt =, or r 1 =. r = ±1. Quiz 4c Solutions 1 Find the area of the surface obtained by rotating the curve x = a coshy/a, a y a about the y-axis. We have dx Therefore, the surface area is π a a a cosh y a cosh y a = 4πa a = sinh y a = cosh y a. = πa cosh y = πa a y + a sinh y a a a cosh y a = πa 1 sinh. Solve the differential equation y = y + 1 with the initial condition y = y. T = e t. e t y = 1e t = y = 5 + ce t. y = y gives that c = y 5. Consequently, the solution is yt = 5 + y 5e t.

3 3 Determine the values of r for which the differential equation t y + 4ty + y = has solutions of the form y = t r for t >. y = t r gives that y = rt r 1 and y = rr 1t r. t rr 1t r + 4trt r t r = or = rr 4r + = r 3r + = r 1r. r = 1 or r =. Quiz 4d Solutions 1 Find the length of the curve y = lncos x, x π/3. In the original print out of the quiz the formula of the curve was y = lncosx which is a mis-print. The intended problem is as given here. You will be given full credit so long as you set up the integral correctly. We have = sin x = sec x. dx cos x Therefore, the arc length is π/3 1 sec xdx = ln sec x + tanx π/3 = ln + 3. Solve the differential equation y =.5y 45 with the initial condition y = 85. T 1 = e t/. e t/ y = 45e t/ = y = 45 + ce t/. y = 85 gives that c = 13. Consequently, the solution is yt = e t/. 3 Determine the values of r for which the differential equation y + y 3y = has solutions of the form y = e rt. y = e rt gives that y = re rt and y = r e rt. We must have r e rt + re rt 3e rt =, or r + r 3 = r + 3r 1 =. r = 1 or r = 3. Quiz 4e Solutions

4 1 Find the area of the surface obtained byg rotating the curve x = y, 1 y about the x-axis. We have Hence the surface area is π 1 y 16y = π 16 dx = 16y. 16y + 1 3/ = π Solve the differential equation y = y 5 with the initial condition y = y. 1 = e t. e t y = 5e t = y = 5 + ce t. y = y gives that c = y + 5. Consequently, the solution is yt = 5 + y + 5e t/. 3 Determine the values of r for which the differential equation y + y 6y = has solutions of the form y = e rt. y = e rt gives that y = re rt and y = r e rt. We must have r e rt + re rt 6e rt =, or r + r 6 = r + 3r =. r = or r = 3. Quiz 4f Solutions 1 Find the length of the curve y = x / lnx/4, x 4. We have y = x 1/4x and we see that dx = x x = x x Therefore, the arc length is 4 L = 1 x dx = 4x + ln x 4 = 6 + ln 4 4. Solve the differential equation y = y 5 with the initial condition y = y. = e t.

5 e t y = 5e t = y = 5 + ce t. y = y gives that c = y + 5. Consequently, the solution is yt = 5 + y + 5e t/. 3 Determine the values of r for which the differential equation y 13y + 1y = has solutions of the form y = e rt. y = e rt gives that y = re rt and y = r e rt. We must have r e rt 13re rt + 1e rt =, or r 13r + 1 = r 1r 1 =. r = 1 or r = 1. Quiz 4g Solutions 1 Find the area of the surface obtained by rotating the curve y = 1 x, x 1 about the y-axis. We have y = x. Therefore, the surface area is πx 4x dx = π 4x 1 = π 5 1. Solve the differential equation y = y 1 with the initial condition y = y. = e t. e t y = 1e t = y = ce t. y = y gives that c = y + 1. Consequently, the solution is yt = y + 1e t/. 3 Determine the values of r for which the differential equation t y 4ty + 4y = has solutions of the form y = t r for t >. y = t r gives that y = rt r 1 and y = rr 1t r. t rr 1t r 4trt r 4t r = or = rr 1 4r + 4 = r 5r + 4 = r 1r 4. r = 1 or r = 4. Quiz 4h Solutions

6 1 Find the length of the curve y = e x, x 1. We have y = e x. Therefore, the arc length is ex dx = 1+e u u 1 du. The above arrives from the change of variables u = e x. Now we have u u 1 du = 1 1 u + 1 du = x + 1 u 1 ln u 1 u c. Therefore, the arc length in question is e + 1 ln e e Solve the differential equation y = 9.8 y/5 with the initial condition y =. 1 5 = e t/5. e t/5 y = 9.8e t/5 = y = ce t/5. y = gives that c = 9.8. Consequently, the solution is yt = e t/. 3 Determine the values of r for which the differential equation y 3y + y = has solutions of the form y = e rt. y = e rt gives that y = re rt, y = r e rt and y = r e rt. We must have r 3 e rt 3r e rt + re rt =, or r 3 3r + r = rr 1r =. r =, r = 1 or r =.