Assignment 5 Solutions

Assignment 5 Solutions - MATH 138 Calculus 2 Solutionss to...

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MATH 138 Calculus 2, Solutionss to Assignment 5 1: (a) Find the length of the curve y = 4 x - x 2 with 0 x 3. Solution: We have y = 2 - x 4 x - x 2 so the arclength is given by L = 3 0 1 + 2 - x 4 x - x 2 2 dx = 3 0 1 + 4 - 4 x + x 2 4 x - x 2 dx = 3 0 4 4 x - x 2 dx = 3 0 2 dx 4 x - x 2 . Note that 4 x - x 2 = 4 - ( x - 2) 2 . Let 2 sin θ = x - 2 so that 2 cos θ = 4 x - x 2 and 2 cos θ dθ = dx . Then L = 3 0 2 dx 4 x - x 2 = π/ 6 - π/ 2 4 cos θ dθ 2 cos θ = π/ 6 - π/ 2 2 = 2 θ π/ 6 - π/ 2 = π 3 + π = 4 π 3 . (b) Find the length of the curve y = 3 x 2 / 3 with 0 x 8. Solution: We have y = 2 x - 1 / 3 , so the arclength is given by L = 8 0 1 + (2 x - 1 / 3 ) 2 dx = 8 0 1 + 4 x 2 / 3 dx = 8 0 x 2 / 3 + 4 x 1 / 3 dx . Let u = x 2 / 3 + 4 so that du = 2 3 x - 1 / 3 dx , that is 3 2 du = 1 x 1 / 3 dx . Then we have L = 8 0 x 2 / 3 + 4 x 1 / 3 dx = 8 4 u · 3 2 du = 8 4 3 2 u 1 / 2 du = u 3 / 2 8 4 = 16 2 - 8 . 2: (a) Find the area of the surface which is obtained by revolving the curve y = x with 0 x 2 about the x -axis. Solution: We have y = 1 2 x so the surface area is A = 2 0 2 π x 1 + 1 4 x dx = 2 0 π 4 x + 1 dx = π 6 (4 x + 1) 3 / 2 2 0 = π 6 (27 - 1) = 13 π 3 . (If you cannot solve the integral π 4 x + 1 dx by inspection, then try the substitution u = 4 x + 1). (b) Find the area of the surface which is obtained by revolving the curve y = cos x with - π 2 x π 2 about the x -axis. Solution: We have y = - sin x . By symmetry, the surface area is given by A = 2 π/ 2 0 2 π cos x 1 + sin 2 x dx = π/ 2 0 4 π cos x 1 + sin 2 x dx = 1 0 4 π 1 + u 2 du , where u = sin x so du = cos x dx . Now let tan θ = u so that sec θ = 1 + u 2 and sec 2 θ dθ = du . Then A = π/ 4 0 4 π sec 3
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