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 0 = 2 - x 4 x - x 2 so the arclength is given by L = Z 3 0 s 1 + ± 2 - x 4 x - x 2 ² 2 dx = Z 3 0 r 1 + 4 - 4 x + x 2 4 x - x 2 dx = Z 3 0 r 4 4 x - x 2 dx = Z 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 = Z 3 0 2 dx 4 x - x 2 = Z π/ 6 - π/ 2 4 cos θ dθ 2 cos θ = Z π/ 6 - π/ 2 2 = h 2 θ i 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 0 = 2 x - 1 / 3 , so the arclength is given by L = Z 8 0 q 1 + (2 x - 1 / 3 ) 2 dx = Z 8 0 r 1 + 4 x 2 / 3 dx = Z 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 = Z 8 0 x 2 / 3 + 4 x 1 / 3 dx = Z 8 4 u · 3 2 du = Z 8 4 3 2 u 1 / 2 du = h u 3 / 2 i 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 0 = 1 2 x so the surface area is A = Z 2 0 2 π x r 1 + 1 4 x dx = Z 2 0 π 4 x + 1 dx = h π 6 (4 x + 1) 3 / 2 i 2 0 = π 6 (27 - 1) = 13 π 3 . (If you cannot solve the integral Z 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
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This note was uploaded on 09/18/2008 for the course MATH MAth 138 taught by Professor Marshman during the Spring '08 term at Waterloo.

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

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