APPM 1360 Midterm: appm1360fall2015exam2_sol

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turquoisegnu128

31 Jan 2019

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Solution: (a)(10 pts) note that y = sin(x)/ cos(x) = tan(x) and recall that 1 + tan2(x) = sec2(x), thus, L =z b a p1 + (dy/dx)2 dx = z /4. | sec(x)| dx = ln| sec(x) + tan(x)|(cid:12)(cid:12)(cid:12)(cid:12) = ln(cid:16) 2 + 1(cid:17) ln (1) = ln(cid:16) 2 + 1(cid:17) 2 (c)(5 pts) here v = [r2 r2] y where r = arccos(y) and r = /4 so we have v =z 2/2. Further, suppose that the mass of the plate is m = 4e 8. Solution: (a)(10 pts) note that x = y3/3 = g(y) and g (y) = y2, and so, in terms of y, we have. 2 g(y)p1 + g (y)2 dy (in terms of x, we have f (x) = (3x)1/3 and f (x) = (3x) 2/3 and so u3/2(cid:12)(cid:12)(cid:12)(cid:12) 32/3 z 8/3 and now if we let u = (3x)4/3 + 1 then we see, as before, that sa = (173/2 1)/9. )

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