math105samplefinalsolution

math105samplefinalsolution - The University of British...

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Unformatted text preview: The University of British Columbia Sample Final Examination - 2009 Solutions Mathematics 105 1 . Short-Answer Questions . (a) The level curve is given by x 2 + y 2 = c for some constant c . Since 1 2 +2 2 = c , we get x 2 + y 2 = 5. Thus, the radius is √ 5. (b) ∂f ∂y =- sin x y- x y 2 = x y 2 sin x y . Hence, we get ∂f ∂y ( π, 2) = π 2 2 sin π 2 = π 4 . (c) Solving the system ∂z ∂x = 2 x- 2 y + 12 = 0 ∂z ∂y =- 2 x- 2 y = 0 , we get x =- 3 and y = 3. Hence, the point is (- 3 , 3). (d) lim h → f ( x + h,y )- f ( x,y ) h = ∂f ∂x = 2 2 x- 3 y . (e) C ( x ) = Z (0 . 3 x 2 + 2 x ) dx = 0 . 3 x 3 3 + x 2 + c . Since 2000 = C (0) = c , we have C ( x ) = . 1 x 3 + x 2 + 2000. It follows that C (20) = 0 . 1(20 3 ) + 20 2 + 2000 = 3200. (f) Let u = ln5 x . Then du = 5 5 x dx = dx x and Z ln(5 x ) x dx = Z udu = u 2 2 + c = (ln5 x ) 2 2 + c . (g) Let u = sin- 1 x and dv = dx . Then du = 1 √ 1- x 2 dx and v = x . Therefore, Z sin- 1 xdx = x sin- 1 x- Z x √ 1- x 2 dx . Let w = √ 1- x 2 . We have dw =- x √ 1- x 2 dx . Thus, Z sin- 1 xdx = x sin- 1 x- Z- dw = x sin- 1 x + w + c = x sin- 1 x + p 1- x 2 + c . (h) Z 1 πe- 2 x dx =- π 2 e- 2 x 1 = π 2 1- 1 e 2 . (i) Average price = 1 30- 10 Z 30 10 50 e- . 05 x dx = 5 2 Z 30 10 e- . 05 x dx = 5 2 e- . 05 x- . 05 30 10 = 50( e- . 5- e- 1 . 5 ). (j) 5 = √ 2 Z 6 3 p f ( x ) dx . Hence, Z 6 3 p f ( t ) dt = 5 √ 2 . 1 (k) A ( t ) = Z t 1 2 dt = 2 3 t 3 2 . The integrating factor is e 2 3 t 3 2 . (l) 1 = Z ∞ cxe- 2 x dx = c Z ∞ xe- 2 x dx = c 1 4 . Hence, c = 4. (m) The area is Z π/ 2 (sin x- sin 3 x ) dx = Z π/ 2 sin x (1- sin 2 x ) dx = = Z π/ 2 sin x cos 2 xdx = Z 1- u 2 du =- u 3 3 1 = 1 3 . ↑ u = cos x du =- sin xdx (n) 0 . 5 = Z b 2 1 4 e- t/ 4 dt = 1 4 (- 4) e- t/ 4 b 2 =- e- b/ 4 + e- 1 / 2 . Hence, e- b/ 4 = e- 1 / 2- . 5,- b 4 = ln e- 1 / 2- . 5 or b =- 4 ln e- 1 / 2- . 5 ....
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math105samplefinalsolution - The University of British...

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