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Exam2_S2009 Solutions

Thus the area of the region is given by

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Thus the area of the region is given by integraldisplay 1 1 bracketleftbiggintegraldisplay 1 x 2 ye 5 x x 5 dy bracketrightbigg dx. Do the inside integral first: integraldisplay 1 x 2 ye 5 x x 5 dy = e 5 x x 5 integraldisplay 1 x 2 y dy (since e 5 x x 5 is constant with respect to y ) = e 5 x x 5 1 2 y 2 vextendsingle vextendsingle vextendsingle y =1 y = x 2 = e 5 x x 5 parenleftbigg 1 2 (1) 2 1 2 ( x 2 ) 2 parenrightbigg = 1 2 (1 x 4 ) e 5 x x 5 Now we substitute this into the outside integral: integraldisplay 1 1 1 2 (1 x 4 ) e 5 x x 5 dx Let u = 5 x x 5 . Then du = 5 5 x 4 dx or 1 5 du = 1 x 4 dx . Then integraldisplay 1 1 1 2 (1 x 4 ) e 5 x x 5 dx = 1 10 integraldisplay ? ? e u du = 1 10 e 5 x x 5 vextendsingle vextendsingle vextendsingle x =1 x = 1 = 1 10 parenleftBig e 5(1) (1) 5 e 5( 1) ( 1) 5 parenrightBig = e 4 e 4 10 2
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4. Let f ( x, y ) = radicalbig y x 2 . a. Find the domain of f . Since we can’t take an even root of a negative number, the domain of f is the set of points ( x, y ) such that y x 2 . The domain can be viewed as the following: -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -4 -3 -2 -1 1 2 3 4 5 6 7 8 b. Sketch the level curves for f corresponding to z = 0 , 1 , 2. For the level curve z = 0, we observe that radicalbig y x 2 = 0 only when y = x 2 . Thus the level curve is the graph of x 2 . Similarly, we know that radicalbig y x 2 = 1 only when y = x 2 + 1, and radicalbig y x 2 = 2 only when y = x 2 + 4. Thus the level curves are: -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 -4 -3 -2 -1 1 2 3 4 5 6 7 8 z=0 z=1 z=2 3
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5. Which of the following functions are solutions to the differential equation xy ′′ + y = 0? a. y = 2 sin x + e b. y = ln x c. y = x 3 x 2 a. If y = 2 sin x + e then y = 2 cos x and y ′′ = 2 sin x . Then xy ′′ + y = x ( 2 sin x ) + (2 cos x ) = 2 x sin x + 2 cos x negationslash = 0 . Thus (a) is not a solution. b. If y = ln x then y = 1 x and y ′′ = 1 x 2 . Then xy ′′ + y = x parenleftbigg 1 x 2 parenrightbigg + parenleftbigg 1 x parenrightbigg = 1 x + 1 x = 0 .
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