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Unformatted text preview: Math 2220 Prelim I Solutions Spring 2010 1. Let f ( x, y ) = x 3 y x 6 + y 2 for ( x, y ) negationslash = (0 , 0). (4 pts) (a) Show that the cubic curves y = λx 3 are level curves of the function f . Solution. Substituting y = λx 3 in f ( x, y ) yields f ( x, λx 3 ) = x 3 ( λx 3 ) x 6 + ( λx 3 ) 2 = λx 6 x 6 + λ 2 x 6 = λ 1 + λ 2 . This is a constant, not depending on x . Therefore the curve y = λx 3 is part of the level curve f ( x, y ) = λ 1 + λ 2 . (4 pts) (b) Draw the level curves of f which pass through (1 , 1 2 ) and (1 , − 3). Solution. The curve y = λx 3 passing through (1 , 1 2 ) is y = 1 2 x 3 and the curve y = λx 3 passing through (1 , − 3) is y = − 3 x 3 . So you might think that this is all there is to the answer, but there is actually something tricky going on here. The value λ = 1 2 for the curve y = 1 2 x 3 gives f (1 , 1 2 ) = λ 1+ λ 2 = 2 5 . However, the equation λ 1+ λ 2 = 2 5 actually has another solution besides λ = 1 2 , namely λ = 2, as one can see by simplifying the equation λ 1+ λ 2 = 2 5 to a quadratic equation. This means that the level set f ( x, y ) = 2 5 consists of not one but two cubic curves, y = 1 2 x 3 and y = 2 x 3 . Similarly for the level curve y = − 3 x 3 passing through (1 , − 3) we have the equation λ 1+ λ 2 = − 3 10 which has the solution λ = − 1 3 in addition to the solution λ = − 3, so the level set f ( x, y ) = − 3 10 consists of the two cubic curves y = − 3 x 3 and y = − 1 3 x 3 . You can easily draw these cubic curves, so we won’t do this here. (6 pts) (c) Use these level curves to determine whether lim ( x,y ) → (0 , 0) f ( x, y ) exists. Solution. The limit does not exist because the point (0 , 0) can be approached via several different level curves, for example y = x 3 which is part of the level curve f ( x, y ) = 1 2 and y = 2 x 3 which is part of the level curve f ( x, y ) = 2 5 . As ( x, y ) approaches (0 , 0) along the curve y = x 3 the value of f ( x, y ) is the constant 1 2 so the limit is 1 2 , while if ( x, y ) approaches (0 , 0) along the curve y = 2 x 3 the limit is...
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 Spring '08
 PARKINSON
 Math, Multivariable Calculus, 5 pts, 6 pts, 4 pts, Gradient, 7 pts

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