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aq2 - 1 Now the second derivatives work out to be f xx =-2...

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PRINT NAME: SOLUTIONS Calculus IV [2443–002] Quiz II Q1]... State the second derivative test for functions of two variables. Ans: Let ( a, b ) satisfy f x ( a, b ) = 0 and f y ( a, b ) = 0. Define D ( x, y ) = ( f xx )( f yy ) - ( f xy ) 2 If D ( a, b ) > 0 and f xx ( a, b ) > 0, then ( a, b ) is a local minimum point. If D ( a, b ) > 0 and f xx ( a, b ) < 0, then ( a, b ) is a local maximum point. If D ( a, b ) < 0 and f xx ( a, b ) > 0, then ( a, b ) is neither a local max nor a local min point [Saddle]. (If D ( a, b ) = 0, the test is inconclusive.) Q2]... Find and test the critical points of the function f ( x, y ) = xye - ( x 2 + y 2 ) / 2 . Soln: We compute the first derivatives using the product and chain rules. f x = (1 - x 2 ) ye - ( x 2 + y 2 ) / 2 f y = (1 - y 2 ) xe - ( x 2 + y 2 ) / 2 . Since e to any power is always positive, we see that f x = 0 = f y if and only if (1 - x 2 ) y = 0 = (1 - y 2 ) x , and that these equations are true if and only if ( x, y ) is one of the following five points: (0
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Unformatted text preview: 1). Now, the second derivatives work out to be f xx =-2 xye-( x 2 + y 2 ) / 2-xy (1-x 2 ) e-( x 2 + y 2 ) / 2 , and f yy =-2 xye-( x 2 + y 2 ) / 2-xy (1-y 2 ) e-( x 2 + y 2 ) / 2 , and f xy = (1-x 2 )(1-y 2 ) e-( x 2 + y 2 ) / 2 . Thus, we can evaluate the second derivatives at the critical points to get. • D (0 , 0) = 0 2-1 2 =-1 < 0 implies a saddle point at (0 , 0). • D (-1 , 1) = D (1 ,-1) = (2 e-1 ) 2-2 > 0, and f xx (-1 , 1) = f xx (1 ,-1) = 2 e-1 > 0 implies a local minimum at (-1 , 1) and at (1 ,-1). • D (-1 ,-1) = D (1 , 1) = (-2 e-1 ) 2-2 > 0, and f xx (-1 ,-1) = f xx (1 , 1) =-2 e-1 < 0 implies a local maximum at (-1 ,-1) and at (1 , 1)....
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