2122, Final Exam - Solution

# 2122, Final Exam - Solution - MAT2122 Final Exam-Solution...

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MAT2122, Final Exam-Solution Fall 2015. Instructor : Mohammad Bardestani. 1

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2 Question 1 : [ 5 points ] Let f ( x , y , z ) = ( y - z , x 2 + y 2 ) and g ( u , v ) = (sin( uv ) , e u - v , u + v - 1). Find the derivative and the Jacobian of the map f g at the point (0 , 0). 5 points: 1 - formulas for D f , D g , 1 - explicit calculation of D f , D g , 1 - correct derivative, 1 - correct Jacobian, 1 - absence of minor mistakes. Solution: The derivative of g is D g = v cos( uv ) u cos( uv ) e u - v - e u - v 1 1 , so that D g (0 , 0) = 0 0 1 - 1 1 1 . The derivative of f is D f = 0 1 - 1 2 x 2 y 0 ! , so that for the point g (0 , 0) = (0 , 1 , - 1) D f (0 , 1 , - 1) = 0 1 - 1 0 2 0 ! , whence D ( f g )(0 , 0) = D f ( g (0 , 0)) D g (0 , 0) = 0 1 - 1 0 2 0 ! 0 0 1 - 1 1 1 = 0 - 2 2 - 2 ! and Jac( f g )(0 , 0) = det D ( f g )(0 , 0) = 0 - 2 2 - 2 = 4 .
3 Question 2 : [ 5 points ] Let f ( x , y ) = e xy + 2 - ln( x + y 3 ). Find the equation of the tangent plane to the graph of f at the point (2 , - 1). By using second order Taylor’s formula find an approximate value of f (2 . 01 , - 0 . 98). 5 points: 1-tangent plane, 1-second derivatives, 1-Taylor’s formula, 1-approximate value, 1-absence of minor mistakes. Solution: The value of the function f at the point ( x 0 , y 0 ) = (2 , - 1) is z 0 = f ( x 0 , y 0 ) = 1 . The first derivatives of f are f 0 x ( x , y ) = ye xy + 2 - 1 x + y 3 , f 0 y ( x , y ) = xe xy + 2 - 3 y 2 x + y 3 , whence f 0 x ( x 0 , y 0 ) = - 2 , f 0 y ( x 0 , y 0 ) = - 1 , so that the equation of the tangent plane ( z - z 0 ) = f 0 x ( x 0 , y 0 )( x - x 0 ) + f 0 y ( x 0 , y 0 )( y - y 0 ) takes the form ( z - 1) = - 2( x - 2) - ( y + 1) , or z = - 2 x - y + 4 . The second derivatives of f are f 00 xx = y 2 e xy + 2 + 1 ( x + y 3 ) 2 , f 00 xy = e xy + 2 + xye xy + 2 + 3 y 2 ( x + y 3 ) 2 , f 00 yy = x 2 e xy + 2 - 3 2 y · ( x + y 3 ) - y 2 · 3 y 2 ( x + y 3 ) 2 = x 2 e xy + 2 - 3 2 xy - y 4 ( x + y 3 ) 2 , whence f 0 xx ( x 0 , y 0 ) = 2 , f 0 xy ( x 0 , y 0 ) = 2 , f 0 yy ( x 0 , y 0 ) = 19 . Thus, the second order Taylor approximation of f at the point (2 , - 1) is f ( x , y ) f ( x 0 , y 0 ) + f 0 x ( x 0 , y 0 )( x - x 0 ) + f 0 y ( x 0 , y 0 )( y - y 0 ) + 1 2 f 00 xx ( x 0 , y 0 )( x - x 0 ) 2 + f 00 xy ( x 0 , y 0 )( x - x 0 )( y - y 0 ) + 1 2 f 00 yy ( x 0 , y 0 )( y - y 0 ) 2 = 1 - 2( x - 2) - ( y + 1) + ( x - 2) 2 + 2( x - 2)( y + 1) + 19 2 ( y + 1) 2 = 1 - 2 · 0 . 01 + 0 . 02 + 0 . 01 2 - 2 · 0 . 02 · 0 . 01 + 19 2 0 . 02 2 = 1 . 0035 .

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4 Question 3 : [ 5 points ] Find and classify all critical points of the function f ( x , y ) = x 3 - 3 x + 2 y 3 - 3 y 2 - 12 y + 1 . 5 points: 1-critical points, 1-Hessian, 1-determinant test, 1-diagonal test, 1-absence of minor mistakes. Solution: Since f 0 x ( x , y ) = 3 x 2 - 3 = 3( x 2 - 1) , f 0 y ( x , y ) = 6 y 2 - 6 y - 12 = 6( y 2 - y - 2) , the critical points are obtained by combining the solutions x 1 = - 1 , x 2 = 1 and y 1 = - 1 , y 2 = 2 of the equations x 2 - 1 = 0 and y 2 - y - 2 = 0, respectively, i.e., there are four critical points P 1 = ( - 1 , - 1) , P 2 = ( - 1 , 2) , P 3 = (1 , - 1) , P 4 = (1 , 2) .
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