MAT2122 final solutions

# MAT2122 final solutions - MAT2122 Multivariable...

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MAT2122 Multivariable Calculus (Fall 2016) Final solutions (the total is 80 points) 1 (8 points). Find the distance from the point P = (0 , 1 , 1) to the tangent line to the path c ( t ) = ( e t - 1 , ln t, t 2 3 t ) at time t 0 = 1 . Solution: The velocity vector of the path c is c ( t ) = ( e t 1 , 1 /t, 2 t 3) , c whence the velocity at time t 0 (i.e., the direction of the tangent line) is u = c ( t 0 ) = c (1) = (1 , 1 , 1) . c The position at time t 0 is Q = c ( t 0 ) = c (1) = (1 , 0 , 2) . c so that −→ PQ = (1 , 1 , 3). c The projection of the vector onto vector u is α u , where α = · u b u b 2 = 1 1 + 3 3 = 1 , c whence the vector v = α u = (1 , 1 , 3) (1 , 1 , 1) = (0 , 2 , 2) c is perpendicular to the tangent line. Therefore, the distance from the point P to the tangent line to the path c at time t 0 = 1 is the length of the vector v , i.e., 2 2. c Absence of minor mistakes. c 2 (7 points). Let g ( x, y ) = p cos( x + y ) , e xy +1 , x y P and f ( u, v, w ) = ( u ln v, w 2 ) . Find the derivative (by using the chain rule) and the Jacobian of the map f g at the point (1 , 1) . Solution: The derivative of the function g is D g ( x, y ) = ( cos( x + y ) ) x ( cos( x + y ) ) y ( e xy +1 ) x ( e xy +1 ) y ( x y ) x ( x y ) y = sin( x + y ) sin( x + y ) ye xy +1 xe xy +1 1 y x y 2 , whence D g (1 , 1) = 0 0 1 1 1 1 . c

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2 The derivative of the function f is D f ( u, v, w ) = ( u ln v ) u ( u ln v ) v ( u ln v ) w ( w 2 ) u ( w 2 ) v ( w 2 ) w = ln v u v 0 0 0 2 w , so that its value at the point g (1 , 1) = (1 , 1 , 1) c is D f (1 , 1 , 1) = 0 1 0 0 0 2 . c Therefore, by the chain rule D f g (1 , 1) = D f ( g (1 , 1)) D g (1 , 1) = D f (1 , 1 , 1) D g (1 , 1) c = 0 1 0 0 0 2 0 0 1 1 1 1 = 1 1 2 2 c whence Jac f g (1 , 1) = det 1 1 2 2 = 4 . c Absence of minor mistakes. c 3 (9 points). Find the partial derivatives up to the second order of the function f ( x, y ) = e x - y sin(2( x + y )) . Find the second order Taylor expansion of the function f at the point (0 , 0) and use it to ±nd an approximate value of f (0 . 2 , 0 . 1) . Solution: To begin with, f (0 , 0) = 0. Further, f x ( x, y ) = e x y sin(2( x + y )) + 2 e x y cos(2( x + y )) , f x (0 , 0) = 2 , c f y ( x, y ) = e x y sin(2( x + y )) + 2 e x y cos(2( x + y )) f y (0 , 0) = 2 , c and f ′′ xx ( x, y ) = 3 e x y sin(2( x + y )) + 4 e x y cos(2( x + y )) , f ′′ xx (0 , 0) = 4 , c f ′′ yy ( x, y ) = 3 e x y sin(2( x + y )) 4 e x y cos(2( x + y )) , f ′′ yy (0 , 0) = 4 , c f ′′ xy ( x, y ) = 5 e x y sin(2( x + y )) , f ′′ xy (0 , 0) = 0 .
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