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# ass6 answr - Assignment 6 solutions Total=5 points 1. Find...

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Assignment 6 solutions Total=5 points 1. Find the tangent plane to the graph of the function f ( x,y ) = 2 π - 3 cos 4 x + 3 sin 2 y at the point (0 ). Solution: The tangent plane of f at a point ( a,b ) is given by the equation z = f ( a,b ) + grad f ( a,b ) ± x - a y - b ² . Here ( a,b ) = (0 ). The x -partial derivative of f , ∂f ∂x , is obtained by treating y as a constant and taking the derivative with respect to x : here it is given by 0 + 12 sin 4 x + 0 = 12 sin 4 x . Similarly ∂f ∂y = 6 cos 2 y , and so grad f = ± ∂f ∂x , ∂f ∂y ² = (12 sin 4 x, 6 cos 2 y ) . For the tangent plane (or linear approximation), we evaluate the gradient at the speciﬁed point (0 ), that is, we plug in 0 for x and π for y : grad f ( a,b ) = grad f (0 ) = (0 , 6) . So the tangent plane is given by z = f ( a,b ) + grad f ( a,b ) ± x - a y - b ² = f (0 ) + (0 , 6) ± x y - π ² =(2 π - 3 * 1 + 3 * 0) + (0 , 6) ± x y - π ² =(2 π - 3) + 0 * x + (6)( y - π ) = - 4 π - 3 + 6 y. Alternatively we could have used the equivalent formula z = f ( a,b ) + ∂f ∂x ( a,b )( x - a ) + ∂f ∂y ( a,b )( y - b ) for the tangent plane. 1

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2. Find the (2,2)-entry of the Jacobian matrix of the function F ( x,y ) = ± x 2 e y + 2 x sin( x y ) sin( x 2 ) - 3 ye - x ² at the point (2 , 1) . Solution: The Jacobian of F = ³ f g ´ is given by J = ∂f ∂x ∂f ∂y ∂g ∂x ∂g ∂y ! . The (2,2) entry means the second row, second column (( i,j ) th entry is the entry in the i th row and j th column), that is, ∂g ∂y . Here g , the second row of F , is the function sin( x 2 ) - 3 ye - x . Its y -partial derivative is ∂g ∂y = ∂y (sin( x 2 ) - 3 ye - x ) = 0 - 3 e - x . We are asked to evaluate this at the point (2
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## This note was uploaded on 09/09/2011 for the course MAT 1332 taught by Professor Munteanu during the Winter '07 term at University of Ottawa.

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ass6 answr - Assignment 6 solutions Total=5 points 1. Find...

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