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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 , π ). 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 . Here ( a, b ) = (0 , π ). 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 = (12 sin 4 x, 6 cos 2 y ) . For the tangent plane (or linear approximation), we evaluate the gradient at the specified point (0 , π ), that is, we plug in 0 for x and π for y : grad f ( a, b ) = grad f (0 , π ) = (0 , 6) . So the tangent plane is given by z = f ( a, b ) + grad f ( a, b ) x - a y - b = f (0 , π ) + (0 , 6) x y - π =(2 π - 3 * 1 + 3 * 0) + (0 , 6) x y - π =(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 at the point (2 , 1) . Solution: The Jacobian of F = 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 , 1); in other words, plug in 2 for x and 1 for y (although there isn’t a y value in this case): ∂g ∂y (2 , 1) = - 3 e - 2 = - 3 e - 2 .
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