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# calc3_gradientstest - 18.02 Spring 2008 Practice Test 2...

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18.02, Spring 2008 Practice Test 2 Solutions 1. Find the equation of the tangent plane to the surface x 3 y + z 2 = 3 at the point ( - 1 , 1 , 2). Solution. The surface is defined by the equation f ( x, y, z ) = x 3 y + z 2 = 3. The normal to the tangent plane is given by f = 3 x 2 y ˆ i + x 3 ˆ j + 2 z ˆ k , and f ( - 1 , 1 , 2) = (3 , - 1 , 4). The equation of the plane is thus 3( x + 1) - ( y - 1) + 4( z - 2) = 0 . 2. Let w = f ( u, v ), where u = xy and v = x/y . Using the chain rule, express ∂w ∂x and ∂w ∂y in terms of x , y , f u , and f v . Solution. We have ∂w ∂x = f u ∂u ∂x + f v ∂v ∂x = yf u + 1 y f v and ∂w ∂y = f u ∂u ∂y + f v ∂v ∂y = xf u - x y 2 f v . 3. Let f ( x, y ) = x 2 y 2 - x . (a) Find f at (2 , 1). Solution. We have f = (2 xy 2 - 1 , 2 x 2 y ) so f (2 , 1) = (3 , 8). (b) Write the equation for the tangent plane to the graph of f at (2 , 1 , 2). Solution. The normal to the tangent plane is (3 , 8 , - 1), so the plane has equation z - 2 = 3( x - 2) + 8( y - 1) , which we can write as 3 x + 8 y - z = 12 . (c) Use a linear approximation to find the approximate value of f (1 . 9 , 1 . 1).

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calc3_gradientstest - 18.02 Spring 2008 Practice Test 2...

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