am1 - PRINT NAME: SOLUTIONS Calculus IV [2443002] Midterm I...

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Unformatted text preview: PRINT NAME: SOLUTIONS Calculus IV [2443002] Midterm I Q1]... Find an equation for the tangent plane to the graph of f ( x,y ) = x 2 +2 xy- y 2 at the point (2 , 1 , 7) . Ans: Equation is given by ( z- z ) = f x (2 , 1)( x- x ) + f y (2 , 1)( y- y ). We have f x = 2 x + 2 y and f y = 2 x- 2 y which gives us f x (2 , 1) = 6 and f y (2 , 1) = 2. Thus, our equation becomes ( z- 7) = 6( x- 2) + 2( y- 1) which simplifies to z = 6 x + 2 y- 7 . The graph of f ( x,y ) = x 2 + 2 xy- y 2 and the vertical plane y = 1 intersect in a curve. Find a parametric equation for the tangent line to this curve of intersection at the point (2 , 1 , 7) . Ans: We have a point on the line; namely (2 , 1 , 7). We only have to find a parallel vector, V . First, note that V lies in the plane y = 1 and so its y-component will be zero (it is perpendicular to the y-axis). So we can write V = h a, ,b i . Now, the ratio b/a is just the slope of the tangent vector to the curve of intersection in the y = 1 plane. This is just= 1 plane....
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am1 - PRINT NAME: SOLUTIONS Calculus IV [2443002] Midterm I...

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