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# h11 - moore(jwm2685 – HW11 – gilbert –(55485 1 This...

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Unformatted text preview: moore (jwm2685) – HW11 – gilbert – (55485) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find an equation for the plane passing through the origin that is parallel to the tan- gent plane to the graph of z = f ( x, y ) = 4 x 2 − 2 y 2 + 2 x + 3 y at the point (1 , − 1 , f (1 , − 1)). 1. z − 10 x + 7 y + 16 = 0 2. z − 10 x − 7 y = 0 correct 3. z − 10 x + 7 y = 0 4. z + 10 x − 7 y = 0 5. z + 10 x + 7 y − 4 = 0 6. z + 10 x − 7 y − 18 = 0 Explanation: Parallel planes have parallel normals. On the other hand, the tangent plane to the graph of z = f ( x, y ) at the point ( a, b, f ( a, b )) has normal n = (− f x ( a, b ) , − f y ( a, b ) , 1 ) . But when f ( x, y ) = 4 x 2 − 2 y 2 + 2 x + 3 y we see that f x = 8 x + 2 , f y = − 4 y + 3 , and so when a = 1 , b = − 1, n = (− 10 , − 7 , 1 ) . Thus an equation for the plane through the origin with normal parallel to n is ( x, y, z )· n = ( x, y, z )·(− 10 , − 7 , 1 ) = 0 , which after evaluation becomes z − 10 x − 7 y = 0 . keywords: 002 10.0 points Find an equation for the tangent plane to the graph of f ( x, y ) = radicalbig 8 − 2 x 2 + y 2 at the point P (2 , 1 , f (2 , 1)). 1. x − 4 y + z − 6 = 0 2. x + 4 y − z − 8 = 0 3. 4 x − y + z − 8 = 0 correct 4. 4 x − y − z − 6 = 0 5. 4 x + y + z − 10 = 0 6. x + 4 y − z − 10 = 0 Explanation: The equation of the tangent plane to the graph of z = f ( x, y ) at the point P ( a, b, f ( a, b )) is given by z = f ( a, b ) + ∂f ∂x vextendsingle vextendsingle vextendsingle ( a, b ) ( x − a ) + ∂f ∂y vextendsingle vextendsingle vextendsingle ( a, b ) ( y − b ) . Now when f ( x, y ) = radicalbig 8 − 2 x 2 + y 2 , we see that ∂f ∂x = − 2 x radicalbig 8 − 2 x 2 + y 2 , while ∂f ∂y = y radicalbig 8 − 2 x 2 + y 2 . moore (jwm2685) – HW11 – gilbert – (55485) 2 Thus at P , f (2 , 1) = 1 , while ∂f ∂x vextendsingle vextendsingle vextendsingle (2 , 1) = − 4 , ∂f ∂y vextendsingle vextendsingle vextendsingle (2 , 1) = 1 ....
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h11 - moore(jwm2685 – HW11 – gilbert –(55485 1 This...

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