M408m-Luecke_HW7 - romasko(qrr58 Assignment 7 luecke(55035...

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romasko (qrr58) – Assignment 7 – luecke – (55035) 1 This print-out should have 18 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 tangent plane to the graph of f ( x, y ) = radicalbig 3 + x 2 3 y 2 at the point P (2 , 1 , f (2 , 1)). 1. 2 x 3 y 2 z + 3 = 0 correct 2. 3 x 2 y + 2 z 3 = 0 3. 2 x + 3 y 2 z 3 = 0 4. 3 x + 2 y 2 z 5 = 0 5. 2 x 3 y + 2 z 5 = 0 6. 3 x 2 y + 2 z + 3 = 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 3 + x 2 3 y 2 , we see that ∂f ∂x = x radicalbig 3 + x 2 3 y 2 , while ∂f ∂y = 3 y radicalbig 3 + x 2 3 y 2 . Thus at P , f (2 , 1) = 2 , while ∂f ∂x vextendsingle vextendsingle vextendsingle (2 , 1) = 1 , ∂f ∂y vextendsingle vextendsingle vextendsingle (2 , 1) = 3 2 . So at P the tangent plane has equation z = 2 + ( x 2) 3 2 ( y 1) , which after rearrangement becomes 2 x 3 y 2 z + 3 = 0 . keywords: tangent plane, partial derivative, radical function, square root function, 002 10.0 points Find the linearization, L ( x, y ), of f ( x, y ) = x y at the point ( 2 , 4). 1. L ( x, y ) = 2 + 1 2 x + 2 y 2. L ( x, y ) = 2 1 2 x + 2 y 3. L ( x, y ) = 2 + 2 x + 1 2 y 4. L ( x, y ) = 2 + 2 x 1 2 y correct 5. L ( x, y ) = 4 + 1 2 x y 6. L ( x, y ) = 4 + x + 1 2 y Explanation: The linearization of f = f ( x, y ) at a point ( a, b ) is given by L ( x, y ) = f ( a, b )+( x a ) ∂f ∂x vextendsingle vextendsingle vextendsingle ( a,b ) +( y b ) ∂f ∂y vextendsingle vextendsingle vextendsingle ( a,b ) . But when f ( x, y ) = x y , ∂f ∂x = y , ∂f ∂y = x 2 y ;
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romasko (qrr58) – Assignment 7 – luecke – (55035) 2 thus when ( a, b ) = ( 2 , 4), ∂f ∂x vextendsingle vextendsingle vextendsingle ( a,b ) = 2 , ∂f ∂y vextendsingle vextendsingle vextendsingle ( a,b ) = 1 2 , while f ( a, b ) = 4. Consequently, L ( x, y ) = 2 + 2 x 1 2 y . keywords: 003 10.0 points Find the differential of u = e 6 t sin 4 θ . 1. du = 6 e 6 t sin(4 θ ) dt + 4 e 6 t cos(4 θ ) cor- rect 2. du = 4 e 6 t sin(4 θ ) dt + 4 e 6 t cos(4 θ ) 3. du = 6 e 6 t sin(4 θ ) dt + 4 e 6 t sin(4 θ ) 4. du = 4 e 6 t sin(4 θ ) dt + 4 e 4 t cos(4 θ ) 5. du = 6 e 6 t sin(4 θ ) dt + 4 e 4 t cos(4 θ ) Explanation: The differential of u is given by du = ∂u ∂t dt + ∂u ∂θ dθ . Thus, since ∂u ∂t = 6 e 6 t sin(4 θ ) and ∂u ∂θ = 4 e 6 t cos(4 θ ) it follows that du = 6 e 6 t sin(4 θ ) dt + 4 e 6 t cos(4 θ ) . keywords: 004 10.0 points Find the linearization, L ( x, y ), of the func- tion f ( x, y ) = tan 1 ( x + 8 y ) at the point ( 15 , 2) . 1. L ( x, y ) = x + 4 y + π 4 + 1 2 2. L ( x, y ) = x + 4 y 1 2 3. L ( x, y ) = 1 2 x + 4 y 1 2 4. L ( x, y ) = 1 2 x + 4 y + π 4 1 2 correct 5. L ( x, y ) = x + 4 y + π 4 1 2 Explanation: The linearization of f = f ( x, y ) at a point ( a, b ) is given by L ( x, y ) = f ( a, b ) + ( x a ) ∂f ∂x vextendsingle vextendsingle vextendsingle ( a,b ) +( y b ) ∂f ∂y vextendsingle vextendsingle vextendsingle ( a,b ) .
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