Exam 11: Partial Derivatives

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Given f(x,y)=xe−2yf ( x , y ) = x e ^ { - 2 y } , at the point (4,0)( 4,0 ) find the equation of the tangent plane to the graph of ff . Also, find a normal vector to the graph of ff .

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Use a linear approximation to estimate 1.0221.983+1\frac { 1.02 ^ { 2 } } { 1.98 ^ { 3 } + 1 } .

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Let f(x,y,z)=(x−y)(y−z)(z−x)f ( x , y , z ) = ( x - y ) ( y - z ) ( z - x ) . Compute δfδx+δfδy+δfδz\frac { \delta f } { \delta x } + \frac { \delta f } { \delta y } + \frac { \delta f } { \delta z } .

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Optimize f(x,y)=x2+y2+2f ( x , y ) = x ^ { 2 } + y ^ { 2 } + 2 subject to xy=4x y = 4 .

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The function f(x,y)=x2+y2+3xyf ( x , y ) = x ^ { 2 } + y ^ { 2 } + 3 x y has one critical point. Determine its location and type.

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Find a normal vector to the surface with parametric equations x=2uv,y=u2−v2,z=u2+v2x = 2 u v , y = u ^ { 2 } - v ^ { 2 } , z = u ^ { 2 } + v ^ { 2 } at the point (1,0,1)( 1,0,1 ) .

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Let f(x,y,z)=x2y3−xz+exln⁡yf ( x , y , z ) = x ^ { 2 } y ^ { 3 } - \frac { x } { z } + e ^ { x } \ln y . Find fxf _ { x } , fyf _ { y } , and fzf _ { z } .

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In which direction does the directional derivative for f(x,y)=x2yf ( x , y ) = x ^ { 2 } y at (1,1)( 1,1 ) have value 1? Value −2- 2 ? Is there a direction for which the value is 4?

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Show that there does not exist any function  Show that there does not exist any function   with continuous second partial derivatives in  \mathbb { R } ^ { 2 }  such that  \frac { \delta f } { \delta x } = x ^ { 2 } y  and  \frac { \partial f } { \partial y } = x  . with continuous second partial derivatives in R2\mathbb { R } ^ { 2 } such that δfδx=x2y\frac { \delta f } { \delta x } = x ^ { 2 } y and ∂f∂y=x\frac { \partial f } { \partial y } = x .

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Evaluate lim⁡(x,y,z)→(1,0,−2)exy−z2\lim _ { ( x , y , z ) \rightarrow ( 1,0 , - 2 ) } e ^ { x y - z ^ { 2 } }

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Let z=f(x,y)z = f ( x , y ) such that ∇f(−1,2)=(3,−4)\nabla f ( - 1,2 ) = ( 3 , - 4 ) . Find an equation of the tangent line to the level curve of f(x,y)f ( x , y ) that passes through point (−1,2)( - 1,2 ) .

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Evaluate lim⁡(x,y)→(2,2)tan⁡((x−2)2+(y−2)2)(x−2)2+(y−2)2\lim _ { ( x , y ) \rightarrow ( 2,2 ) } \frac { \tan \left( ( x - 2 ) ^ { 2 } + ( y - 2 ) ^ { 2 } \right) } { ( x - 2 ) ^ { 2 } + ( y - 2 ) ^ { 2 } }

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Given f(x,y)=14x4y3f ( x , y ) = \frac { 1 } { 4 } x ^ { 4 } y ^ { 3 } , at the point (−1,2)( - 1,2 ) . Find (a) the maximum value of the directional derivative and (b) the unit vector in the direction in which the directional derivative takes on its maximum value.

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Let f(x,y,z)=zxyf ( x , y , z ) = z ^ { x y } , z>0z > 0 . Find the value of the partial derivative fy(2,1,e)f _ { y } ( 2,1 , e ) .

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