Consider the surface given as the graph of $z = f ( x , y )$ .Let $\vec { u } = \frac { 1 } { \sqrt { 26 } } ( \vec { i } + 5 \vec { j } )$ and let $\vec { v } = \frac { 1 } { \sqrt { 20 } } ( 4 \vec { i } - 2 \vec { j } )$ .Suppose that $f_{0}(1,3)=\frac{21}{\sqrt{26}} ; f_{0}(1,3)=\frac{-4}{\sqrt{20}} ; f(1,3)=-7$ .Find the equation of the tangent plane to the surface at the point $( 1,3 )$ .

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The Answer of Consider the surface given as the graph of $z = f ( x , y )$ .Let $\vec { u } = \frac { 1 } { \sqrt { 26 } } ( \vec { i } + 5 \vec { j } )$ and let $\vec { v } = \frac { 1 } { \sqrt { 20 } } ( 4 \vec { i } - 2 \vec { j } )$ .Suppose that $f_{0}(1,3)=\frac{21}{\sqrt{26}} ; f_{0}(1,3)=\frac{-4}{\sqrt{20}} ; f(1,3)=-7$ .Find the equation of the tangent plane to the surface at the point $( 1,3 )$ .

Consider the Surface Given as the Graph Of z=f(x,y)z = f ( x , y )z=f(x,y)

Consider the Surface Given as the Graph Of $z = f ( x , y )$