Question

Find the equation tangent plane to the graph z = (x-4)^2 + (y-4)^2 at p=(5, 6, 5)...?

Answer 1

let \[f(x,y,z)=(x-4)^2+(y-4)^2-z\]
we have the following partial derivatives:
\[\partial f/\partial x=2x-8 \]
\[\partial f/\partial y=2y-8\]
\[\partial f/\partial z=-1\]
\[grad f(x,y,z)= (2x-8)i+(2y-8)j-k \]
\[grad f(5,6,5)=2i+4j-k\]
hence an equation of the tangent line at P(5,6,5) is given by
\[2(x-5)+4(y-6)-(z-5)=0 \implies 2x+4y-z=29\]
hope it's right :)

Answer 2

Thanks

Answer 3

Wow, that was two months ago :D .. You're welcome anyways!!