Question

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

Answer 1

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Answer 2

All rright, here we go!

let
f(x,y,z)=(x−4)2+(y−4)2−z

we have the following partial derivatives:
∂f/∂x=2x−8

∂f/∂y=2y−8

∂f/∂z=−1

gradf(x,y,z)=(2x−8)i+(2y−8)j−k

gradf(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⟹2x+4y−z=29

Answer 3

dw= dz + dx + dy

Answer 4

\[\partial{z}/\partial{x}|_5 = 2(x - 4)|_5 = 2\]
\[\partial{z}/\partial{y}|_6 = 2(y - 4)|_6 = 4\]
Hence the tangent plane has gradient (2, 4).
Its equation is in the form \[z = 2x + 4y + p\]
And it passes through (5, 6, 5), so
\[5 = 2*5 + 4*6 + p\]
\[p = -29\]
Result: \[z = 2x + 4y - 28\]
How about that...