Equation of tangent plane at (2, -1, ln 7): z = ln 7 + (4/7)(x-2) - (6/7)(y+1)

[JFonseka]

Just when I thought I got the hang of tangent planes and surfaces there comes a question I haven't quite seen before

z = ln (x$$^{2}$$+3y$$^{2}$$)

Find a normal vector n and the equation of the tangent plane to the surface at the point
(2, -1, ln 7)

So keeping the cartesian equation in mind:

z = z$$_{0}$$ + F$$_{x}$$(x,y)(x - x$$_{0}$$) + F$$_{y}$$(x,y)(y - y$$_{0}$$)

Partial derivative with respect to x: $$\frac{2x}{x^{2} + 3y^{2}}$$
Evaluate with the values and I get 4/11

Partial derivate with respect to y: $$\frac{6y}{x^{2} + 3y^{2}}$$
Evaluate with the values and I get -6/11

Have I done this correctly?

So putting it all in the equation I get z = ln 7 + 4x/11 -6y/11 -14/11

I get the feeling I did something wrong somewhere, and of course the normal vector would be (4/11, -6/11, -1) if everything was right.

 

[ice109]

just take the gradient of $ln (x ^{2} +3y ^{2} ) - z$ and evaluate it at the point

 

[JFonseka]

Do you mean dz/dx or dz/dy or something else?

 

[ice109]

Do you mean dz/dx or dz/dy or something else?

have you not learned about gradient? $\nabla=<\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}>$ = the defining vector of the tangent plane.

 

[JFonseka]

Well that's what I did earlier lol.

(4/11, -6/11, -1)

Thanks

 

[HallsofIvy]

Then do the arithmetic again. x²+ 3y²= 4²+ 3(-1)² is NOT 11!

 

[JFonseka]

It should be 2^2 + 3(-1)^2 which is 7...hehe, now where did that rogue 11 come from, lol.