Find the equation for the tangent plane to the surface. Answer doesn't seem right.

**downthesun01** · October 19th 2010, 03:52 PM

$z=ln(3x^{2}+6y^{2}+1)$ , $P(0,0,0)$

Ok, so I found $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$

$\frac{\partial f}{\partial x}=\frac{6x}{3x^{2}+6y^{2}+1}$

$\frac{\partial f}{\partial y}=\frac{12y}{3x^{2}+6y^{2}+1}$

So then,

$f_{x}(0,0)=\frac{0}{1}=0$

$f_{y}(0,0)=\frac{0}{1}=0$

And I have a normal vector $n=(0,0,-1)$ at point $(0,0,0)$

So then,

$0(x-0)+0(y-0)-(z-0)=0$

$=>z=0$

Did I do this correctly? Thanks for looking.

**Plato** · October 19th 2010, 04:11 PM

Originally Posted by downthesun01

$z=ln(3x^{2}+6y^{2}+1)$ , $P(0,0,0)$

You should have looked at $F(x,y,z)=\ln(3x^{2}+6y^{2}+1)-z$ .

Where is your $F_z?$

**downthesun01** · October 19th 2010, 04:37 PM

$f_{z}=-1$

$f_{z}(0,0,0)=-1$

$n=0i+0j-k$ at point $(0,0,0)$

$0(x-0)+0(y-0)-(z-0)=0$

$z=0$

Still getting z=0

And it still seems like an odd answer for some reason.

**earboth** · October 19th 2010, 11:28 PM

Originally Posted by downthesun01

$f_{z}=-1$

$f_{z}(0,0,0)=-1$

$n=0i+0j-k$ at point $(0,0,0)$

$0(x-0)+0(y-0)-(z-0)=0$

$z=0$

Still getting z=0

And it still seems like an odd answer for some reason.

1. Which reasons?

2. I've attached the graph of z and the tangent plane. And it looks as if your calcualtions are OK. So again: which reasons?

**downthesun01** · October 20th 2010, 04:51 AM

Honestly, z=0 just seems kind of boring. I know, not a very good reason. Thanks for the graph though. It makes it really easy to see that the answer is correct. Much appreciated.

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Find the equation for the tangent plane to the surface. Answer doesn't seem right.

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