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

• Oct 19th 2010, 04:52 PM
downthesun01
Find the equation for the tangent plane to the surface. Answer doesn't seem right.
$z=ln(3x^{2}+6y^{2}+1)$, $P(0,0,0)$

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.
• Oct 19th 2010, 05:11 PM
Plato
Quote:

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 $F_z?$
• Oct 19th 2010, 05:37 PM
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.
• Oct 20th 2010, 12:28 AM
earboth
Quote:

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?
• Oct 20th 2010, 05:51 AM
downthesun01
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.