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

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

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

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

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

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

So then,

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

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

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

So then,

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

$\displaystyle =>z=0$

Did I do this correctly? Thanks for looking.

2. Originally Posted by downthesun01
$\displaystyle z=ln(3x^{2}+6y^{2}+1)$, $\displaystyle P(0,0,0)$
You should have looked at $\displaystyle F(x,y,z)=\ln(3x^{2}+6y^{2}+1)-z$.

Where is your $\displaystyle F_z?$

3. $\displaystyle f_{z}=-1$

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

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

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

$\displaystyle z=0$

Still getting z=0

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

4. Originally Posted by downthesun01
$\displaystyle f_{z}=-1$

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

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

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

$\displaystyle z=0$

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?

5. 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.