# Math Help - Partial Derivatives Problem

1. ## Partial Derivatives Problem

Hi all,

This is the following homework problem I am having trouble with:

Consider the surface: 25x^2+25y^2+1z^2=51

P=(1,1,1) on this surface.

So far, I have gotten:

x=1+50t
y=1+50t
z=1+2t

The equation to the plane P must be written as z=, so I wrote:

z-1=50(x-1)+50(y-1)
z=50(x-1)+50(y-1)+1

Whenever I type this into the online homework, it displays "incorrect". Any suggestions?

Thank you in advance!

2. I'm probably not going to be able to help you with this, and I don't know if it's just me, but I can't see either images you attempted to put in your post. Thought i'd give you the heads up.
edit: I can see it now. I'll think about it for a little while to see if i can be of assistance but odds are I won't do much good :/
edit2: Alright, thought about it. Not sure what the question is asking? It just says consider the surface and the point on the surface... what do you need to find?

3. Thanks, wombat!

4. Haha no problem but i'm not really sure what you're thanking me for. Is what you typed in your original post the entire question or is there more?

5. No, this is it...I said thanks because you didn't have to mention anything, lol.

6. What exactly is the question? If you're looking for the tangent plane at (1,1,1), then you can use the formula $f'_x(a,b,c)(x-a)+f'_y(a,b,c)(y-b)+f'_z(a,b,c)(z-c)=0$ where $f(x,y,z)=25x^2+25y^2+1z^2$ and $(a,b,c)=(1,1,1)$

7. Originally Posted by newman611
Hi all,

This is the following homework problem I am having trouble with:

Consider the surface: 25x^2+25y^2+1z^2=51

P=(1,1,1) on this surface.

So far, I have gotten:

x=1+50t
y=1+50t
z=1+2t
For what? These are parametric equations of a line- the line perpendicular to the surface at (1, 1, 1). They do tell you that a vector perpendicular to the surface, and so perpendicular to the tangent plane, is $50\vec{i}+ 50\vec{j}+ 2\vec{z}$ but you don't need the equation of the line to determine that.

The equation to the plane P must be written as z=, so I wrote:

z-1=50(x-1)+50(y-1)
z=50(x-1)+50(y-1)+1
Well, you don't say how you got that from what you did before so I can't make heads or tails of it. The equation of a plane through $(x_0, y_0, z_0)$ with normal vector $A\vec{i}+ B\vec{j}+ C\vec{k}$ is $A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0$.
Here A= B= 50, C= 2, and $x_0= y_0= z_0= 1$ so the equation is 50(x- 1)+ 50(y- 1)+ 2(z- 1)= 0. Solve that for z.

Whenever I type this into the online homework, it displays "incorrect". Any suggestions?

Thank you in advance!