Partial Derivatives Problem

**newman611** · November 27th 2010, 05:49 PM

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!

**wombat** · November 27th 2010, 06:12 PM

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?

**newman611** · November 27th 2010, 06:29 PM

Thanks, wombat!

**wombat** · November 27th 2010, 06:33 PM

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?

**newman611** · November 27th 2010, 06:36 PM

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

**Mondreus** · November 27th 2010, 06:40 PM

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 $(a,b,c)=(1,1,1)$

**HallsofIvy** · November 28th 2010, 03:31 AM

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!

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