# Homework Help: Eq. of tangent plane at a point.

1. Oct 29, 2015

### catch22

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
what is the difference between the form

z - zo = fx(x-xo) + fy(y-yo)

and:

Fx(xo,yo,zo) + Fy(xo,yo,zo) + Fz(xo,yo,zo)=0
____________________________________________________________________________

using the first eq :

while the latter gives:

Am I doing something wrong? shouldn't they give the same answer?

2. Oct 30, 2015

### Student100

You're doing something wrong, when you solve for $z-z_0$, what happens to the term in front of $(z-z_0)$, and what about your signs?

3. Oct 30, 2015

### catch22

are you talking about the first equation z - zo = fx(x-xo) + fy(y-yo)? I thought there aren't suppose to be any terms in front of (z−z0)

actually, in order to get z - zo = fx(x-xo) + fy(y-yo), didn't we have to divide everything by z so we had something like A/C = Fx and B/C = Fy

4. Oct 30, 2015

### catch22

I think I found the problem.

if I use gradient, the Fz has to equate to -1, right?

5. Oct 30, 2015

### catch22

is this correct?

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6. Oct 30, 2015

### HallsofIvy

The difference between "z= f(x, y)" and "F(x, y, z)= 0" is that you are taking F(x, y, z)= f(x, y)- z. Taking the gradient $\nabla F= <f_x, f_y, -1>[$.

It is correct that $F(1, -1, 4\sqrt{2})= \frac{1}{4}+ \frac{1}{4}+ \frac{32}{64}= 1$ but that does NOT say "z= 1". You are told that at this point $z= 4\sqrt{2}$. The fact that $F(1, -1, 4\sqrt{2})= 1$ just verifies that the given point is on the given surface.

The tangent plane is $z- z_0= f_x(x_0, y_0)(x- x_0)+ f_y(x_0, y_0)(y- y_0)$ only applies if you are given z= f(x, y) which is NOT the case here. You are given, rather, $F(x, y, z)= x^2/4+ y^2/4+ z^2/64= 1$. In order to convert that to the "z= f(x,y)" form you need to solve for z: $z^2/64= 1- x^2/4- y^2/4$, $z/8= \sqrt{1- x^2/4- y^2/4}$, $z= 8\sqrt{1- x^2/4- y^2/4}$.

Much simpler is to use the fact that if F(x, y, z)= 1 (or any constant) then $\nabla F$ is perpendicular to the surface at any point. Since for any point (x, y, z) on the tangent plane to the surface at $(x_0, y_0, z_0)$, the vector $(x- x_0)\vec{i}+ (y- y_0)\vec{j}+ (z- z_0)\vec{k}$ is in the tangent plane, we must have $\nabla F\cdot [(x- x_0)\vec{i}+ (y- y_0)\vec{j}+ (z- z_0)\vec{k}]= 0$ which is the same as $F_x(x_0, y_0, z_0)(x- x_0)+ F_y(x_0, y_0, z_0)(y- y_0)+ F_z(x_0, y_0, z_0)(z- z_0)= 0$ as the equation of the tangent plane.

7. Oct 30, 2015

### catch22

looks correct?

8. Oct 30, 2015

### HallsofIvy

What are $x_0$, $y_0$ and $z_0$?

9. Oct 30, 2015

### catch22

I used r⋅n = n ⋅ (1,-1,4sqrt(2))

r being (x,y,z)

10. Oct 30, 2015

### HallsofIvy

So you do not know what $x_0$, $y_0$, and $z_0$ are? Then you need to talk to your teacher.

11. Oct 30, 2015

### catch22

they are the point given, (1,-1,4sqrt(2))

12. Nov 1, 2015

### HallsofIvy

Yes. I asked that in post #8 and you did not answer. Now, my question is "why do you not have (x- 1), (y+ 1), and $(z- 4\sqrt{2})$ in your answer"?

13. Nov 3, 2015

### catch22

that was what I initially had