1. ## Directional Derivatives

We're doing gradient vectors, and I'm kind of confused by the applications of them.

Here's a question where I'm not entirely sure where to start:

Find the points on the ellipsoid $\displaystyle x^2+2y^2+3z^2 = 1$ where the tangent plane is parallel to the plane $\displaystyle 3x-y+3z=1$

First, I know that the normal plane to the ellipsoid at these points is going to be <3,-1,3>. I know that to find the gradient vector I take the partials with respect to x and y, but then I'm not sure where to go from there. I don't have points I'm working with, I'm being asked to find points. The whole directional derivative thing is actually very confusing in general--if someone could help me out that would be great!

2. If you let $\displaystyle f(x,y,z)=x^2+2y^2+3z^2$ then $\displaystyle \bigtriangledown f$ will be normal to the level surface $\displaystyle f=1$. Calculate $\displaystyle \bigtriangledown f$ and find where it's parallel to the normal of the given plane.

3. Originally Posted by ojones
If you let $\displaystyle f(x,y,z)=x^2+2y^2+3z^2$ then $\displaystyle \bigtriangledown f$ will be normal to the level surface $\displaystyle f=1$. Calculate $\displaystyle \bigtriangledown f$ and find where it's parallel to the normal of the given plane.
Ok, I found fx = 2x, fy= 4y, and fz= 6z when I took the partials, so $\displaystyle \bigtriangledown f = <2x,4y,6z>$

I'm still confused about how to find where it's parallel to the normal though. I get that the normal vector at both points will be $\displaystyle <3,-1,3>$, I just can't figure out how to approach solving it.

4. OK, you've calculated $\displaystyle \bigtriangledown f$ correctly.

Now, $\displaystyle (3,-1,3)$ is a normal to the given plane (not sure why you're referring to points here). You want to find points $\displaystyle (x,y,z)$ on the surface such that $\displaystyle \bigtriangledown f(x,y,z)=\lambda (3,-1,3)$ for some scalar $\displaystyle \lambda$. Substitute in what you know.

5. Originally Posted by ojones
OK, you've calculated $\displaystyle \bigtriangledown f$ correctly.

Now, $\displaystyle (3,-1,3)$ is a normal to the given plane (not sure why you're referring to points here). You want to find points $\displaystyle (x,y,z)$ on the surface such that $\displaystyle \bigtriangledown f(x,y,z)=\lambda (3,-1,3)$ for some scalar $\displaystyle \lambda$. Substitute in what you know.
Okay, I did that (using n for lambda here) and got that $\displaystyle x=3n/2, y= -n/2, z= n/2$ and by plugging in I got that $\displaystyle n=+-\sqrt{{2/7}}$. Then I calculated the points to be $\displaystyle (3\sqrt{2/7},-\sqrt{2/7},3\sqrt{2/7})$ and $\displaystyle (-3\sqrt{2/7},\sqrt{2/7},-3\sqrt{2/7})$. Is that correct?

Thank you for helping me. If you don't mind me asking, could you explain the logic behind the method used to solve this problem? I'm trying to fully understand it, but I can't figure out how it came down to $\displaystyle \bigtriangledown f(x,y,z)=\lambda (3,-1,3)$ in order to find the solution.

6. Your $\displaystyle y$ value looks incorrect but you have the right method.

The logic behind this is that $\displaystyle \bigtriangledown f(x_0,y_0,z_0)$ is normal to the surface at the point $\displaystyle (x_0,y_0,z_0)$ (well known property of $\displaystyle \bigtriangledown$). Hence it's normal to the tangent plane to the surface at that point. We're simply looking for points where this normal is parallel to the normal of the given plane. Two vectors are parallel if one is a scalar multiple of the other.