# Proving that a curve intersects a surface at a right angle

1. Nov 17, 2013

### fogvajarash

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

a) Show that the curve determined by:

$x=2(t^{3}+2)/3$, $y=2t^{2}$, $z=3t-2$

intersects the surface:

$x^{2}+2y^{2}+3z^{2}=15$

at a right angle at the point (2, 2, 1)

b) Verify that the curve $x^{2}-y^{2}+z^{2}=1$, $xy+xz=2$ is tangent to the surface $xyz-x^{2}-6y+6y=0$ at the point (1, 1, 1)

2. Relevant equations

3. The attempt at a solution

I don't know where to start (they have given us this assignment without explaining completely all of the concepts). I was thinking of taking the gradient of the curve and the surface in a), but I'm not sure how to proceed from there (and the curve will have 3 components in the gradient vector, and the surface will only have 2 components because we have x, y ,z). Could someone guide me in the right direction? As well, is there a general method to solve this type of exercises?

2. Nov 17, 2013

### Dick

Start by trying to get started. What's the tangent to the curve? It's probably a good idea to figure out what the value of t is at the intersection. And why do you think the gradient of the surface will have only two components? Do you know how to find the gradient of a function?

3. Nov 17, 2013

### fogvajarash

The value of t = 1.

I thought that the gradient of a function f(x,y) was found as $∇f=\displaystyle\frac{\partial f}{\partial x}i+\displaystyle\frac{\partial f}{\partial y}j$, in which i and j are the unit vectors. However when dealing with a surface (implicit), I'm not sure on how to start. I tried differentiating the expression respect to x on one side, and respect to y on the other side.

4. Nov 17, 2013

### Dick

That's in two dimensions for f(x,y). Your surface is in three dimensions f(x,y,z). There's also a $\frac{\partial f}{\partial z}$ part.

5. Nov 17, 2013

### fogvajarash

Alright so from what i get we have that $\displaystyle\frac{\partial z}{\partial x}=\displaystyle\frac{-x}{3z}$, however, how can we find $\displaystyle\frac{\partial f}{\partial z}$? Or maybe I am misunderstanding how the gradient works. How can we find it in this case at least for $\displaystyle\frac{\partial f}{\partial x}$? Is there a general method to find gradients in this case for surfaces (or implicit functions)?

6. Nov 17, 2013

### Dick

You can write your surface as f(x,y,z)=0 where $f(x,y,z)=x^{2}+2y^{2}+3z^{2}-15$. The gradient is $∇f=\displaystyle\frac{\partial f}{\partial x}i+\displaystyle\frac{\partial f}{\partial y}j+\displaystyle\frac{\partial f}{\partial z}k$. You are reading complications into this that aren't there.

7. Nov 18, 2013

### fogvajarash

That's much better, I had never thought of doing that. Thank you! I'll reply again when i get the time to work on the problem.

8. Nov 18, 2013

### fogvajarash

So now i did your method and found that the "tangent" (direction vector) was:

2i + 4j + 3k

4i + 8j + 6k

Then i argued, as these two vectors are parallel (scalar multiples of each other), we should have that the direction vector is parallel to the normal vector of the surface. However, this normal vector will be normal to the surface in question, so this means that the surface itself will be at a right angle with the curve (as it is at a right angle with the tangent line). Is my reasoning acceptable?

And on the other hand, how can we know when to apply that notation of F(x, y, z) = 0? When we have a surface or something similar? (and how in some functions of f(x, y), the gradient vector is only given in two dimensions but it is a surface as well? Or in this case, are we supposing that they are level surfaces?). Thank you Dick for your help.

9. Nov 18, 2013

### Dick

That's fine. The curve tangent is parallel to the normal direction of the surface so the curve is perpendicular to the surface. And yes, if you can write the surface in the form F(x,y,z)=0 you can apply the gradient to F to get the normal direction. If your function where x^2+2y^2-15, then it's still a surface and you can still use the gradient, but since there is no z there will be no k in the normal vector.