# Normal vector of curved surface

1. Mar 15, 2015

### Flotensia

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

2. Relevant equations

3. The attempt at a solution
I can understand it intuitively, but can't prove mathematically...Can you help me??

2. Mar 15, 2015

### Ray Vickson

PF rules forbid us from offering help until you have demonstrated an effort and shown your work.

3. Mar 15, 2015

### Flotensia

I'm so sorry. It was my first time to write in this forum. I know that cross product makes perpendicular vectors. But in this problem, I don't understand how we explain three dimension by using two parameter, u and v. I searched in internet and thought it is related to gradient. Is it right?

4. Mar 15, 2015

### Brian T

If you use 3 parameters to parametrize a region of space in ℝ3, what you get is a region covering some volume.
However, if you only use 2 parameters to parametrize a region of space in ℝ3, you get a "2D" curved surface located in that 3D space.

5. Mar 15, 2015

### Flotensia

Ahh, I got what 2 parameters mean. Thanks. then could you help me more to solve that problem??

6. Mar 15, 2015

### Dick

You know that the partial derivatives are tangent vectors, right? And you know that the cross product is orthogonal to the two vectors, I hope. And dot products are related to cosines of the included angle and cross products are related to sine? Put all of those ingredients together.

7. Mar 15, 2015

### Flotensia

I can explain that in word and can image in mind, but i can't in numerical expression... That's my problem...

8. Mar 15, 2015

### Dick

Look at the denominator in terms of the included angle $\theta$. Use $1-\cos^2(\theta)=\sin^2(\theta)$. Now does that help?

9. Mar 15, 2015

### Flotensia

I was so silly. It helps me a lot. Thanks for your help!!

10. Mar 15, 2015

### Dick

A nudge is a good as a wink. And you are welcome!