- #1
Flotensia
- 15
- 0
Homework Statement
Homework Equations
The Attempt at a Solution
I can understand it intuitively, but can't prove mathematically...Can you help me??
In summary, the student is asking for help with a homework problem, but is not able to express the problem in numerical terms. The tutor offers help, and the student is grateful.
I can understand it intuitively, but can't prove mathematically...Can you help me??
- #2
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
- 10,704
- 1,723
Flotensia said:Homework Statement
Homework Equations
The Attempt at a Solution
I can understand it intuitively, but can't prove mathematically...Can you help me??
PF rules forbid us from offering help until you have demonstrated an effort and shown your work.
- #3
Flotensia
- 15
- 0
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
Brian T
- 130
- 31
Flotensia said:
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
Flotensia
- 15
- 0
Ahh, I got what 2 parameters mean. Thanks. then could you help me more to solve that problem??
- #6
Dick
Science Advisor
Homework Helper
- 26,258
- 623
Flotensia said:
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
Flotensia
- 15
- 0
I can explain that in word and can image in mind, but i can't in numerical expression... That's my problem...
- #8
Dick
Science Advisor
Homework Helper
- 26,258
- 623
Flotensia said:
Look at the denominator in terms of the included angle ##\theta##. Use ##1-\cos^2(\theta)=\sin^2(\theta)##. Now does that help?
- #9
Flotensia
- 15
- 0
I was so silly. It helps me a lot. Thanks for your help!
- #10
Dick
Science Advisor
Homework Helper
- 26,258
- 623
Flotensia said:
A nudge is a good as a wink. And you are welcome!
FAQ: Normal vector of curved surface
What is a normal vector of a curved surface?
A normal vector is a vector that is perpendicular to a curved surface at a specific point. It is used to determine the direction of the surface at that point.
How is a normal vector calculated for a curved surface?
A normal vector can be calculated by taking the cross product of two tangent vectors on the surface. Tangent vectors are vectors that lie on the surface and are parallel to the direction of the surface at a specific point.
Why is the normal vector important for curved surfaces?
The normal vector is important because it helps us understand the orientation and direction of a curved surface. It is also used in many mathematical and physical applications, such as calculating surface area and determining the direction of forces on a curved surface.
How does the normal vector change along a curved surface?
The normal vector changes along a curved surface based on the shape and curvature of the surface. At different points on the surface, the normal vector will have different directions and magnitudes.
Can the normal vector be negative for a curved surface?
Yes, the normal vector can be negative for a curved surface. This means that the vector is pointing in the opposite direction of the positive normal vector, which is typically defined as pointing away from the surface.