Calculating Normal Vector Derivative in Differential Geometry Using Curve Basis

Click For Summary
The discussion focuses on calculating the matrix for the derivative of the normal vector, dN, in the context of differential geometry using the curve a(u,v) = (u,v,uv). The user successfully derived the partial derivatives a_u and a_v, leading to the normal vector calculation. However, confusion arises regarding the transition from the normal vector N to its derivative dN, as the textbook lacks clarity on this process. The user considers using the Jacobian matrix for the derivative but is uncertain if this approach is correct. Clarification is sought on whether to factor out components when applying the Jacobian for the normal vector's derivative.
metzky
Messages
2
Reaction score
0

Homework Statement


Using the curve \vec{a}(u,v)= (u,v,uv) for all (u,v) ε R^2

Find the matrix for d\vec{N} in the basis of {\vec{a}_{u},\vec{a}_{v}}

Homework Equations


Well first off i found the partial derivatives
\vec{a}_{u} which is 1,0,v, while \vec{a}_{v} is 0,1,u
Then using those i found the normal vector which i calculated as 1/\sqrt{v^{2}+u^{2}+1} (-v,-u,1)

The Attempt at a Solution

Now this is where i get lost. Our book does not explain this very well at all. It just shows going fron N to dN with no explanation. I tried using the jacobian matrix to calculate the derivative but I'm not sure if this is the right approach. Most of the examples don't have a matrix from so i Know I'm doing something wrong.

The problem is a set from
kbw0RwT.png
for reference. I need dN to move on to find the second fundamental forms and so forth
 
Physics news on Phys.org
Actually could i bring out the 1/sq u^2... out of the jacobian matrix then use the matrix to find the vector or no? so only the -v,-u,1 would be getting partially derived in the matrix?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
7
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K