Calculating Normal Vector Derivative in Differential Geometry Using Curve Basis

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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
 
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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?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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