# vector calculus examples

the following hieroglyph

$\left(\begin{array}{c}V\\ W\end{array}\right)\stackrel{\phi}\to\left(\begin{array}{c} \cos V\cos W\\ \cos V\sin W\\ \sin V\end{array}\right)$

represent a mapping $\mathbb{R}^2\to\mathbb{R}^3$. It is enough to take

$-\frac{\pi}{2}< V<\frac{\pi}{2}\quad,\quad 0< W<2\pi$

to G.P.S -ing (almost) every point in $\mathbb{R}^3$ which are at distance one from the origin. That is the sphere.

Now the partial derivatives $\partial_1=\frac{\partial\phi}{\partial V}$ and $\partial_2=\frac{\partial\phi}{\partial W}$ are pictorially as:

These derivative are:

$\partial_1=\left(\begin{array}{c}-\sin V\cos W\\ -\sin V\sin W\\ \cos V\end{array}\right)$ and $\partial_2=\left(\begin{array}{c}-\cos V\sin W\\ \cos V\cos W\\ 0\end{array}\right)$

An easy calculation give that for the inner products $\langle\phi,\partial_1\rangle=\langle\phi,\partial_2\rangle=0$, so the $\partial_i$ are orthogonal to the position $\phi$. Then the product $\partial_1\times\partial_2$, which is orthogonal to the plane determined by the couple of vectors $\{\partial_1,\partial_2\}$, hence colinear to the position $\phi$. In fact, after normalization $N=\frac{\partial_1\times\partial_2}{||\partial_1\times\partial_2||}$.

A good exercise is to unfold the same program for the torus by employing the parameterization given by:

$\left(\begin{array}{c}V\\ W\end{array}\right)\stackrel{\phi}\to\left(\begin{array}{c} (2+\cos V)\cos W\\ (2+\cos V)\sin W\\ \sin V\end{array}\right)$

Filed under differential geometry, geometry, math

### 5 responses to “vector calculus examples”

1. Agreed,i got the same results that janzlev :)

• XD buena onda, you’re bearing and grasping the idea :P

2. Error in line 4, what I meant to say was: only $\partial _{1}$ is orthogonal to the position $\Phi$

3. Employing the parametrization given by:
$\left(\begin{array}{c}v\\w\end{array}\right)\overset{\Phi}{\rightarrow}\left(\begin{array}{c} (a+b\cos v)\cos w\\(a+b\cos v)\sin w\\b\sin w\end{array}\right)$
The partial derivatives:

$\partial _{1}=\left(\begin{array}{c} -b\sin v\cos w\\-b\sin v\sin w\\b\cos v\end{array}\right)$ and $\partial _{2}=\left(\begin{array}{c} -\sin w(a+b\cos v)\\-\cos w(a+b\cos v)\\{0}\end{array}\right)$

Then we calculate the inner products:
$\left\langle \Phi,\partial_{1}\right\rangle = -b a\sin v$ while for $\left\langle \Phi,\partial_{2}\right\rangle =0$
But only $\partial _{2}=0$ so only $\partial _{1}$ is orthogonal to the position $\Phi$
Then the product $\partial _{1}\times\partial _{2}$ is orthogonal to the plane determined by the couple of vectors ${\partial _{1}\partial _{2}}$

$\partial _{1}\times\partial _{2}=\left(\begin{array}{c}-b\cos v\cos w(a+b\cos v)\\-b\cos v\sin w(a+bcos v)\\-bsin v(a+b\cos v)\end{array}\right)$
but is not collinear to the position $\Phi$

$\left\|\partial _{1}\times\partial _{2}\right\|=b(a+b\cos v)$

$N=\frac{\partial _{1}\times\partial _{2}}{\left\|\partial _{1}\times\partial _{2}\right\|}=\left(\begin{array}{c}-\cos v\cos w\\-\cos v\sin w\\-\sin v\end{array}\right)$ unlike the sphere $\Phi\neq-N$

• ah!! : ) ok,ok… your are totally right, this is precisely the type of nuances we are seeking to confer, to contrast