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)

5 Comments

Filed under differential geometry, geometry, math

5 responses to “vector calculus examples

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

  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

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