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Math Help - direction of second differential vector

  1. #1
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    direction of second differential vector

    How do I show that the direction of second differential vector is normal to the surface?
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  2. #2
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    Re: direction of second differential vector

    The position vector is a function of two variables. The "second differential" with respect to which variable? If, for example, we take, as parametric equations for the unit sphere, \vec{r}= cos(\theta)sin(\phi)\vec{i}+ sin(\theta)sin(\phi)\vec{j}+ cos(\phi)\vec{k} then \vec{r}_\theta= -sin(\theta)sin(\phi)\vec{i}+ cos(\theta)sin(\phi)\vec{j} and then the mixed derivative is -sin(\theta)cos(\theta)\vec{i}+ cos(\theta)cos(\phi)\vec{j} and the dot product of that with \vec{r} is (cos(\theta)sin(\phi))(-sin(\theta)cos(\theta)+ (sin(\theta)sin(\phi))(cos(\theta)cos(\phi)= 0 so the mixed second derivative is perpendicular to the surface.

    But the second derivative with respect to \theta both times is -cos(\theta)sin(\phi)\vec{i}- sin(\theta)sin(\phi)\vec{j} and that is NOT perpendicular to the surface.
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