# direction of second differential vector

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• Jul 8th 2013, 03:32 PM
shyamalshukla
direction of second differential vector
How do I show that the direction of second differential vector is normal to the surface?
• Jul 8th 2013, 03:57 PM
HallsofIvy
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.