How do I show that the direction of second differential vector is normal to the surface?

Printable View

- Jul 8th 2013, 03:32 PMshyamalshukladirection 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 PMHallsofIvyRe: 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, $\displaystyle \vec{r}= cos(\theta)sin(\phi)\vec{i}+ sin(\theta)sin(\phi)\vec{j}+ cos(\phi)\vec{k}$ then $\displaystyle \vec{r}_\theta= -sin(\theta)sin(\phi)\vec{i}+ cos(\theta)sin(\phi)\vec{j}$ and then the**mixed**derivative is $\displaystyle -sin(\theta)cos(\theta)\vec{i}+ cos(\theta)cos(\phi)\vec{j}$ and the dot product of that with $\displaystyle \vec{r}$ is $\displaystyle (cos(\theta)sin(\phi))(-sin(\theta)cos(\theta)+ (sin(\theta)sin(\phi))(cos(\theta)cos(\phi)= 0$ so the mixed second derivative is perpendicular to the surface.

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