Derivative with respect to the normal vector and with respect to the radius

Question

Suppose a function $f$ is defined on the boundary of a ball of radius $r$ in $ \mathbb{R}^{k} $ ($k\geq2$), and let $n_{i}$ be the inward normal vector. What is the relation between derivatives with respect to $n_{i}$ and with respect to $r$? Is it true to say that: $$\dfrac{\partial f}{\partial n_{i}} =- \dfrac{\partial f}{\partial r}?$$

Answer 1

Assume that $D$ is the ball centered at the origin and of radius $r_0$. Then for every $x\in\partial D$ you have that the inward normal is given by $n_i=-x/r_0$ and so the normal derivative is given by $$\frac{\partial f}{\partial n_i}(x)=\nabla f(x)\cdot n_i=-\nabla f(x)\cdot \frac{x}{r_0}.$$ Now assume that $k=2$ for simplicity and consider the function $f$ in polar coordinates, that is $$g(r,\theta):=f(r \cos\theta,r \sin\theta).$$ Then by the chain rule \begin{align}\frac{\partial g}{\partial r}(r,\theta)&=\frac{\partial f}{\partial x_1}(r \cos\theta,r \sin\theta)\cos\theta+\frac{\partial f}{\partial x_2}(r \cos\theta,r \sin\theta)\sin\theta\\ &=\nabla f(r \cos\theta,r \sin\theta)\cdot\frac1r(r \cos\theta,r \sin\theta)\end{align} and so at the point $x=(r_0 \cos\theta,r_0 \sin\theta)$ you do have $$\frac{\partial g}{\partial r}(r_0,\theta)=-\frac{\partial f}{\partial n_i}(r_0 \cos\theta,r_0 \sin\theta).$$ In higher dimension the proof is probably similar, just messier.