13 - Directional Derivatives

Directional Derivatives

Let $f:{\mathbb{R}}^n\to \mathbb{R}$ be a differentiable scalar function
Let $\mathbf{u}$ be a unit vector then

The directional derivatives of $f$ at $\mathbf{a}$ in the direction $\mathbf{u}$ is

$$\underset{t\to 0}{\lim }\left(\frac{f(\mathbf{a}+t\mathbf{u})-f(\mathbf{a})}{t}\right)$$

Same as the rate of change of $f$ at $\mathbf{a}$ in the direction of $\mathbf{u}$

Relation between Directional Derivatives and Gradient Vectors

The directional derivative of a function $f$ at point of $\mathbf{a}$ in the direction of $\mathbf{u}$ is

$$\nabla f(\mathbf{a})\cdot \mathbf{u}$$

Proof

Let

$$F(t)=f(\mathbf{a}+t\mathbf{u})=f(a_1+t{u}_1,\cdots ,a_n+t{u}_n)$$

Then

$$\underset{t\to 0}{\lim }\left(\frac{f(\mathbf{a}+t\mathbf{u})-f(\mathbf{a})}{t}\right)=\underset{t\to 0}{\lim }\frac{F(t)-F(0)}{t}=F^{\prime }(0)$$

By Chain Rule

$$\begin{array}{rl}{F}^{\prime }(0)& =\frac{dF}{dt}{\mid }_{t=0}\\ & =\frac{\partial f}{\partial x_1}(\mathbf{a})\frac{d{x}_1}{dt}+\cdots +\frac{\partial f}{\partial x_n}(\mathbf{a})\frac{d{x}_n}{dt}\\ & =\frac{\partial f}{\partial x_1}(\mathbf{a})u_1+\cdots +\frac{\partial f}{\partial x_n}(\mathbf{a})u_n\\ & =\nabla f(\mathbf{a})\cdot \mathbf{u}\end{array}$$

Maximising the rate of change corollary

The rate of change of $f$ is greatest in the direction of $\nabla f$

Hence when $u=\frac{\nabla f}{|\nabla f|}$ then the (maximum) rate of change is

$$|\nabla f|$$

Gradient Vector is normal to the Tangent Plane

Given a surface $S\subseteq {\mathbb{R}}^3$ with equation

$$f(x,y,z)=c$$

and point

$$\mathbf{p}\in S$$

Then

$$\nabla f(\mathbf{p})\text{ is normal to }S\text{ at }\mathbf{p}$$

Proof

Let $u,v$ be coordinates near $\mathbf{p}$
Let $\mathbf{r}:(u,v)\to \mathbf{r}(u,v)$ be a parametrisation of part of $S$

The normal to $S$ at $\mathbf{p}$ is in the direction of

$$\frac{\partial \mathbf{r}}{\partial u}\wedge \frac{\partial \mathbf{r}}{\partial v}$$

Noting that $f(\mathbf{r}(u,v))=c$ and so $\frac{\partial f}{\partial u}=\frac{\partial f}{\partial v}=0$
Writing $\mathbf{r}(u,v)=(x(u,v),y(u,v),z(u,v))$ then

$$\begin{array}{rl}\nabla f\cdot \frac{\partial \mathbf{r}}{\partial u}& =\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right)\cdot \left(\frac{\partial x}{\partial u},\frac{\partial y}{\partial u},\frac{\partial z}{\partial u}\right)\\ & =\frac{\partial f}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial u}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial u}\\ & =\frac{\partial f}{\partial u}\text{ by Chain Rule}\\ & =0\end{array}$$

Similarly

$$\nabla f\cdot \frac{\partial \mathbf{r}}{\partial v}=0$$

Hence $\nabla f$ is in the direction of $\frac{\partial \mathbf{r}}{\partial u}\wedge \frac{\partial \mathbf{r}}{\partial v}$, hence normal to surface $S$