Directional Derivative

The directional derivative is a Vector gives the rate of change of a function at a given point.

Definition §

Let $\hat{\mathbf{u}}$ be a unit vector in the x-y plane. The rate of change of $f$ at a point $p=(x_0,y_0)$ in the direction $\hat{\mathbf{u}}$ is the directional derivative, written as any of the following:

$${\nabla }_{\hat{\mathbf{u}}}f(p)=D_{\hat{\mathbf{u}}}f(p)=\hat{\mathbf{u}}\cdot \nabla f(p)$$

This gives the gradient of the surface of $f$ at the point $P$, in the direction $\hat{\mathbf{u}}$

Info

The del operator (or nabla symbol) $\nabla$, in differential calculus, gives the Gradient Vector of a function: $\nabla f=(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y})$

Vector Properties §

Since the directional derivative is essentially the dot product of two vectors, we can rewrite it as:

$\nabla f\cdot \hat{\mathbf{u}}=|\nabla f||\hat{\mathbf{u}}|\cos (\theta )$ $=|\nabla f|\cos (\theta )$ (Since the magnitude of a unit vector is 1)

Hence, the gradient, $|\nabla f|$ is a maximum when $\cos (\theta )=1$ , i.e. $\theta =0$. Similarly, it is the lowest when $\cos (\theta )=-1$, i.e. $\theta =\pi$.

This gives a geometrical interpretation of $\nabla f:$

• the direction of $\nabla f$ is the direction of steepest ascent of $f$ . • the length of $\nabla f,||\nabla f||$, is the slope of the surface in the direction of steepest ascent. • the direction of $-\nabla f$ is the direction of steepest descent of f . • $\nabla f$ is perpendicular to the level curves of f .

Examples §

1: Simple directional derivative

Find the directional derivative of $f(x,y)=x{e}^y$ at $(2,0)$ in the direction from $(2,0)$ towards $\left(\frac{1}{2},2\right)$

First, we need to find the directional unit vector, $\hat{\mathbf{u}}$. $\mathbf{u}=\left(\frac{1}{2},2\right)-(2,0)=(-\frac{3}{2},2)$ $|\mathbf{u}|=\sqrt{{\left(\frac{3}{2}\right)}^2+2^2}=\sqrt{\frac{25}{4}}=\frac{5}{2}$ $\hat{\mathbf{u}}=\frac{1}{5/2}(-\frac{3}{2}\mathbf{i}+2\mathbf{j})$ $=\frac{2}{5}\left(-\frac{3}{2}\mathbf{i}+2\mathbf{j}\right)=-\frac{3}{5}\mathbf{i}+\frac{4}{5}\mathbf{j}$

Now we need to find the gradient of f, $\nabla f$ at $(2,0)$: $\frac{\partial f}{\partial x}=e^y$ $\frac{\partial f}{\partial y}=x{e}^y$ $\nabla f(2,0)=(\frac{\partial f}{\partial x}|(2,0),\frac{\partial f}{\partial y}(2,0))$ $\frac{\partial f}{\partial x}|(2,0)=e^0=1$ $\frac{\partial f}{\partial y}|(2,0)=2e^0=2$ $\nabla f(2,0)=(1,2)$

Therefore, the directional derivative is: $(1,2)\cdot (-\frac{3}{5},\frac{4}{5})$ $=-\frac{3}{5}+\frac{8}{5}$ $=1$