Directional Derivative

The Directional Derivative

The directional derivative measures the rate of change of $f$ in a specific direction $\mathbf{u}$:

$D_{\mathbf{u}} f(x, y) = \nabla f \cdot \hat{\mathbf{u}}$

Where $\hat{\mathbf{u}} = (u_x, u_y)$ is a unit vector ($|\hat{\mathbf{u}}| = 1$).

This is the dot product of the gradient with the direction vector.

Expanding the Formula

$D_{\mathbf{u}} f = \frac{\partial f}{\partial x} u_x + \frac{\partial f}{\partial y} u_y$

Why Unit Vectors?

Using a unit vector normalizes the measurement. If we doubled the direction vector, we'd get twice the rate — but we want a pure measure of slope in that direction, independent of how long the direction vector is.

Special Cases

• $\mathbf{u} = (1, 0)$: directional derivative $= \partial f / \partial x$
• $\mathbf{u} = (0, 1)$: directional derivative $= \partial f / \partial y$
• $\mathbf{u} = \nabla f / |\nabla f|$: maximum directional derivative $= |\nabla f|$
• $\mathbf{u} = -\nabla f / |\nabla f|$: minimum (most negative) directional derivative $= -|\nabla f|$

Example

For $f(x,y) = x^2 + y^2$ at $(1, 1)$, direction $\mathbf{u} = (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$:

• $\nabla f = (2, 2)$
• $D_{\mathbf{u}} f = 2 \cdot \frac{1}{\sqrt{2}} + 2 \cdot \frac{1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \approx 2.8284$

Your Task

Implement double directional_deriv(double (*f)(double, double), double x, double y, double ux, double uy, double h) that computes the directional derivative using central differences and the dot product formula.

Note: assume $(u_x, u_y)$ is already a unit vector.