Question

The directional derivative of $f = x^{3} + 4y^{2} + z^{2}$ at the point $P(2,1,3)$ in the direction of the vector $\vec{V} = 3\hat{\imath} - 4\hat{k}$ is ............. (rounded to one decimal place).

Hint:

Directional derivative $D_{\hat{u}}f$ is the dot product of the gradient with the {unit} direction vector—always normalize the given direction first.

Answer 1

Step 1: Gradient of $f$.
$\nabla f=(\partial f/\partial x,\partial f/\partial y,\partial f/\partial z)=(3x^{2},\,8y,\,2z).$
At $P(2,1,3)$: $\nabla f=(12,\,8,\,6).$
Step 2: Unit direction vector.
$\vec{V}=(3,0,-4)$, $|\vec{V}|=\sqrt{3^{2}+0^{2}+(-4)^{2}}=5 \Rightarrow \hat{u}=\left(\frac{3}{5},0,-\frac{4}{5}\right).$
Step 3: Directional derivative.
$D_{\hat{u}}f=\nabla f\cdot\hat{u}=12\cdot\frac{3}{5}+8\cdot 0+6\cdot\left(-\frac{4}{5}\right)=\frac{36}{5}-\frac{24}{5}=\frac{12}{5}=2.4.$