Question

The directional derivative of $f(x,y,z)=4x^{2}+2y^{2}+z^{2}$ at the point (1,1,1) in the direction of the vector $\vec{v}=\hat{i}-\hat{k}$ is \(\underline{\hspace{2cm}}\) (rounded off to two decimal places).

Hint:

Directional derivatives require the gradient dotted with a unit direction vector. Always normalize the direction vector first.

Answer 1

Correct Answer: 4.2 - 4.3

The gradient of the function is:
$\nabla f = (8x) \hat{i} + (4y) \hat{j} + (2z) \hat{k}$
At point (1,1,1):
$\nabla f = 8\hat{i} + 4\hat{j} + 2\hat{k}$
Given direction vector: $\vec{v} = \hat{i} - \hat{k}$
Normalize it:
$|\vec{v}| = \sqrt{1^{2} + 1^{2}} = \sqrt{2}$
Unit vector: $\hat{u} = \dfrac{1}{\sqrt{2}}(\hat{i} - \hat{k})$
Directional derivative:
$D_{\hat{u}} f = \nabla f \cdot \hat{u}$
Compute dot product:
$= (8\hat{i} + 4\hat{j} + 2\hat{k}) \cdot \dfrac{1}{\sqrt{2}}(\hat{i} - \hat{k})$
$= \dfrac{1}{\sqrt{2}} (8 - 2)$
$= \dfrac{6}{\sqrt{2}} = 3\sqrt{2} = 4.2426$
Rounded to two decimals: $4.24$