Question

calculate the directional derivative of scalar field f(x,y,z)=(x ²+yz+z²)at (1,2,1) along n^=1/3(2i-j+k^)

Answer 1

The directional derivative in the direction of $\mathbf{n} = \dfrac23\mathbf{i}-\dfrac13\mathbf{j} + \dfrac13\mathbf{k}$ is given as follows:

$$\nabla_{\mathbf{n}}f = \nabla f\cdot \dfrac{\mathbf{n}}{|\mathbf{n}|}$$

where $\nabla f$ is the gradient of $f$. Thus, obtain:

$$\nabla f = 2x\mathbf{i} + z\mathbf{j}+(y+2z)\mathbf{k}\\ |\mathbf{n}| = \sqrt{\dfrac49 + \dfrac19 + \dfrac19} = \dfrac{\sqrt{6}}{3}\\ \dfrac{\mathbf{n}}{|\mathbf{n}|} = \dfrac{1}{\sqrt{6}}(2\mathbf{i}-\mathbf{j} + \mathbf{k})$$

Finally,

$$\nabla_{\mathbf{n}}f = (2x\mathbf{i} + z\mathbf{j}+(y+2z)\mathbf{k})\cdot \dfrac{1}{\sqrt{6}}(2\mathbf{i}-\mathbf{j} + \mathbf{k})=\\ =\dfrac{1}{\sqrt{6}}(4x-z+y+2z) = \dfrac{1}{\sqrt{6}}(4x+y+z)$$

Substituting point (1,2,1), obtain:

$$\dfrac{1}{\sqrt{6}}(4\cdot 1+2+1) = \dfrac{7}{\sqrt{6}}$$

Answer. $\dfrac{7}{\sqrt{6}}$.