Question

Find the directional derivative of the function at the given point in the direction of the vector v. f(x, y, z) = xe^y + ye^z + ze^x, (0, 0, 0), v = 6, 3, −3

Answer 1

Answer:

$D_{\vec{u}}f(0,0,0)=\frac{6}{\sqrt{54}}$

Step-by-step explanation:

We need to find the directional derivative of the function at the given point in the direction of the vector v.

$f(x, y, z)=xe^{y} + ye^{z} + ze^{x}$ ,point (0, 0, 0) and $v=<6, 3, -3>$

 

By Theorem: If f is a differentiable function of x , y and z , then f has a directional derivative for any unit vector $\overrightarrow{v} =<v_1,v_2,v_3>$ and

$D_{\overrightarrow{u}}f(x,y,z)=f_{x}(x,y,z)u_1+f_{y}(x,y,z)u_2+f_{z}(x,y,z)u_3$

where $\overrightarrow{u}=\frac{\overrightarrow{v}}{||v||}$

since,  $v=<6, 3, -3>$

then $\overrightarrow{u}=\frac{\overrightarrow{v}}{||v||}$

$\overrightarrow{u}=< \frac{6}{\sqrt{6^{2}+3^{2}+(-3)^{2}}},\frac{3}{\sqrt{6^{2}+3^{2}+(-3)^{2}}},\frac{-3}{\sqrt{6^{2}+3^{2}+(-3)^{2}}} >$

$\overrightarrow{u}=< \frac{6}{\sqrt{54}},\frac{3}{\sqrt{54}},\frac{-3}{\sqrt{54}} >$

The partial derivatives are

$f_{x}(x,y,z)=e^{y}+ze^{x}$  

$f_{y}(x,y,z)=xe^{y}+e^{z}$

$f_{z}(x,y,z)=ye^{z}+e^{x}$

Then the directional derivative is

$D_{\vec{u}}f(x,y,z)=(e^{y}+ze^{x})(\frac{6}{\sqrt{54}})+(xe^{y}+e^{z})(\frac{3}{\sqrt{54}})+(ye^{z}+e^{x})(\frac{-3}{\sqrt{54}})$

so, directional derivative at point (0,0,0)

$D_{\vec{u}}f(0,0,0)=(e^{0}+0e^{0})(\frac{6}{\sqrt{54}})+(0e^{0}+e^{0})(\frac{3}{\sqrt{54}})+(0e^{0}+e^{0})(\frac{-3}{\sqrt{54}})$

$D_{\vec{u}}f(0,0,0)=\frac{6}{\sqrt{54}}+\frac{3}{\sqrt{54}}+\frac{-3}{\sqrt{54}}$

$D_{\vec{u}}f(0,0,0)=\frac{6+3-3}{\sqrt{54}}$

$D_{\vec{u}}f(0,0,0)=\frac{6}{\sqrt{54}}$