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

`D_{\vec{u}}f(0,0,0)=(6)/(√(54))`

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}}`

since,
`v=<6, 3, -3>`

then
`\overrightarrow{u}=\frac{\overrightarrow{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}=< (6)/(√(54)),(3)/(√(54)),(-3)/(√(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))((6)/(√(54)))+(xe^(y)+e^(z))((3)/(√(54)))+(ye^(z)+e^(x))((-3)/(√(54)))`

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

`D_{\vec{u}}f(0,0,0)=(e^(0)+0e^(0))((6)/(√(54)))+(0e^(0)+e^(0))((3)/(√(54)))+(0e^(0)+e^(0))((-3)/(√(54)))`

`D_{\vec{u}}f(0,0,0)=(6)/(√(54))+(3)/(√(54))+(-3)/(√(54))`

`D_{\vec{u}}f(0,0,0)=(6+3-3)/(√(54))`

`D_{\vec{u}}f(0,0,0)=(6)/(√(54))`