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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 = 4, 3, −2

User Sachin D
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1 Answer

6 votes

Answer:


D_{\vec{u}}f(0,0,0)=(5)/(√(29))

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=<4, 3, -2>

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=<4, 3, -2>

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


\overrightarrow{u}=< \frac{4}{\sqrt{4^(2)+3^(2)+(-2)^(2)}},\frac{3}{\sqrt{4^(2)+3^(2)+(-2)^(2)}},\frac{-2}{\sqrt{4^(2)+3^(2)+(-2)^(2)}} >


\overrightarrow{u}=< (4)/(√(29)),(3)/(√(29)),(-2)/(√(29) )>

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))((4)/(√(29)))+(xe^(y)+e^(z))((3)/(√(29)))+(ye^(z)+e^(x))((-2)/(√(29)))

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


D_{\vec{u}}f(0,0,0)=(e^(0)+0e^(0))((4)/(√(29)))+(0e^(0)+e^(0))((3)/(√(29)))+(0e^(0)+e^(0))((-2)/(√(29)))


D_{\vec{u}}f(0,0,0)=(4)/(√(29))+(3)/(√(29))+(-2)/(√(29))


D_{\vec{u}}f(0,0,0)=(4+3-2)/(√(29))


D_{\vec{u}}f(0,0,0)=(5)/(√(29))

User Shagglez
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6.3k points