Vector calculus

Calculus Level 4

The directional derivative of x 2 + y 2 + 4 x z x^2+y^2+4xz at ( 1 , 2 , 2 ) (1,-2,2) in the direction of the vector 2 i 2 j k 2i-2j-k is


The answer is 8.

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1 solution

Kartik Sharma
Sep 25, 2015

Directional Derivative, as it sounds, is the derivative of a function in E n E^n space in the direction of a specified vector in space.

Or mathematically,

v ( w ) = ( w ) v ^ \displaystyle \nabla_{\overrightarrow{v}}(w) = \nabla(w) \cdot \hat{v}

= ( w ) v v \displaystyle = \nabla(w) \cdot \frac{\overrightarrow{v}}{|\overrightarrow{v}|}

Now, w = x 2 + y 2 + 4 x z w = x^2 + y^2 + 4xz , v = 2 i ^ 2 j ^ k ^ v = 2\hat{i} - 2\hat{j} - \hat{k}

( w ) ( 1 , 2 , 2 ) 2 i ^ 2 j ^ k ^ 3 \displaystyle \nabla(w)|_{(1,-2,2)} \cdot \frac{2\hat{i} - 2\hat{j} - \hat{k}}{3}

And we know that

( w ) = ( w ) x i ^ + ( w ) y j ^ + ( w ) z k ^ \displaystyle \nabla(w) = \frac{\partial(w)}{\partial x} \hat{i} + \frac{\partial(w)}{\partial y} \hat{j} + \frac{\partial(w)}{\partial z} \hat{k}

( w ) = 20 i ^ + 8 j ^ 4 k ^ \displaystyle \nabla(w) = 20\hat{i} + 8 \hat{j} - 4\hat{k}

Coming back to our problem , we substitute and get

24 3 = 8 \displaystyle \frac{24}{3} = 8

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