The directional derivative of at point (2,1,1) in the direction of vector $^i - 2^j + 3^k$ is ___
$f (x, y, z) = x y^{3} + y z^{3}$
-1.07

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Solution

The correct option is A -1.07
We have,

$V f = (y^{3})^i + (3 x y^{2} + z^{3})^j + (3 x z^{2})^k a t (2, 1, 1)$

$V f =^i + 7^j + 3^k$

So, direction derivative in direction of $^i + 7^j + 3^k$ is,

$= (^i + 7^j + 3^k) \frac{(^i - 2^j + 3^k)}{\sqrt{1 + 4 + 9}}$

$= \frac{1 - 14 + 9}{\sqrt{14}}$

$= \frac{- 4}{\sqrt{14}} = - 1.069 ≃ - 1.07$

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