Question

if w=xy+yz+z find directional derivative of w at (1,-2,0)in the directions at (3,6,9)​

Answer 1

Explanation:

The answer is -0.8.Go n crack the mcq

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

$\sf{ \:w = xy + yz + z \: }$

TO DETERMINE

The directional derivative of w at (1,-2,0) in the directions at (3,6,9)

CALCULATION

Here

$\sf{ \:w = f(x,y,z) = xy + yz + z \: }$

Now

$\displaystyle \sf{ \: \frac{ \partial w}{ \partial x} = y \: }$

$\displaystyle \sf{ \: \frac{ \partial w}{ \partial y} = x + z\: }$

$\displaystyle \sf{ \: \frac{ \partial w}{ \partial z} = y + 1\: }$

$\sf{Again \: the \: unit \: vector \: in \: the \: direction \: \vec{r} = 3 \hat{i} + 6 \hat{j} + 9 \hat{k} \: is}$

$\displaystyle \: \sf{ \frac{ \vec{r} }{ | \vec{r}| } =\: \frac{1}{ \sqrt{126} } ( \: 3\hat{i} + 6\hat{j} + 9\hat{k}\: )\: }$

Again

$\sf{ \nabla{w} = y \hat{i} + (x + z) \hat{j} + (y + 1) \hat{k}\: }$

So at ( 1,-2,0)

$\sf{ \nabla{w} = - 2\hat{i} + (1 + 0) \hat{j} + ( - 2+ 1) \hat{k}\: }$

$\implies \: \sf{ \nabla{w} = - 2\hat{i} + \hat{j} - \hat{k}\: }$

Hence the required answer is

$= \sf{( - 2\hat{i} + \hat{j} - \hat{k}\: )}. \: \displaystyle \: \sf{ \: \frac{1}{ \sqrt{126} } ( \: 3\hat{i} + 6\hat{j} + 9\hat{k}\: )\: }$

$= \displaystyle\sf{ \: \frac{ - 6 + 6 - 9}{ \sqrt{126} } }$

$= \displaystyle\sf{ \: \frac{ - 9}{ \sqrt{126} } }$