Answer to Question #163599 in Calculus for asad

Question

Find the directional derivative of  at point  in the direction of vector. Where

 ,  and

F(x,y,z)=z-x/z+y, p (3,4,-5) and u= 2i+6j-k

Answer 1

Directional derivative of function $F(x,y,z)=\dfrac{z-x}{z+y}$ along direction of vector $\bold{u}=2\bold{i}+6\bold{j}-\bold{k}$ at the point $P(3,4,-5)$ is

$\dfrac{\partial F(P)}{\partial\bold{u}}=$ $\dfrac{\partial F(P)}{\partial x}\cos\alpha+\dfrac{\partial F(P)}{\partial y}\cos\beta+\dfrac{\partial F(P)}{\partial z}\cos\gamma$ ,

where

$\dfrac{\partial F}{\partial x}=$ $\dfrac{\partial}{\partial x}\left(\dfrac{z-x}{z+y}\right)=$ $\dfrac{1}{z+y}\cdot\dfrac{\partial}{\partial x}\left(z-x\right)=$ $\dfrac{1}{z+y}\cdot\left(-1\right)=$ $-\dfrac{1}{z+y}$ ,

$\dfrac{\partial F(P)}{\partial x}=-\dfrac{1}{-5+4}=-\dfrac{1}{-1}=1$ ,

$\dfrac{\partial F}{\partial y}=$ $\dfrac{\partial}{\partial y}\left(\dfrac{z-x}{z+y}\right)=$ $(z-x)\cdot\dfrac{\partial}{\partial y}\left(\dfrac{1}{z+y}\right)=$ $(z-x)\cdot\left(-\dfrac{1}{(z+y)^2}\right)=$

$-\dfrac{z-x}{(z+y)^2}$ ,

$\dfrac{\partial F(P)}{\partial y}=-\dfrac{-5-3}{(-5+4)^2}=-\dfrac{-8}{(-1)^2}=\dfrac{8}{1}=8$ ,

$\dfrac{\partial F}{\partial z}=$ $\dfrac{\partial}{\partial z}\left(\dfrac{z-x}{z+y}\right)=$ $\dfrac{\left(\dfrac{\partial}{\partial z}(z-x)\right)\cdot(z+y)-(z-x)\cdot\dfrac{\partial}{\partial z}(z+y)}{(z+y)^2}=$

$\dfrac{1\cdot(z+y)-(z-x)\cdot1}{(z+y)^2}=$ $\dfrac{z+y-z+x}{(z+y)^2}=\dfrac{y+x}{(z+y)^2}$ ,

$\dfrac{\partial F(P)}{\partial z}=$ $\dfrac{4+3}{(-5+4)^2}=\dfrac{7}{(-1)^2}=\dfrac{7}{1}=7$ ,

$|\bold{u}|=\sqrt{2^2+6^2+(-1)^2}=\sqrt{4+36+1}=\sqrt{41}$

$\cos\alpha=\dfrac{2}{\sqrt{41}}$ , $\cos\beta=\dfrac{6}{\sqrt{41}}$ , $\cos\gamma=-\dfrac{1}{\sqrt{41}}$ .

So,

$\dfrac{\partial F(P)}{\partial\bold{u}}=1\cdot\dfrac{2}{\sqrt{41}}+8\cdot\dfrac{6}{\sqrt{41}}-7\cdot\dfrac{1}{\sqrt{41}}=$ $\dfrac{2+48-7}{\sqrt{41}}=\dfrac{43}{\sqrt{41}}$ .