Answer to Question #163601 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

find the partial derivatives:

$\frac{{\partial F}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {z - \frac{x}{z} + y} \right) = - \frac{1}{z} \Rightarrow \frac{{\partial F}}{{\partial x}}(P) = - \frac{1}{{ - 5}} = \frac{1}{5}$

$\frac{{\partial F}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {z - \frac{x}{z} + y} \right) = 1 \Rightarrow \frac{{\partial F}}{{\partial y}}(P) = 1$

$\frac{{\partial F}}{{\partial z}} = \frac{\partial }{{\partial z}}\left( {z - \frac{x}{z} + y} \right) = 1 + \frac{x}{{{z^2}}} \Rightarrow \frac{{\partial F}}{{\partial z}}(P) = 1 + \frac{3}{{{{( - 5)}^2}}} = \frac{{28}}{{25}}$

find the vector module:

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

Then the direction cosines are

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

from where

$\frac{{dF}}{{du}} = \frac{{\partial F}}{{\partial x}}(P)\cos \alpha + \frac{{\partial F}}{{\partial y}}(P)\cos \beta + \frac{{\partial F}}{{\partial z}}(P)\cos \gamma = \frac{{\frac{1}{5} \cdot 2 + 1 \cdot 6 - \frac{{28}}{{25}} \cdot 1}}{{\sqrt {41} }} = \frac{{132}}{{25\sqrt {41} }}$

Answer: $\frac{{132}}{{25\sqrt {41} }}$