Q. 27 - Calculus

Question

In Exercises 21–28, find the directional derivative of the given

function at the specified point $P$ and in the direction of the

given unit vector$u.$

Step-by-Step Solution

Answer

The directional derivative of function is $-\frac{36}{5}$

Step 1: Given data

The  function is $f(x,y,z)=x^2+y^2-z^3$ 

The given points is$P=\left(x_0,y_0,z_0\right)=(2,-2,2)\text{ and }u=(\alpha ,\beta ,\gamma )=\left(\frac{3}{5},0,\frac{4}{5}\right)$   

Step 2: Solution

Consider directional derivative

$D_{u}f\left(x_0,y_0,z_0\right)={\mathrm{Lim}}_{h\to 0}\frac{f\left(x_0+\alpha h,y_0+\beta h,z_0+\gamma h\right)-f\left(x_0,y_0,z_0\right.}{h}$ 

$D_{u}f(2,-2,2)={\mathrm{Lim}}_{h\to 0}\frac{f\left(2+\frac{3}{5}h,-2+0h,2+\frac{4}{5}h\right)-f(2,-2,2)}{h}$       $\left(1\right)$

Therefore,

$f\left(2+\frac{3}{5}h,-2+0h,2+\frac{4}{5}h\right)={\left(2+\frac{3}{5}h\right)}^2+(-2+0h{)}^2-{\left(2+\frac{4}{5}h\right)}^3$ 

$=4+\frac{12}{5}h+\frac{9}{25}{h}^2+4-8-\frac{48}{5}h-\frac{96}{25}{h}^2-\frac{64}{125}{h}^3$ 

$=-\frac{36}{5}h-\frac{87}{25}{h}^2-\frac{64}{125}{h}^3$          $\left(\left(2\right)\right)$

And

$f\left(x_0,y_0,z_0\right)=f(2,-2,2)=2^2+\left(-2^2\right)-2^3$ 

$f(2,-2,2)=0$     $\left(3\right)$

Step 3: Substitute

Substituting 

$D_{w}f(2,-2,2)={\mathrm{Lim}}_{h\to 0}\frac{-\frac{36}{5}h-\frac{87}{25}{h}^2-\frac{64}{125}{h}^3-0}{h}$ 

$={\mathrm{Lim}}_{h\to 0}-\frac{36}{5}-\frac{87}{25}h-\frac{64}{125}{h}^2$ 

$D_{w}f(2,-2,2)=-\frac{36}{5}$