Q. 28 - 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. 

$\begin{array}{r}f(x,y,z)=x^2+y^2-z^3\text{ at }P=(2,-2,2)\\ \mathbf{u}=\frac{2}{3}i-\frac{2}{3}j+\frac{1}{3}k\end{array}$ 

Step-by-Step Solution

Answer

The directional derivative of the 

function is $-\frac{24}{3}$

Step 1: Given data

The function is $\begin{array}{r}f(x,y,z)=x^2+y^2-z^3\end{array}$

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

Step 2: Solution

Consider directional derivative

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

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

Therefore,

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

$=4+\frac{8}{3}h+\frac{4}{9}{h}^2+4-\frac{8}{3}h+\frac{4}{9}{h}^2-8$ 

$=-\frac{24}{3}h-24h^2-\frac{8}{27}{h}^3$ 

$=-\frac{24}{3}h-\frac{208}{9}{h}^2-\frac{8}{27}{h}^3$ 

And

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

$f(2,-2,2)=0$

Step 3: substitute

Substituting

$D_{u}f(2,-2,2)={\mathrm{Lim}}_{h\to 0}\frac{-\frac{24}{3}h-\frac{208}{9}{h}^2-\frac{8}{27}{h}^3-0}{h}$

$={\mathrm{Lim}}_{h\to 0}-\frac{24}{3}-\frac{208}{9}h-\frac{8}{27}{h}^2$ 

$D_{u}f(2,-2,2)=-\frac{24}{3}$