Q. 38 - Calculus

Question

In Exercises $35-38$ find the directional derivative of the given

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

given vector v.

$\begin{array}{r}f(x,y,z)=x^2+y^2-z^3\text{ at }P=(0,-2,5)\\ v=-i+3j+5k\end{array}$

Step-by-Step Solution

Answer

The directional derivative of function is  $-\frac{137}{\sqrt{35}}$

Step 1: Given data

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

$P=\left(x_0,y_0,z_0\right)=(0,-2,5)\text{ and }v=-i+3j+5k$

Step 2: Solution

Consider directional derivative

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

Therefore

 

$\parallel v\parallel =\sqrt{35}$   

$\therefore u=(\alpha ,\beta ,\gamma )=\left(\frac{-1}{\sqrt{35}},\frac{3}{\sqrt{35}},\frac{5}{\sqrt{35}}\right)$ 

Consider

$f\left(x_0+\alpha h,y_0+\beta h,z_0+\gamma h\right)=f\left(0-\frac{1}{\sqrt{35}}h,-2+\frac{3}{\sqrt{35}}h,5+\frac{5}{\sqrt{35}}h\right)$

$={\left(0-\frac{1}{\sqrt{35}}h\right)}^2+{\left(-2+\frac{3}{\sqrt{35}}h\right)}^2-{\left(5+\frac{5}{\sqrt{35}}h\right)}^3$

  

$=\frac{1}{35}{h}^2+\left(4-\frac{12}{\sqrt{35}}h+\frac{9}{35}{h}^2\right)-\left(125+\frac{375}{\sqrt{35}}h+\frac{375}{35}{h}^2+\frac{125}{(\sqrt{35}{)}^3}{h}^3\right)$

$=\frac{1}{35}{h}^2+4-\frac{12}{\sqrt{35}}h+\frac{9}{35}{h}^2-125-\frac{375}{\sqrt{35}}h-\frac{375}{35}{h}^2-\frac{125}{(\sqrt{35}{)}^3}{h}^3$

$=-121-\frac{365}{35}{h}^2-\frac{137}{\sqrt{35}}h-\frac{125}{(\sqrt{35}{)}^3}{h}^3$   

And

$f\left(x_0,y_0,z_0\right)=0^2+(-2{)}^2-5^3=-121$

  

Step 3: substitute

Substituting

$D_{w}f(0,-2,5)={\mathrm{Lim}}_{h\to 0}\frac{-121-\frac{137}{\sqrt{35}}h-\frac{365}{35}{h}^2-\frac{125}{(\sqrt{35}{)}^3}{h}^3-121}{h}$

$={\mathrm{Lim}}_{h\to 0}-\frac{137}{\sqrt{35}}-\frac{365}{35}h-\frac{125}{(\sqrt{35}{)}^3}{h}^2$

$D_{w}f(0,-2,5)=-\frac{137}{\sqrt{35}}$