Q. 23 - 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 $u.$

 $f(x,y)=\frac{x}{y^2}\text{ at }P=(-2,1),\mathbf{u}=\left\{\frac{\sqrt{10}}{10},-\frac{3\sqrt{10}}{10}\right.>$

Step-by-Step Solution

Answer

The directional derivative of the  function$f(x,y)=\frac{x}{y^2}$ is$D_{n}f(-2,1)=\frac{-11\sqrt{10}}{10}$.$\frac{-11\sqrt{10}}{10}$

Step 1: Given data

The function is$f(x,y)=\frac{x}{y^2}$ 

The given points is $P=\left(x_0,y_0\right)=(-2,1)\text{ and }u=(\alpha ,\beta )=\left(\frac{\sqrt{10}}{10},-\frac{3\sqrt{10}}{10}\right)$   

Step 2: Solution

Consider directional derivative

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

$D_{w}f(-2,1)={\mathrm{Lim}}_{h\to 0}\frac{f\left(-2+\frac{\sqrt{10}}{10}h,1-\frac{3\sqrt{10}}{10}h\right)-f(-2,1)}{h}$ Equation $\left(1\right)$

Therefore,

$f\left(-2+\frac{\sqrt{10}}{10}h,1-\frac{3\sqrt{10}}{10}h\right)=\frac{-2+\frac{\sqrt{10}}{10}h}{{\left(1-\frac{3\sqrt{10}}{10}h\right)}^2}$ 

$=\frac{-2+\frac{h}{\sqrt{10}}}{{\left(1-\frac{3}{\sqrt{10}}h\right)}^2}$ 

$\left.=\frac{-2\sqrt{10}+h}{\left(1-\frac{6}{\sqrt{10}}h+\frac{9}{10}\right.}{h}^2\right)$      

Step 3: Solve

$=\frac{\sqrt{10}}{\left(\frac{10-6\sqrt{10}h+9h^2}{10}\right)}$ 

 $=\frac{\sqrt{10}(-2\sqrt{10}+h)}{10-6\sqrt{10}h+9h^2}$ 

$=\frac{-2\times 10+h\sqrt{10}}{10-6\sqrt{10}h+9h^2}$    Equation $\left(2\right)$

And

$f\left(x_0,y_0\right)=f(-2,1)=\frac{-2}{1^2}=-2$       Equation $\left(3\right)$

Step 4: Substitute

Substituting Equation $\left(2\right) ,\left(3\right)$ in$\left(1\right)$ 

$D_{n}f(-2,1)={\mathrm{Lim}}_{h\to 0}\frac{\frac{-2\ast 10+h\sqrt{10}}{10-6\sqrt{10}h+9h^2}-(-2)}{h}$ 

$={\mathrm{Lim}}_{h\to 0}\frac{-20+h\sqrt{10}+20-12\sqrt{10}h+18h^2}{h\left(10-6\sqrt{10}h+9h^2\right)}$ 

$={\mathrm{Lim}}_{h\to 0}\frac{-11h\sqrt{10}+18h^2}{h\left(10-6\sqrt{10}h+9h^2\right)}$ 

$={\mathrm{Lim}}_{h\to 0}\frac{h(-11\sqrt{10}+18h)}{h\left(10-6\sqrt{10}h+9h^2\right)}$ 

$={\mathrm{Lim}}_{h\to 0}\frac{(-11\sqrt{10}+18h)}{\left(10-6\sqrt{10}h+9h^2\right)}=\frac{-11\sqrt{10}}{10}$ 

$D_{n}f(-2,1)=\frac{-11\sqrt{10}}{10}$