Q. 24 - 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.

$f(x,y)=\frac{x}{y^2}\text{ at }P=(-2,1),\mathbf{u}=-\frac{5}{13}\mathbf{i}+\frac{12}{13}\mathbf{j}$  

Step-by-Step Solution

Answer

The directional derivative of the  

$f(x,y)=\frac{x}{y^2}$ function is

Step 1: Given data

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

$P=\left(x_0,y_0\right)=(-2,1) u=(\alpha i+\beta j)=\left(\frac{-5}{13}i+\frac{12}{13}j\right)$ 

  

Step 2: Solution

Consider directional derivative

 $D_{k}f\left(x_e,y_e\right)={\mathrm{Lim}}_{h\to 0}\frac{f\left(x_e+\alpha h+y_0+\beta h\right)-f\left(x_{e+}{y}_e\right)}{h}$ 

$D_{k}f(-2,1)={\mathrm{Lim}}_{k\to 0}\frac{f\left(-2-\frac{5}{13}h,1+\frac{12}{13}h\right)-f(-2,1)}{h}$ 

Therefore

$f\left(-2-\frac{5}{13}h,1+\frac{12}{13}h\right)=\frac{-2-\frac{5}{13}h}{{\left(1+\frac{12}{13}h\right)}^2}$ 

$=\frac{-2-\frac{5}{13}h}{\left(1+\frac{24}{13}h+\frac{144}{169}{h}^2\right)}$ 

$=\frac{-26-5h}{\left(1+\frac{24}{13}h+\frac{144}{169}{h}^2\right)}$ 

$=\frac{-26-5h}{13\ast \left(1+\frac{24}{13}h+\frac{144}{169}{h}^2\right)}$     

Step 3

$=\frac{-26-5h}{13\ast \left(\frac{169+312h+144h^2}{169}\right)}$ 

$=\frac{-338-65h}{169+312h+144h^2}$ 

and

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