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 given

function is$\frac{43}{13}$ 

Step 1: Given data

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

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

Step 2: Solution

Consider directional derivative

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

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

Therefore

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

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

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

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

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

$=\frac{-338-65h}{169+312h+144h^2}$

   And

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

Step 3: Substitute

Substituting

$D_{w}f(-2,1)={\mathrm{Lim}}_{h\to 0}\frac{\frac{-338-65h}{169+312h+144h^2}-(-2)}{h}$

$={\mathrm{Lim}}_{h\to 0}\frac{-338-65h+338+624h+288h^2}{h\left(169+312h+144h^2\right)}$

$={\mathrm{Lim}}_{h\to 0}\frac{559h+288h^2}{h\left(169+312h+144h^2\right)}$

$={\mathrm{Lim}}_{h\to 0}\frac{h(559+288h)}{h\left(169+312h+144h^2\right)}$

$={\mathrm{Lim}}_{h\to 0}\frac{(559+288h)}{\left(169+312h+144h^2\right)}=\frac{559}{169}$
$D_{w}f(-2,1)=\frac{43}{13}$