Q. 22 - 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 $u.$

 $f(x,y)=x^2-y^2\text{ at }P=(2,3),\mathbf{u}=\frac{3}{5}\mathbf{i}-\frac{4}{5}\mathbf{j}\mathbf{ }$ 

Step-by-Step Solution

Answer

 The directional derivative of the $f(x,y)=x^2-y^2$function is $\frac{36}{5}$ 

Step 1: Given data

The objective to determine the directional derivative function

$f(x,y)=x^2-y^2$

$P=(2,3),\mathbf{u}=\frac{3}{5}\mathbf{i}-\frac{4}{5}\mathbf{j}$ 

Step 2: Solution

Consider directional derivative

$D_{n}f\left(x_0,y_6\right)={\mathrm{Lim}}_{h\to 0}\frac{f\left(x_0+\alpha h,y_0+\beta h\right)-f\left(x_0,y_0\right)}{h}$ 

$D_{n}f(2,3)={\mathrm{Lim}}_{a\to 0}\frac{f\left(2+\frac{3}{5}h,3-\frac{4}{5}h\right)-f(2,3)}{h}$     Equation $1$

Therefore

$f\left(2+\frac{3}{5}h,3-\frac{4}{5}h\right)={\left(2+\frac{3}{5}h\right)}^2-{\left(3-\frac{4}{5}h\right)}^2$ 

$=\left(4+\frac{12}{5}h+\frac{9}{25}{h}^2\right)-\left(9-\frac{24}{5}h+\frac{16}{25}{h}^2\right)$ 

$=-5+\frac{36}{5}h-\frac{7}{25}{h}^2$     Equation $2$ 

and

$f\left(x_0,y_0\right)=2^2-3^2=-5$   Equation $3$

Step 3: Substitute

Substituting

 $D_{N}f(2,3)={\mathrm{Lim}}_{h\to 0}\frac{-5+\frac{36}{5}h-\frac{7}{25}{h}^2+5}{h}$ 

$={\mathrm{Lim}}_{h\to 0}\frac{36}{5}-\frac{7}{25}h$ 

$D_{N}f(2,3)=\frac{36}{5}$