Q. 21 - 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 $u$.

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

Step-by-Step Solution

Answer

The directional derivative of the given

function  is $-\sqrt{2}$

Step 1: Given data

$f(x,y)=x^2-y^2$ 

$P=\left(x_0,y_0\right)=(2,3)u=(\alpha ,\beta )=\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$ 

  

Step 2: Solution

Consider directional derivative

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

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

Therefore 

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

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

         Equation $2$

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

 

Step 3: Substitute

Substituting equation $2$ and $3$ 

$D_{v}f(2,3)={\mathrm{Lim}}_{h\to 0}\frac{-5-\sqrt{2}h+5}{h}$ 

$={\mathrm{Lim}}_{b\to 0}-\sqrt{2}$ 

$D_{v}f(2,3)=-\sqrt{2}$