Q.29 - Calculus

Question

In Exercises 29–34, find the equation of the line tangent to the

surface at the given point P and in the direction of the given

unit vector u. Note that these are the same functions, points,

and vectors as in Exercises 21–26.

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

Step-by-Step Solution

Answer

The equation of the line tangent to the

surface is$x=2+\frac{\sqrt{2}}{2}t,y=3+\frac{\sqrt{2}}{2}t,z=-5-\sqrt{2}t$  

Step 1: Given data

Given   $f(x,y)=x^2-y^2$ 

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

Step 2: Solution

The equation of line of tangent is 

$\begin{array}{r}{f}_x\left(x_0,y_0\right)\left(x-x_0\right)+f_y\left(x_0,y_0\right)\left(y-y_0\right)=z-f\left(x_0,y_0)\right.\end{array}$

$f_x(2,3)(x-2)+f_y(2,3)(y-3)=z-f(2,3)$ 

Assume

$f_x\left(x_0,y_0\right)={\left.\frac{d}{dx}f(x,y)\right|}_{\left(x_0,y_0\right)}={\left.\frac{d}{dx}\left(x^2-y^2\right)\right|}_{\left(x_0,y_0\right)}$ 

$f_x(2,3)={\left.2x\right|}_{(2,3)}$

$f_x(2,3)=4$                     $\left(2\right)$

Step: 3

$f_y\left(x_0,y_0\right)={\left.\frac{d}{dy}f(x,y)\right|}_{\left(x_0,y_0\right)}={\left.\frac{d}{dy}\left(x^2-y^2\right)\right|}_{\left(x_0,y_0\right)}$ 

$f_y(2,3)=-{\left.2y\right|}_{(2,3)}$ 

$f_y(2,3)=-6$                          $\left(3\right)$

$\begin{array}{r}f\left(x_0,y_0\right)=f(2,3)=\left(2^2-3^2\right)\end{array}$ 

$\begin{array}{r}f(2,3)=-5\end{array}$   $\left(4\right)$

Step 4: Substitute

Substituting $\left(2\right),\left(3\right),\left(4\right)$ in equation $\left(1\right)$ 

$4(x-2)-6(y-3)=z+5$ 

$4x-8-6y+18=z+5$ 

$4x-6y-z=5-18+8$ 

$4x-6y-z=-5$

Step 5: Equation

Here

Consider the equation of normal line is

 

Now

$x=x_0+\alpha t,y=y_0+\beta t,z=z_0+\gamma t$ 

Where 

$z_0=f\left(x_0+\alpha t\right), y=y_0+\beta z=z_0+\gamma t$ 

Consider directional derivative

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

$\begin{array}{r}\therefore D_{n}f(2,3)={\mathrm{Lim}}_{h\to 0}\frac{f\left(2+\frac{\sqrt{2}}{2}h,3+\frac{\sqrt{2}}{2}h\right)-f(2,3)}{h}\end{array}$

Step 6

therefore

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

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

$=-5-\sqrt{2}h$ 

$f\left(x_0,y_0\right)=2^2-3^2=-5$ 

Substituting

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

$={\mathrm{Lim}}_{h\to 0}-\sqrt{2}$ 

$D_{n}f(2,3)=-\sqrt{2}$ 

$x=2+\frac{\sqrt{2}}{2}t,y=3+\frac{\sqrt{2}}{2}t,z=-5-\sqrt{2}t$