Q. 30 - 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.

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

Step-by-Step Solution

Answer

The equation of the line tangent  is $x=2+\frac{3}{5}t,y=3-\frac{4}{5}t,z=-5+\frac{36}{5}t$.

Step 1: Given data

$f(x,y)=x^2-y^2$ 

At point $P=\left(x_0,y_0\right)=(2,3)\text{ and }u=(\alpha ,\beta )=\frac{3}{5}i-\frac{4}{5}j$ 

Step 2: Solution

The equation of line tangent

$\begin{array}{r}{f}_x(2,3)(x-2)+f_y(2,3)(y-3)=z-f(2,3)\end{array}$       $\left(1\right)$

Consider

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

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

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

Step 3

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

$\begin{array}{r}{f}_y(2,3)=-{\left.2y\right|}_{(2,3)}\end{array}$ 

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

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

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

Step 4: Substitute

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

$\begin{array}{r}4(x-2)-6(y-3)=z+5\end{array}$ 

$\begin{array}{r}4x-8-6y+18=z+5\end{array}$ 

$\begin{array}{r}4x-6y-z=5-18+8\end{array}$ 

$\begin{array}{r}4x-6y-z=-5\end{array}$

Step 5: equation of normal line

The equation of normal line is 

 

Now

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

Where

$z_0=f\left(x_0,y_0\right)\text{ and }\gamma =D_{w}f\left(x_0,y_0\right)$ 

Consider directional derivative

$\begin{array}{r}{D}_{v}f\left(x_0,y_0\right)={\mathrm{Lim}}_{h\to 0}\frac{f\left(x_0+\alpha h,y_0+\beta h\right)-f\left(x_0,y_0\right)}{h}\end{array}$ 

$\begin{array}{r}{D}_{v}f(2,3)={\mathrm{Lim}}_{h\to 0}\frac{f\left(2+\frac{3}{5}h,3-\frac{4}{5}h\right)-f(2,3)}{h}\end{array}$    $\left(5\right)$

Step 6

Therefore,

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

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

$\begin{array}{r}=-5+\frac{36}{5}h-\frac{7}{25}{h}^2\end{array}$       $\left(6\right)$

And

$f\left(x_0,y_0\right)=2^2-3^2=-5$          $\left(7\right)$

Step 7: Substitute

Substituting $\left(6\right),\left(7\right)$ in $\left(1\right)$ 

$\begin{array}{r}{D}_{w}f(2,3)={\mathrm{Lim}}_{h\to 0}\frac{-5+\frac{36}{5}h-\frac{7}{25}{h}^2+5}{h}\end{array}$ 

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

$\begin{array}{r}{D}_{v}f(2,3)=\frac{36}{5}\end{array}$

$x=2+\frac{3}{5}t,y=3-\frac{4}{5}t,z=-5+\frac{36}{5}t$