Q. 32 - 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$21–26.$

$f(x,y)=\frac{x}{y^2}\text{ at }P=(-2,1),\mathbf{u}=-\frac{5}{13}\mathbf{i}+\frac{12}{13}\mathrm{j}$ 

Step-by-Step Solution

Answer

The equation of the line tangent is  $x=-2-\frac{5}{13}t,y=1+\frac{12}{13}t,z=-2+\frac{43}{13}t$ 

Step 1: Given data

$f(x,y)=\frac{x}{y^2}$ 

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

Step 2: solution

The equation of line tangent is 

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

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

Assume

$f_x\left(x_0,y_0\right)={\left.\frac{d}{dx}f(x,y)\right|}_{\left(x_0{5}_0\right)}={\left.\frac{d}{dx}\frac{x}{y^2}\right|}_{\left(x_0,v_0\right)}$ 

$f_x(-2,1)={\left.\frac{1}{y^2}\right|}_{(-2,1)}$ 

$f_x(-2,1)=1$   $\left(2\right)$ 

Step 3

 

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

 

$f_y(-2,1)=4$ 

fy(−2,1)=4

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

 $f(-2,1)=-2$

Step 4: Substitute

Substituting

$1(x+2)+4(y-1)=z+2$ 

$x+2+4y-4=z+2$ 

$x+4y-z=2-2+4$ 

$x+4y-z=4$ 

Step 5

Consider the equation of normal line

 

here

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

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

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

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

Step 6

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

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

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

 

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

$=\frac{-338-65h}{169+312h+144h^2}$         $\left(6\right)$ 

and 

$f\left(x_0,y_0\right)=f(-2,1)=\frac{-2}{1^2}=-2$     $\left(7\right)$ 

Step 7

  

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

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

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

 

$D_{N}f(-2,1)=\frac{43}{13}$   

Therefore, 

$x=-2-\frac{5}{13}t,y=1+\frac{12}{13}t,z=-2+\frac{43}{13}t$