Q. 33

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 $21 - 26 .$

$f (x, y) = \sqrt{\frac{y}{x}} at P = (4, 9), u = (- \frac{\sqrt{17}}{17}, - \frac{4 \sqrt{17}}{17})$

Step-by-Step Solution

Verified

Answer

The equation of the tangent line is $x = 4 - \frac{\sqrt{17}}{17} t, y = 9 - \frac{4 \sqrt{17}}{17}, t, z = \frac{3}{2} - \frac{7 \sqrt{17}}{816} t$

1Step 1: Given Information

Given line target is,

f (x, y) = \sqrt{\frac{y}{x}}

Point of, $P = (x_{0}, y_{0}) = (4, 9)$ and $u = (α, β) = (- \frac{\sqrt{17}}{17}, - \frac{4 \sqrt{17}}{17})$

2Step 2: Value of function

The equation of line of tangent is

$\begin{aligned} f_{x} (x_{0}, y_{0}) (x - x_{0}) + f_{y} (x_{0}, y_{0}) (y - y_{0}) = z - f (x_{0}, y_{0}) \end{aligned}$

$f_{x} (4, 9) (x - 4) + f_{y} (4, 9) (y - 9) = z - f (4, 9)$

Regarded,

$f_{x} (x_{0}, y_{0}) = {\frac{d}{d x} f (x, y)|}_{(x_{0}, 0_{0})} = {\frac{d}{d x} \sqrt{\frac{y}{x}}|}_{(x_{0}, y_{0})}$

$f_{x} (4, 9) = {\frac{1}{2 \sqrt{\frac{y}{x}}} \frac{d}{d x} (\frac{y}{x})|}_{(4)} = {\frac{\frac{- y}{x^{2}}}{2 \sqrt{\frac{y}{x}}}|}_{(40)}$

$f_{x} (4, 9) = \frac{- 9}{4^{2}}$

$f_{x} (4, 9) = - \frac{9}{\frac{9}{4}}$

3Step 3: Value of function

$f_{y} (x_{0}, y_{0}) = {\frac{d}{d y} f (x, y)|}_{(x_{0}, y_{0})} = {\frac{d}{d y} \sqrt{\frac{y}{x}}|}_{(x_{0}, y_{0})}$

$f_{y} (4, 9) = {\frac{1}{2 \sqrt{\frac{y}{x}}} \frac{d}{d x} (\frac{y}{x})|}_{(4)} = \frac{\frac{1}{x}}{{2 \sqrt{\frac{y}{x}}|}_{(49)}}$

$f_{x} (4, 9) = \frac{1}{2 \sqrt{\frac{9}{4}}}$

$f_{x} (4, 9) = \frac{1}{12}$ $(3)$

$f (x_{0}, y_{0}) = f (4, 9) = \sqrt{\frac{9}{4}}$

$f (- 2, 1) = \frac{3}{2}$ $(4)$

4Step 4: Result of function

Substituting

$\frac{- 9}{48} (x - 4) + \frac{1}{12} (y - 9) = z - \frac{3}{2}$

$\frac{- 9}{48} x + \frac{36}{48} + \frac{1}{12} y - \frac{8}{12} = z - \frac{3}{2}$

$\frac{- 3}{16} x + \frac{3}{4} + \frac{1}{12} y - \frac{2}{3} = z - \frac{3}{2}$

$\frac{- 3}{16} x + \frac{1}{12} y - z = - \frac{3}{2} - \frac{3}{4} + \frac{2}{3}$

Multiply $48$ on both sides

$- 9 x + 4 y - 48 z = 48 (\frac{- 19}{12})$

$- 9 x + 4 y - 48 z = - 76$

5Step 5

Consider equation of normal line

$\vec{r} (t) = <x_{0}, y_{0}, z_{0}> + t \nabla f (x_{0}, y_{0}, z_{0})$

Now,

$x = x_{0} + α t, y = y_{0} + β t, z = z_{0} + γ t$

where

$z_{0} = f (x_{0}, y_{0}) and γ = \nabla f (P) \cdot u$

Directional derivative of function at point $P$ with directional derivative given by

6Step 6

$\nabla f (P) \cdot u = \nabla f (4, 9) \cdot u = ({\frac{d f}{d x}|}_{(4, 9)} i + {\frac{d f}{d y}|}_{(4.9)} j) \cdot (- \frac{\sqrt{17}}{17} i - \frac{4 \sqrt{17}}{17} j)$

$= [{(\frac{1}{2 \sqrt{\frac{y}{x}}} \frac{d}{d x} (\frac{y}{x}))}_{(4.9)} i + {(\frac{1}{2 \sqrt{\frac{y}{x}}} \frac{d}{d x} (\frac{y}{x}))}_{(4, 9)} j] \cdot (- \frac{\sqrt{17}}{17} i - \frac{4 \sqrt{17}}{17} j)$

$= [{(\frac{\frac{- y}{x^{2}}}{2 \sqrt{\frac{y}{x}}})}_{(4, 9)} i + {(\frac{\frac{1}{x}}{2 \sqrt{\frac{y}{x}}})}_{(4, 9)} j \cdot (- \frac{\sqrt{17}}{17} i - \frac{4 \sqrt{17}}{17} j)$

style="width:30%" width="333" height="78" style="max-width: none; vertical-align: -33px;" $= [(\frac{\frac{- 9}{4^{2}}}{2 \sqrt{\frac{9}{4}}}) i + (\frac{\frac{1}{4}}{2 \sqrt{\frac{9}{4}}}) j \cdot (- \frac{\sqrt{17}}{17} i - \frac{4 \sqrt{17}}{17} j)$

7Step 7

$= [(\frac{- \frac{9}{16}}{2 \cdot \frac{3}{2}}) i + (\frac{\frac{1}{4}}{2 \cdot \frac{3}{2}}) j \cdot (- \frac{\sqrt{17}}{17} i - \frac{4 \sqrt{17}}{17} j)$