Q. 33 - Calculus

Question

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. 

Step-by-Step Solution

Answer

The equation of the line tangent to the surface at the given point P and in the direction of the given unit vector u is $x=4-\frac{\sqrt{17}}{17}t,y=9-\frac{4\sqrt{17}}{17},t,z=\frac{3}{2}-\frac{7\sqrt{17}}{816}t$

Step 1: Given information

The same functions, points, and vectors is 

Step 2: Calculation

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

At point: $P=\left(x_0,y_0\right)=(4,9)$ and $u=(\alpha ,\beta )=\left(-\frac{\sqrt{17}}{17},-\frac{4\sqrt{17}}{17}\right)$

The tangent line's equation is

$$\begin{gathered} f_x\left(x_0,y_0\right)\left(x-x_0\right)+f_y\left(x_0,y_0\right)\left(y-y_0\right)=z-f\left(x_0,y_0\right)f_x(4,9)(x-4)+f_y(4,9)(y-9) \\ =z-f(4,9) \end{gathered}$$

Assume,

$$\begin{gathered} f_x\left(x_0,y_0\right)={\left.\frac{d}{dx}f(x,y)\right|}_{\left(x_0,y_0\right)} \\ ={\left.\frac{d}{dx}\sqrt{\frac{y}{x}}\right|}_{\left(x_0,y_0\right)}{f}_x(4,9) \\ ={\left.\frac{1}{2\sqrt{\frac{y}{x}}}\frac{d}{dx}\left(\frac{y}{x}\right)\right|}_{\left(49\right)} \\ ={\left.\frac{-y}{\frac{x^2}{2\sqrt{\frac{y}{x}}}}\right|}_{\left(49\right)}{f}_x(4,9)=\frac{4^2}{2\sqrt{\frac{9}{4}}}{f}_x(4,9)=-\frac{9}{48} \end{gathered}$$

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

$\begin{array}{rcl}{f}_y(4,9)& =& {\left.\frac{1}{2\sqrt{\frac{y}{x}}}\frac{d}{dx}\left(\frac{y}{x}\right)\right|}_{(+9)}\\ & =& \frac{\frac{1}{x}}{{\left.2\sqrt{\frac{y}{x}}\right|}_{\left(\alpha 9\right)}}{f}_x(4,9)\\ & =& \frac{\frac{1}{4}}{2\sqrt{\frac{9}{4}}}{f}_x(4,9)\\ & =& \frac{1}{12}\end{array}$

$$\begin{gathered} f\left(x_0,y_0\right)=f(4,9)=\sqrt{\frac{9}{4}} \\ f(-2,1)=\frac{3}{2} \end{gathered}$$

Step 3: Further calculation

Equation (1) is solved by substituting equations (2), (3), and (4).

$$\begin{gathered} \frac{-9}{48}(x-4)+\frac{1}{12}(y-9)=z-\frac{3}{2} \\ \frac{-9}{48}x+\frac{36}{48}+\frac{1}{12}y-\frac{8}{12}=z-\frac{3}{2} \\ \frac{-3}{16}x+\frac{3}{4}+\frac{1}{12}y-\frac{2}{3}=z-\frac{3}{2} \\ \frac{-3}{16}x+\frac{1}{12}y-z=-\frac{3}{2}-\frac{3}{4}+\frac{2}{3} \end{gathered}$$

Multiply both sides by 48 to obtain

$$\begin{gathered} -9x+4y-48z=48\left(\frac{-19}{12}\right) \\ -9x+4y-48z=-76 \end{gathered}$$

Consider the following equation for the normal line:

Here

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

Where, $z_0=f\left(x_0,y_0\right)$ and

$\gamma =\nabla f\left(P\right)·u$

Step 4: Calculation

The directional derivative of a function at point P with directional unit vector u is calculated as follows:

$$\begin{gathered} \nabla f\left(P\right)·u=\nabla f(4,9)·u=\left({\left.\frac{df}{dx}\right|}_{(4,9)}i+{\left.\frac{df}{dy}\right|}_{(4,9)}j\right)·\left(-\frac{\sqrt{17}}{17}i-\frac{4\sqrt{17}}{17}j\right) \\ =\left[{\left(\frac{1}{2\sqrt{\frac{y}{x}}}\frac{d}{dx}\left(\frac{y}{x}\right)\right)}_{(4,9)}i+{\left(\frac{1}{2\sqrt{\frac{y}{x}}}\frac{d}{dx}\left(\frac{y}{x}\right)\right)}_{(4,9)}j\right]·\left(-\frac{\sqrt{17}}{17}i-\frac{4\sqrt{17}}{17}j\right) \end{gathered}$$

$$\begin{gathered} =\left[{\left(\frac{\frac{-y}{x^2}}{2\sqrt{\frac{y}{x}}}\right)}_{(4,9)}i+{\left(\frac{\frac{1}{x}}{2\sqrt{\frac{y}{x}}}\right)}_{(4,9)}j\right]·\left(-\frac{\sqrt{17}}{17}i-\frac{4\sqrt{17}}{17}j\right) \\ =\left[\left(\frac{\frac{-9}{4^2}}{2\sqrt{\frac{9}{4}}}\right)i+\left(\frac{\frac{1}{4}}{2\sqrt{\frac{9}{4}}}\right)j·\left(-\frac{\sqrt{17}}{17}i-\frac{4\sqrt{17}}{17}j\right)\right. \\ =\left[\left(\frac{-\frac{9}{16}}{2·\frac{3}{2}}\right)i+\left(\frac{\frac{1}{4}}{2·\frac{3}{2}}\right)j\right]·\left(-\frac{\sqrt{17}}{17}i-\frac{4\sqrt{17}}{17}j\right) \\ =\left(-\frac{9}{48}i+\frac{1}{12}j\right)·\left(-\frac{\sqrt{17}}{17}i-\frac{4\sqrt{17}}{17}j\right) \\ =\left(\frac{9·\sqrt{17}}{48·17}-\frac{4\sqrt{17}}{12·17}\right) \end{gathered}$$

$$\begin{gathered} =\left(\frac{9·\sqrt{17}}{16·3·17}-\frac{\sqrt{17}}{3·17}\right) \\ =\frac{1}{3·17}\left(\frac{9·\sqrt{17}}{16}-\frac{\sqrt{17}}{1}\right) \\ =\frac{1}{3·17}\left(\frac{9·\sqrt{17}-16\sqrt{17}}{16}\right) \\ \nabla f\left(P\right)·u=-\frac{7}{816}\sqrt{17} \end{gathered}$$

Therefore

$x=4-\frac{\sqrt{17}}{17}t,y=9-\frac{4\sqrt{17}}{17},t,z=\frac{3}{2}-\frac{7\sqrt{17}}{816}t$