Q 34. - 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 tangent line is $x=4-\frac{15}{17}t,y=9+\frac{8}{17},t,z=\frac{3}{2}-\frac{103}{816}t$.

Step 1. Given information

  Equation of function is,

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

Point of $P=\left(x_0,y_0\right)=(4,9)$ and $\mathrm{u}=\frac{15}{17}\mathrm{i}+\frac{8}{17}\mathrm{j}$

Step 2. Function of x and y

The equation of tangent of line is  

$\begin{array}{r}{f}_x(4,9)(x-4)+f_y(4,9)(y-9)=z-f(4,9)\end{array}$

Regarded,

$$\begin{gathered} f_x\left(x_0,y_0\right)={\left.\frac{d}{dx}f(x,y)\right|}_{\left(x_0{v}_0\right)}={\left.\frac{d}{dx}\sqrt{\frac{y}{x}}\right|}_{\left(x_0,y_0\right)} \\ {f}_x(4,9)={\left.\frac{1}{2\sqrt{\frac{y}{x}}}\frac{d}{dx}\left(\frac{y}{x}\right)\right|}_{(4,9)}={\left.\frac{\frac{-y}{x^2}}{2\sqrt{\frac{y}{x}}}\right|}_{(4{)}^2} \left(1\right) \\ {f}_x(4,9)=\frac{\frac{-9}{4^2}}{2\sqrt{\frac{9}{4}}} \\ {f}_x(4,9)=-\frac{9}{48} \left(2\right) \end{gathered}$$

Now for $y$,

$$\begin{gathered} f_y\left(x_0,y_0\right)={\left.\frac{d}{dy}f(x,y)\right|}_{\left(x_0,y_0\right)}={\left.\frac{d}{dy}\sqrt{\frac{y}{x}}\right|}_{\left(x_0,y_0\right)} \\ {f}_y(4,9)={\left.\frac{1}{2\sqrt{\frac{y}{x}}}\frac{d}{dx}\left(\frac{y}{x}\right)\right|}_{(4\cdot )}={\left.\frac{\frac{1}{x}}{2\sqrt{\frac{y}{x}}}\right|}_{(4,9)} \\ {f}_x(4,9)=\frac{1}{2\sqrt{\frac{9}{4}}} \\ {f}_x(4,9)=\frac{1}{12} \left(3\right) \\ f\left(x_0,y_0\right)=f(4,9)=\sqrt{\frac{9}{4}} \\ f(-2,1)=\frac{3}{2} \left(4\right) \end{gathered}$$

Step 3. Value of function

Substituting all equation on first equation,

We get,

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

Multiply $48$ on both sides

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

Step 4. Directional derivative function

The equation of normal line is,

where,

$z_0=f\left(x_0,y_0\right)$and $\mathrm{\gamma }=\mathrm{\nabla }\mathrm{f}\left(\mathrm{P}\right)\times \mathrm{u}$

 Directional derivative of function is,

$\mathrm{\nabla }f\left(P\right)\times u=\mathrm{\nabla }f(4,9)\times u=\left({\left.\frac{df}{dx}\right|}_{(4,9)}i+{\left.\frac{df}{dy}\right|}_{(4,9)}j\right)\times \left(\frac{15}{17}i+\frac{8}{17}j\right)$

Step 5. Calculation

So,

$$\begin{gathered} \left.\mathrm{\nabla }f\left(P\right)\times u={\left[\frac{\frac{-y}{x^2}}{2\sqrt{\frac{y}{x}}}\right)}_{\left(4.9\right)}i+{\left(\frac{\frac{1}{x}}{2\sqrt{\frac{y}{x}}}\right)}_{\left(4.9\right)}j\right]\times \left(\frac{15}{17}i+\frac{8}{17}j\right) \\ =\left[\left(\frac{\frac{-9}{4^2}}{2\sqrt{\frac{9}{4}}}\right)i+\left(\frac{\frac{1}{4}}{2\sqrt{\frac{9}{4}}}\right)j\right]\times \left(\frac{15}{17}i+\frac{8}{17}j\right) \\ =\left[\left(\frac{-\frac{9}{16}}{2\times \frac{3}{2}}\right)i+\left(\frac{\frac{1}{4}}{2\times \frac{3}{2}}\right)j\right]\times \left(\frac{15}{17}i+\frac{8}{17}j\right) \\ =\left(-\frac{9}{48}i+\frac{1}{12}j\right)\times \left(\frac{15}{17}i+\frac{8}{17}j\right) \\ =\left(-\frac{9}{48}i+\frac{1}{12}j\right)\times \left(\frac{15}{17}i+\frac{8}{17}j\right) \\ =\left(\frac{-9\times 15}{48\times 17}+\frac{8}{12\times 17}\right) \\ =\left(\frac{-135}{16\times 3\times 17}+\frac{2}{3\times 17}\right) \\ =\frac{1}{3\times 17}\left(\frac{-135}{16}+2\right) \\ \mathrm{\nabla }\mathrm{f}\left(\mathrm{P}\right)\times \mathrm{u}=-\frac{103}{816} \end{gathered}$$