Q. 34 - 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 $\mathrm{u}.$ Note that these are the same functions, points,

and vectors as in Exercises $21 -26$

$f(x,y)=\sqrt{\frac{y}{x}}$at$P=(4,9),\mathbf{u}=\frac{15}{17}\mathbf{i}+\frac{8}{17}\mathbf{j}$ 

Step-by-Step Solution

Answer

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

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

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,

$f_x\left(x_0,y_0\right)={\left.\frac{d}{dx}f(x,y)\right|}_{\left(x_0{v}_0\right)}={\left.\frac{d}{dx}\sqrt{\frac{y}{x}}\right|}_{\left(x_0,y_0\right)}$ 

$f_x(4,9)={\left.\frac{1}{2\sqrt{\frac{y}{x}}}\frac{d}{dx}\left(\frac{y}{x}\right)\right|}_{(4,9)}={\left.\frac{\frac{-y}{x^2}}{2\sqrt{\frac{y}{x}}}\right|}_{(4{)}^2}$    $\left(1\right)$

$f_x(4,9)=\frac{\frac{-9}{4^2}}{2\sqrt{\frac{9}{4}}}$ 

$f_x(4,9)=-\frac{9}{48} \left(2\right)$       

Step 3: Function of y

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

$f_y(4,9)={\left.\frac{1}{2\sqrt{\frac{y}{x}}}\frac{d}{dx}\left(\frac{y}{x}\right)\right|}_{(4\cdot )}={\left.\frac{\frac{1}{x}}{2\sqrt{\frac{y}{x}}}\right|}_{(4,9)}$ 

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

$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)$        

Step 4: Value of function

Substituting all equation on first equation,

We get,

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

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

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

Multiply$48$on both sides

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

$-9x+4y-48z=-76$ 

Step 5: 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,

$\mathrm{\nabla }f\left(P\right)\times u=\mathrm{\nabla }f(4,9)\times u=\left({\left.\frac{df}{dx}\right|}_{(4,9)}i+{\left.\frac{df}{dy}\right|}_{(4,9)}j\right)\times \left(\frac{15}{17}i+\frac{8}{17}j\right)$ $=\left[{\left(\frac{1}{2\sqrt{\frac{y}{x}}}\frac{d}{dx}\left(\frac{y}{x}\right)\right)}_{(4,9)}i+{\left(\frac{1}{2\sqrt{\frac{y}{x}}}\frac{d}{dx}\left(\frac{y}{x}\right)\right)}_{(4,9)}j\right]\cdot \left(\frac{15}{17}i+\frac{8}{17}j\right)$ 

Step 6: Calculation for equation of line tangent

$\left.\mathrm{\nabla }f\left(P\right)\times u={\left[\frac{\frac{-y}{x^2}}{2\sqrt{\frac{y}{x}}}\right)}_{\left(4.9\right)}i+{\left(\frac{\frac{1}{x}}{2\sqrt{\frac{y}{x}}}\right)}_{\left(4.9\right)}j\right]\times \left(\frac{15}{17}i+\frac{8}{17}j\right)$ 

$=\left[\left(\frac{\frac{-9}{4^2}}{2\sqrt{\frac{9}{4}}}\right)i+\left(\frac{\frac{1}{4}}{2\sqrt{\frac{9}{4}}}\right)j\right]\times \left(\frac{15}{17}i+\frac{8}{17}j\right)$ 

$=\left[\left(\frac{-\frac{9}{16}}{2\times \frac{3}{2}}\right)i+\left(\frac{\frac{1}{4}}{2\times \frac{3}{2}}\right)j\right]\times \left(\frac{15}{17}i+\frac{8}{17}j\right)$   

 $=\left(-\frac{9}{48}i+\frac{1}{12}j\right)\times \left(\frac{15}{17}i+\frac{8}{17}j\right)$ 

$=\left(-\frac{9}{48}i+\frac{1}{12}j\right)\times \left(\frac{15}{17}i+\frac{8}{17}j\right)$

$=\left(\frac{-9\times 15}{48\times 17}+\frac{8}{12\times 17}\right)$ 

$=\left(\frac{-135}{16\times 3\times 17}+\frac{2}{3\times 17}\right)$ 

$=\frac{1}{3\times 17}\left(\frac{-135}{16}+2\right)$

$\mathrm{\nabla }\mathrm{f}\left(\mathrm{P}\right)\times \mathrm{u}=-\frac{103}{816}$