Q. 25 - Calculus

Question

In Exercises 21–28, find the directional derivative of the given

function at the specified point P and in the direction of the

given unit vector u.

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

Step-by-Step Solution

Answer

The directional derivative of the given

function is$-\frac{7}{816}\sqrt{17}$  

Step 1: Given data

To find the directional derivative function  $f(x,y)=\sqrt{\frac{y}{x}}$ 

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

Step 2: Solution

the directional derivative of  

point P and in the direction of the given unit vector u given by  

 $\mathrm{\nabla }f\left(P\right)\cdot u=\mathrm{\nabla }f(4,9)\cdot u$ 

$=\left({\left.\frac{df}{dx}\right|}_{(4,9)}i+{\left.\frac{df}{dy}\right|}_{(4,9)}j\right)\cdot \left(-\frac{\sqrt{17}}{17}i-\frac{4\sqrt{17}}{17}j\right)$ 

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

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

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

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

$=\left(-\frac{9}{48}i+\frac{1}{12}j\right)\cdot \left(-\frac{\sqrt{17}}{17}i-\frac{4\sqrt{17}}{17}j\right)$

   

Step 3

$=\left(\frac{9\cdot \sqrt{17}}{48\cdot 17}-\frac{4\sqrt{17}}{12\cdot 17}\right)$ 

$=\left(\frac{9\cdot \sqrt{17}}{16\cdot 3\cdot 17}-\frac{\sqrt{17}}{3\cdot 17}\right)$ 

$=\frac{1}{3\cdot 17}\left(\frac{9\cdot \sqrt{17}}{16}-\frac{\sqrt{17}}{1}\right)$ 

$=\frac{1}{3\cdot 17}\left(\frac{9\cdot \sqrt{17}-16\sqrt{17}}{16}\right)$ 

$\mathrm{\nabla }f\left(P\right)\cdot u=-\frac{7}{816}\sqrt{17}$