Q. 26 - 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.

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

Step-by-Step Solution

Answer

The directional derivative is $=-\frac{103}{816}$

Step 1: Given data

The function is$f(x,y)=\sqrt{\frac{y}{x}}$ 

The given points is $P=\left(x_0,y_0\right)=(4,9)\text{ and }u=(\alpha ,\beta )=\left(\frac{15}{17}i+\frac{8}{17}j\right)$ 

Step 2: Solution

Directional derivative of function 

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

$=\left[{\left(\frac{1}{2\sqrt{\frac{y}{x}}}\frac{d}{dx}\left(\frac{y}{x}\right)\right)}_{(4,9)}i+{\left(\frac{1}{2\sqrt{\frac{y}{x}}}\frac{d}{dx}\left(\frac{y}{x}\right)\right)}_{(4,9)}j\right]\cdot \left(\frac{15}{17}i+\frac{8}{17}j\right)$ 

$={\left[\frac{\frac{-y}{x^2}}{2\sqrt{\frac{y}{x}}}\right)}_{(4,9)}i+{\left(\frac{\frac{1}{x}}{2\sqrt{\frac{y}{x}}}\right)}_{(4,9)}j\cdot \left(\frac{15}{17}i+\frac{8}{17}j\right)$ 

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

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

Step 3

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

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

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

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

$=\frac{1}{3\times 17}\left(\frac{-103}{16}\right)$ 

$=-\frac{103}{816}$