Q. 37 - Calculus

Question

In Exercises $35-38$, find the directional derivative of the given

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

given vector $v$.

$f(x,y)=\sqrt{\frac{y}{x}}\text{ at }P=(1,16),\mathbf{v}=⟨2,-1⟩$

Step-by-Step Solution

Answer

The directional derivative function is $\frac{-33}{40}\sqrt{5}$

Step 1: Given data

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

$\mathrm{P}=\left(x_0,y_0\right)=(1,16)\text{ and }v=(2,-1)$

Step 2: Solution

Therefore

$\parallel v\parallel =\sqrt{2^2+1^2}=\sqrt{5}$

$u=(\alpha ,\beta )=\left(\frac{2\sqrt{5}}{5},\frac{-\sqrt{5}}{5}\right)$

Directional derivative of function is given by

$\mathrm{\nabla }f\left(P\right)\cdot u=\mathrm{\nabla }f(1,16)\cdot u=\left({\left.\frac{df}{dx}\right|}_{(1,16)}i+{\left.\frac{df}{dy}\right|}_{(1,16)}j\right)\cdot \left(\frac{2\sqrt{5}}{5}i-\frac{\sqrt{5}}{5}j\right)$ 

$=\left[{\left(\frac{1}{2\sqrt{\frac{y}{x}}}\frac{d}{dx}\left(\frac{y}{x}\right)\right]}_{(1,16)}i+{\left(\frac{1}{2\sqrt{\frac{y}{x}}\frac{d}{dx}\left(\frac{y}{x}\right)}\right)}_{(1,16)}^j\times \left(\frac{2\sqrt{5}}{5}i-\frac{\sqrt{5}}{5}j\right)\right.$

$=\left[{\left(\frac{\frac{-y}{x^2}}{2\sqrt{\frac{y}{x}}}\right)}_{(1,16)}i+{\left(\frac{\frac{1}{x}}{2\sqrt{\frac{y}{x}}}\right)}_{(1,16)}j\right]\times \left(\frac{2\sqrt{5}}{5}i-\frac{\sqrt{5}}{5}j\right)$

  $=\left[\left(\frac{\frac{-16}{1^2}}{2\sqrt{\frac{16}{1}}}\right)i+\left(\frac{\frac{1}{1}}{2\sqrt{\frac{16}{1}}}\right)j\right]\times \left(\frac{2\sqrt{5}}{5}i-\frac{\sqrt{5}}{5}j\right)$

$=\left[\left(\frac{-16}{8}\right)i+\left(\frac{1}{8}\right)j\right]\times \left(\frac{2\sqrt{5}}{5}i-\frac{\sqrt{5}}{5}j\right)$   

$=\left(-2i+\frac{1}{8}j\right)\times \left(\frac{2\sqrt{5}}{5}i-\frac{\sqrt{5}}{5}j\right)$    

Step 3

$=\left(\frac{-2\times 2\sqrt{5}}{5}-\frac{\sqrt{5}}{8\times 5}\right)$

$=\frac{\sqrt{5}}{5}\left(-4-\frac{1}{8}\right)$

$=\frac{\sqrt{5}}{5}\left(-\frac{33}{8}\right)$

$\mathrm{\nabla }f\left(P\right)\cdot u=\frac{-33}{40}\sqrt{5}$