Q. 37 - Calculus

Question

In Exercise $35-38$, find the directional derivative of the given function at the specified point $P$ and in the direction of the given vector $v$.

$f(x,y)=\sqrt{\frac{y}{x}}at P=(1,16), v=(2,1)$

Step-by-Step Solution

Answer

The directional derivative is $\nabla f\left(P\right).u=\frac{-33}{40}\sqrt{5}$.

Step 1: Directional derivative.

For a given function$P=(x_0,y_0)=(1,16)$ and $v=(2,-1)$, we must find the directional derivative $f(x,y)=\sqrt{\frac{y}{x}}$.

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

Therefore;

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

Step 2: Directional unit vector.

The function's directional derivative at point $P$ with directional unit vector $u$ is,

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

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

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

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

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

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

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

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

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

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