Q. 36 - 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 $v$.

$f(x,y)=\frac{x}{y^2}\text{ at }P=(9,-3),\mathbf{v}=2\mathbf{i}+7\mathbf{j}$

Step-by-Step Solution

Answer

The directional derivative of the function is $\nabla f\left(P\right).u=\frac{44}{9\sqrt{53}}$.

Step 1: Directional derivation.

For a given function $P=(x_0,y_0)=(9,-3)$ and $v=2i+7j$, we must find the directional derivative $f(x,y)= \frac{x}{y^2}$.

$||v||=\sqrt{2^2+7^2 } =\sqrt{53}$

$\therefore u=(\alpha ,\beta )=\left(\frac{2}{\sqrt{53}},\frac{7}{\sqrt{53}}\right)$

Step 2: Directional unit vector.

The directional derivative of a variable at point $P$ with directional unit vector u is calculated as follows:

$\mathrm{\nabla }f\left(P\right)\cdot u=\mathrm{\nabla }f(9,-3)\times u=\left({\left.\frac{df}{dx}\right|}_{(9,-3)}i+{\left.\frac{df}{dy}\right|}_{(9,-3)}j\right)\times \left(\frac{2}{\sqrt{53}}i+\frac{7}{\sqrt{53}}j\right)$

$=\left[{\left(\frac{1}{y^2}\right)}_{(9,-3)}i+{\left(\frac{-2x}{y^3}\right)}_{(9,-3)}j\times \left(\frac{2}{\sqrt{53}}i+\frac{7}{\sqrt{53}}j\right)\right.$

$=\left(\frac{1}{9}i+\frac{18}{27}j\right)\times \left(\frac{2}{\sqrt{53}}i+\frac{7}{\sqrt{53}}j\right)$

$=\left(\frac{1.2}{9.\sqrt{53}}+\frac{18.7}{27.\sqrt{53}}\right)$

$=\left(\frac{2}{9.\sqrt{53}}+\frac{126}{27.\sqrt{53}}\right)$

$=\frac{2}{9.\sqrt{53}}\left(1+\frac{63}{3}\right)$

$\nabla f\left(P\right).u=\frac{44}{9\sqrt{53}}$