Q. 36 - 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 $v$.

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

Step-by-Step Solution

Answer

The directional derivative function is $\frac{44}{9\sqrt{53}}$

Step 1: Given data

The function is  $f(x,y)=\frac{x}{y^2}$

$P=\left(x_0,y_0\right)=(9,-3)\text{ and }v=2i+7j$ 

Step 2: Solution

Therefore

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

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

Direction derivative function of point is given by

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

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

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

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

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

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

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