# What is the equation of the line normal to  f(x)=x^2sinx at  x=pi/3?

Jul 10, 2018

$y = - \frac{18}{6 \cdot \sqrt{3} \cdot \pi + {\pi}^{2}} \cdot x + {\pi}^{2} / \left(6 \cdot \sqrt{3}\right) + \frac{6}{6 \cdot \sqrt{3} + \pi}$

#### Explanation:

We have given

$f \left(x\right) = {x}^{2} \sin \left(x\right)$
then we get by the product rule

$\left(u v\right) ' = u ' v + u v '$
$f ' \left(x\right) = 2 x \sin \left(x\right) + {x}^{2} \cos \left(x\right)$

$f ' \left(\frac{\pi}{3}\right) = 2 \cdot \frac{\pi}{3} \cdot \sin \left(\frac{\pi}{3}\right) + {\left(\frac{\pi}{3}\right)}^{2} \cos \left(\frac{\pi}{3}\right)$
$f ' \left(\frac{\pi}{3}\right) = \frac{\pi}{\sqrt{3}} + {\pi}^{2} / 18$
Note that $\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} , \cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$

so the slope of the normal line is given by

$m = - \frac{18}{6 \cdot \sqrt{3} \cdot \pi + {\pi}^{2}}$
We have
$f \left(\frac{\pi}{3}\right) = {\pi}^{2} / \left(6 \cdot \sqrt{3}\right)$
the searched equation hase the form

$y = m x + n$
We know $m$ and $n$ is given by

${\pi}^{2} / \left(6 \cdot \sqrt{3}\right) + \frac{6}{6 \cdot \sqrt{3} + \pi} = n$