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

Feb 18, 2018

(y-(pi)^2/6)/(x-pi/3)=(18sqrt3)/(pi(18+sqrt3)

#### Explanation:

$\text{Let } y = - 3 {x}^{2} \sin x$

$\text{Let} u = {x}^{2} , v = \sin x$

$y = - u v$

$\text{Differentiating wrt x on both sides}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(y\right) = - \frac{d}{\mathrm{dx}} \left(u v\right)$

$\text{By product rule} , \frac{d}{\mathrm{dx}} \left(u v\right) = u \frac{\mathrm{dv}}{\mathrm{dx}} + v \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{du}}{\mathrm{dx}} = 2 x , \frac{\mathrm{dv}}{\mathrm{dx}} = \cos x$

$\frac{d}{\mathrm{dx}} \left(u v\right) = {x}^{2} \cos x + \left(\sin x\right) \left(2 x\right)$

$\frac{d}{\mathrm{dx}} \left(u v\right) = {x}^{2} \cos x + 2 x \sin x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \left({x}^{2} \cos x + 2 x \sin x\right)$

$x = \frac{\pi}{3}$

$y = - 3 {x}^{2} \sin x = y = - 3 {\left(\frac{\pi}{3}\right)}^{2} \sin \left(\frac{\pi}{3}\right)$

$y = - 3 {\left(\pi\right)}^{2} / 9 \times \frac{1}{2}$

$y = - {\left(\pi\right)}^{2} / 6$

$P \equiv \left(x , y\right) = \left(\frac{\pi}{3} , - {\left(\pi\right)}^{2} / 6\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \left({x}^{2} \cos x + 2 x \sin x\right)$
$= - \left({\left(\frac{\pi}{3}\right)}^{2} \cos \left(\frac{\pi}{3}\right) + 2 \left(\frac{\pi}{3}\right) \sin \left(\frac{\pi}{3}\right)\right)$

$= - \left({\pi}^{2} / 9 \times \frac{1}{2} + \frac{2 \pi}{3} \times \frac{\sqrt{3}}{2}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \left({\left(\pi\right)}^{2} / 18 + \frac{\pi}{\sqrt{3}}\right)$

$\text{Slope of the tangent at "P-=(pi/3,-(pi)^2/6) "is}$

$m = \frac{\mathrm{dy}}{\mathrm{dx}} = - \left({\left(\pi\right)}^{2} / 18 + \frac{\pi}{\sqrt{3}}\right)$

$\text{Slope of the normal m' is given by } m ' = - \frac{1}{m}$

$m ' = \frac{1}{{\left(\pi\right)}^{2} / 18 + \frac{\pi}{\sqrt{3}}} = \frac{1}{\frac{\sqrt{3} {\left(\pi\right)}^{2} + 18 \pi}{18 \sqrt{3}}}$

m'=(18sqrt3)/(sqrt3(pi)^2+18pi)=(18sqrt3)/(pi(sqrt3+18)

m'=(18sqrt3)/(pi(18+sqrt3)

$\text{Equation of the normal passing through the point ", P-=(pi/3,-(pi)^2/6) "and having the slope ", m'=(18sqrt3)/(pi(18+sqrt3) }$

$\text{is given by}$

(y-(pi)^2/6)/(x-pi/3)=(18sqrt3)/(pi(18+sqrt3)