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

Mar 19, 2016

Equation of the line normal to $f \left(x\right)$ at $x = \frac{\pi}{3}$ will be
$\left(y - {\pi}^{2} / 27\right) = \frac{27 \sqrt{3}}{8 {\pi}^{2} - 6 \sqrt{3} \pi} \left(x - \frac{\pi}{3}\right)$

#### Explanation:

As $f \left(x\right) = {x}^{2} {\cot}^{2} x$, $f \left(\frac{\pi}{3}\right) = {\left(\frac{\pi}{3}\right)}^{2} {\cot}^{2} \left(\frac{\pi}{3}\right)$ or

$f \left(\frac{\pi}{3}\right) = {\left(\pi\right)}^{2} / 9 \times {\left(\frac{1}{\sqrt{3}}\right)}^{2} = {\pi}^{2} / 27$

Now $\frac{\mathrm{df}}{\mathrm{dx}} = 2 x {\cot}^{2} x + {x}^{2} \times 2 \cot x \times \left(- {\csc}^{2} x\right)$

= $2 x {\cot}^{2} x - 2 {x}^{2} \cot x {\csc}^{2} x$

At $x = \frac{\pi}{3}$, $\frac{\mathrm{df}}{\mathrm{dx}} = 2 \frac{\pi}{3} {\cot}^{2} \left(\frac{\pi}{3}\right) - 2 {\left(\frac{\pi}{3}\right)}^{2} \cot \left(\frac{\pi}{3}\right) {\csc}^{2} \left(\frac{\pi}{3}\right)$

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

= $2 \frac{\pi}{9} - 2 {\pi}^{2} / 9 \times \left(\frac{4}{3 \sqrt{3}}\right) = 2 \frac{\pi}{9} - \frac{8 {\pi}^{2}}{27 \sqrt{3}}$

= $\frac{6 \sqrt{3} \pi - 8 {\pi}^{2}}{27 \sqrt{3}}$

This will be the slope of tangent and hence slope of normal would be $\frac{- 27 \sqrt{3}}{6 \sqrt{3} \pi - 8 {\pi}^{2}}$ or $\frac{27 \sqrt{3}}{8 {\pi}^{2} - 6 \sqrt{3} \pi}$

Hence equation of normal would be

$\left(y - {\pi}^{2} / 27\right) = \frac{27 \sqrt{3}}{8 {\pi}^{2} - 6 \sqrt{3} \pi} \left(x - \frac{\pi}{3}\right)$