# What is the equation of the line normal to f(x)= x^2sin^2(2x)  at x=pi/6?

Sep 29, 2016

$y - {\pi}^{2} / 48 = - \frac{36}{\sqrt{3} {\pi}^{2} + 9 \pi} \left(x - \frac{\pi}{6}\right)$

#### Explanation:

Let's begin by finding the y coordinate corresponding to the given x coordinate:

$y = {\left(\frac{\pi}{6}\right)}^{2} {\sin}^{2} \left(2 \frac{\pi}{6}\right)$

$y = \left({\pi}^{2} / 36\right) \left(\frac{3}{4}\right) = {\pi}^{2} / 48$

The slope, n, of the normal line is:

$n = - \frac{1}{m}$ where m is the slope of the tangent line.

To obtain the slope of the tangent line, we must compute the first derivative:

$f ' \left(x\right) = 2 x S \in \left(2 x\right) \left(2 x C o s \left(2 x\right) + S \in \left(2 x\right)\right)$

Evaluate at $x = \frac{\pi}{6}$:

$m = f ' \left(\frac{\pi}{6}\right) = 2 \left(\frac{\pi}{6}\right) S \in \left(2 \frac{\pi}{6}\right) \left(2 \left(\frac{\pi}{6}\right) C o s \left(2 \frac{\pi}{6}\right) + S \in \left(2 \frac{\pi}{6}\right)\right)$

$m = 2 \left(\frac{\pi}{6}\right) \left(\frac{\sqrt{3}}{2}\right) \left(\left(\frac{\pi}{6}\right) + \frac{\sqrt{3}}{2}\right)$

$m = \sqrt{3} \left(\frac{\pi}{6}\right) \left(\left(\frac{\pi}{6}\right) + \frac{\sqrt{3}}{2}\right)$

$m = \sqrt{3} {\left(\frac{\pi}{6}\right)}^{2} + \frac{\pi}{4}$

$m = \frac{\sqrt{3} {\pi}^{2} + 9 \pi}{36}$

$n = - \frac{36}{\sqrt{3} {\pi}^{2} + 9 \pi}$

Using the point-slope form of the equation of a line:

$y - {y}_{1} = n \left(x - {x}_{1}\right)$

$y - {\pi}^{2} / 48 = - \frac{36}{\sqrt{3} {\pi}^{2} + 9 \pi} \left(x - \frac{\pi}{6}\right)$