# What is the equation of the line that is normal to f(x)= -xsin^2x at  x=(4pi)/3 ?

Aug 19, 2017

$y + \pi = \frac{4 \sqrt{3}}{3 \sqrt{3} + 16 \pi} \left(x - \frac{4 \pi}{3}\right)$

#### Explanation:

The line intersects the function at:

$f \left(\frac{4 \pi}{3}\right) = - \frac{4 \pi}{3} {\sin}^{2} \left(\frac{4 \pi}{3}\right) = - \frac{4 \pi}{3} {\left(- \frac{\sqrt{3}}{2}\right)}^{2} = - \pi$

To find the slope of the normal line, first find the slope of the tangent line by differentiating the function. We'll need first the product rule, then the chain rule.

$f ' \left(x\right) = - \left(\frac{d}{\mathrm{dx}} x\right) {\sin}^{2} x - x \left(\frac{d}{\mathrm{dx}} {\sin}^{2} x\right)$

$\textcolor{w h i t e}{f ' \left(x\right)} = - {\sin}^{2} x - x \left(2 \sin x\right) \left(\frac{d}{\mathrm{dx}} \sin x\right)$

$\textcolor{w h i t e}{f ' \left(x\right)} = - {\sin}^{2} x - 2 x \sin x \cos x$

So the slope of the tangent line is:

$f ' \left(\frac{4 \pi}{3}\right) = - {\sin}^{2} \left(\frac{4 \pi}{3}\right) - 2 \left(\frac{4 \pi}{3}\right) \sin \left(\frac{4 \pi}{3}\right) \cos \left(\frac{4 \pi}{3}\right)$

$\textcolor{w h i t e}{f ' \left(\frac{4 \pi}{3}\right)} = - {\left(- \frac{\sqrt{3}}{2}\right)}^{2} - \frac{8 \pi}{3} \left(- \frac{\sqrt{3}}{2}\right) \left(- \frac{1}{2}\right)$

$\textcolor{w h i t e}{f ' \left(\frac{4 \pi}{3}\right)} = - \frac{3}{4} - \frac{4 \pi}{\sqrt{3}} = - \frac{3 \sqrt{3} + 16 \pi}{4 \sqrt{3}}$

The normal line is perpendicular to the tangent line, so their slopes will be opposite reciprocals. Thus, the slope of the normal line is:

$\frac{4 \sqrt{3}}{3 \sqrt{3} + 16 \pi}$

Since we know the slope of the normal line and a point that it passes through, $\left(\frac{4 \pi}{3} , - \pi\right)$, we can write the line's equation:

$y - {y}_{0} = m \left(x - {x}_{0}\right)$

color(blue)(y+pi=(4sqrt3)/(3sqrt3+16pi)(x-(4pi)/3)