# What is the slope of the line normal to the tangent line of f(x) = sin(3x-pi)  at  x= pi/3 ?

Nov 28, 2016

$y = - \frac{1}{3} x + \frac{\pi}{9}$

#### Explanation:

We start by finding the corresponding $y$ coordinate.

$f \left(\frac{\pi}{3}\right) = \sin \left(3 \left(\frac{\pi}{3}\right) - \pi\right)$

$f \left(\frac{\pi}{3}\right) = \sin \left(\pi - \pi\right)$

$f \left(\frac{\pi}{3}\right) = \sin \left(0\right)$

$f \left(\frac{\pi}{3}\right) = 0$

We now differentiate. We let $y = \sin u$ and $u = 3 x - \pi$.

Then $\frac{\mathrm{dy}}{\mathrm{du}} = \cos u$ and $\frac{\mathrm{du}}{\mathrm{dx}} = 3$, since $\pi$ is but a constant.

By the chain rule,

$f ' \left(x\right) = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$

$f ' \left(x\right) = \cos u \times 3$

$f ' \left(x\right) = 3 \cos \left(3 x - \pi\right)$

We now determine the slope of the tangent.

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

$f ' \left(\frac{\pi}{3}\right) = 3 \cos \left(0\right)$

$f ' \left(\frac{\pi}{3}\right) = 3$

The normal line is perpendicular to the tangent, so the slope will be the negative reciprocal.

${m}_{\text{normal}} = - \frac{1}{3}$

We can now determine the equation.

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

$y - 0 = - \frac{1}{3} \left(x - \frac{\pi}{3}\right)$

$y = - \frac{1}{3} x + \frac{\pi}{9}$

Hopefully this helps!