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

Apr 1, 2017

The equation for the normal line of the function at $x = \frac{\pi}{3}$ is $y - \frac{\pi \sqrt{3}}{9} = \frac{9}{4 \pi - 3 \sqrt{3}} \left(x - \frac{\pi}{3}\right)$, as given in point-slope form.

#### Explanation:

To find the equation of the line normal to the function, we first find the slope of the tangent line.

Find the derivative of the function, to give us the slope of the tangent line. Using the quotient rule for derivatives, we get:
$f ' \left(x\right) = \frac{\tan x - x {\sec}^{2} x}{\tan} ^ 2 x$

At $x = \frac{\pi}{3}$, we get:
$f ' \left(\frac{\pi}{3}\right) = \frac{\tan \left(\frac{\pi}{3}\right) - \left(\frac{\pi}{3}\right) {\sec}^{2} \left(\frac{\pi}{3}\right)}{\tan} ^ 2 \left(\frac{\pi}{3}\right)$
$f ' \left(\frac{\pi}{3}\right) = \frac{\sqrt{3} - \frac{4 \pi}{3}}{3}$

$f ' \left(\frac{\pi}{3}\right) = \frac{3 \sqrt{3} - 4 \pi}{9}$, which is the slope of the tangent line at the given point.

To find the slope of the normal line, ${m}_{2}$, we need the negative reciprocal of the slope of the tangent line, ${m}_{1}$.
So, ${m}_{2} = - \frac{1}{m} _ 1$.

Using the above equation, we get that the slope of the normal line at $x = \frac{\pi}{3}$ is:
$\frac{9}{4 \pi - 3 \sqrt{3}}$

We need the point coordinates at $x = \frac{\pi}{3}$:
$y = \frac{\frac{\pi}{3}}{\tan} \left(\frac{\pi}{3}\right) = \frac{\frac{\pi}{3}}{\sqrt{3}} = \frac{\pi \sqrt{3}}{9}$.
So, the point is $\left(\frac{\pi}{3} , \frac{\pi \sqrt{3}}{9}\right)$

Now we can use point-slope form, $y - {y}_{1} = {m}_{2} \left(x - {x}_{1}\right)$.
The equation for the normal line is:
$y - \frac{\pi \sqrt{3}}{9} = \frac{9}{4 \pi - 3 \sqrt{3}} \left(x - \frac{\pi}{3}\right)$