# How do you find the polar equation to the tangent, at theta = pi/12, on r = sin 3theta?

Jan 24, 2017

So, $r = \frac{1}{4 \cos \theta - 2 \sin \theta}$. In cartesian form, this is $4 x - 2 y - 1 = 0$

#### Explanation:

The given equation represents a three-petal rose.

The equation to the tangent

at $P \left(f \left(\alpha\right) , \alpha\right)$, on $r = f \left(\theta\right)$ is

$r \left(\sin \theta - \tan \psi \cos \theta\right) = f \left(\alpha\right) \left(\sin \alpha - \tan \psi \cos \alpha\right)$,

where

$\tan \psi = m =$

$\left(\frac{\mathrm{dr}}{d \theta} \sin \alpha + r \cos \alpha\right)$$/ \left(\frac{\mathrm{dr}}{d \theta} \cos \theta - r \sin \theta\right)$ , at P.

Briefly,

$r \sin \left(\theta - \psi\right) = f \left(\alpha\right) \sin \left(\alpha - \psi\right)$

Here, $r = \sin 3 \theta \mathmr{and} \frac{\mathrm{dr}}{d \theta} = 3 \cos 3 \theta$.

P is$\left(\frac{1}{\sqrt{2}} , \frac{\pi}{12}\right)$,

m at P = $\frac{\frac{3}{2} + \frac{1}{2}}{\frac{3}{2} - \frac{1}{2}} = 2$.

Now, the equation to the tangent at P is

$r \left(\sin \theta - 2 \cos \theta\right) = \frac{1}{\sqrt{2}} \left(\frac{1}{\sqrt{2}} - \frac{2}{\sqrt{2}}\right) = - \frac{1}{2}$

So, $r = 1 / \left(4 \cos \theta - 2 \sin \theta\right)$

In cartesian form, this is 4x-2y-1=0#