# What is the equation of the tangent line of r=tan^2theta - sintheta at theta=pi/4?

Feb 8, 2017

0.1197y-0.2143x-0.0103=0.

#### Explanation:

The period is $2 \pi$. What is seen in the graph is redrawn

periodically.

For $\theta \in \left[0 , \frac{\pi}{2}\right) , \left(\frac{\pi}{2} , \pi\right] , \left[\pi , \frac{3}{2} \pi\right) \mathmr{and} \left(\frac{3}{2} \pi , 2 \pi\right]$, it is

in ${Q}_{1} , {Q}_{2} , {Q}_{3} \mathmr{and} {Q}_{4}$, respectively, Pole is not a node.

The two infinitely long vertical loops touch here.

$\theta = \frac{\pi}{2} \uparrow \mathmr{and} \frac{3}{2} \pi \downarrow$ represent asymptotes.

graph{(x^2(x^2+y^2)-y^2sqrt(x^2+y^2)-x^2y)((x-.207)^2+(y-.207)^2-.02)=0 [-10, 10, -5, 5]}

The equation to the tangent at $P \left(a , \alpha\right)$ is

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

where $\psi$ is the inclination of the tangent to $\theta = 0$,

and is given by,

$m = \tan \psi = \frac{r ' \sin \theta + r \cos \theta}{r ' \cos \theta - r \sin \theta}$,

at $\left(a , \alpha\right) \mathmr{and} r ' = \frac{\mathrm{dr}}{d \theta}$.

Here, the point of contact of the tangent P is

$\left(a , \alpha\right) = \left(1 - \frac{1}{\sqrt{2}} , \frac{\pi}{4}\right) = \left(0.2929 , 0.7854\right)$, nearly..

$r ' = 2 \tan \theta {\sec}^{2} \theta - \cos \theta = 4 - \frac{1}{\sqrt{2}} = 3.293$, nearly.

$\tan \psi = \left(\frac{r ' \sin \theta + r \cos \theta}{r ' \cos \theta - r \sin \theta}\right)$

$= \frac{\left(4 - \frac{1}{\sqrt{2}}\right) \left(\frac{1}{\sqrt{2}}\right) + \left(1 - \frac{1}{\sqrt{2}}\right) \left(\frac{1}{\sqrt{2}}\right)}{\left(4 - \frac{1}{\sqrt{2}}\right) \left(\frac{1}{\sqrt{2}}\right) - \left(1 - \frac{1}{\sqrt{2}}\right) \left(\frac{1}{\sqrt{2}}\right)}$

$= \frac{5 \sqrt{2} - 2}{3 \sqrt{2}} = 1.195$, nearly.

$\psi = {\tan}^{- 1} \left(1.195\right) = {50.08}^{o} = 0.8740$ radian.

So, the equation to the tangent is

rsin(theta-0.8740)=0.2929sin(0.7854-0.8740)

=-0.2929sin(0.0886 radian)=-0.2929sin5.076^o)

$= - 0.0259$, nearly, giving

$r \sin \left(\theta - 0.8740\right) = - 0.0259$, with $\theta$ in radian. or

rsin(theta^o-50.08^o)=-0.0259

Expanding and converting to Cartesian form, this is

0.6417y-0.7889x+0.0259=0.

The Socratic twin graph is for the Cartesian frame.

See the graph below.

graph{(x^2(x^2+y^2)-y^2sqrt(x^2+y^2)-x^2y)(0.6417y-0.7889x+0.0259)((x--.207)^2+(y-.207)^2-.001)=0 [-139.4, 139.1, -66.7, 72.5]}