Question

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

Answer 1

0.1197y-0.2143x-0.0103=0.
I have given my best answer. For this purpose, I had made self-corrections,

Explanation:

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

periodically.

For `theta in [0, pi/2), (pi/2, pi], [pi, 3/2pi) and (3/2pi, 2pi]`, it is

in `Q_1,Q_2, Q_3 and Q_4`, respectively, Pole is not a node.

The two infinitely long vertical loops touch here.

`theta = pi/2 uarr and 3/2pi darr` 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( a, alpha )` is

`rsin(theta-psi)=asin(alpha-psi)`,

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

and is given by,

`m=tanpsi=(r'sintheta+rcostheta)/(r'costheta-rsintheta)`,

at `(a, alpha) and r'=(dr)/(d theta)`.

Here, the point of contact of the tangent P is

`(a, alpha)=(1-1/sqrt2, pi/4)=(0.2929, 0.7854)`, nearly..

`r'=2tantheta sec^2theta-costheta=4-1/sqrt2=3.293`, nearly.

`tanpsi=((r'sintheta+rcostheta)/(r'costheta-rsintheta))`

`= ((4-1/sqrt2)(1/sqrt2)+(1-1/sqrt2)(1/sqrt2))/((4-1/sqrt2)(1/sqrt2)-(1-1/sqrt2)(1/sqrt2))`

`=(5sqrt2-2)/(3sqrt2)=1.195`, nearly.

`psi=tan^(-1)(1.195)=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

`rsin(theta-0.8740)=-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]}