Question

What is the Cartesian form of `r-theta = -2cos^3theta-cot^2theta`?

Answer 1

The presence of `theta` forces a split into three equations. Please see the explanation for the three equations.

Explanation:

Because the cotangent function has a divide by zero problem when `theta = 0 or pi`, this translates to the Cartesian restriction `y!=0`

Add `theta` to both sides:

`r = theta - 2cos^3(theta) - cot^2(theta)`

Write `cot^2(theta)` as `(cos^2(theta))/(sin^2(theta))`

`r = theta - 2cos^3(theta) - (cos^2(theta))/(sin^2(theta))`

Multiply both sides by `r^5sin^2(theta)`:

`r^6sin^2(theta) = thetar^3(r^2sin^2(theta)) - 2r^3cos^3(theta)r^2(sin^2(theta)) - r^5cos^2(theta)`

Substitute y for `rsin(theta)` and x for `rcos(theta)`

`r^4y^2 = thetar^3y^2 - 2x^3y^2 - r^3x^2; y!=0`

Use `(x^2 + y^2)^(1/2) = r` to make substitutions in accordance with the power of r:

`(x^2 + y^2)^(2)y^2 = theta(x^2 + y^2)^(3/2)y^2 - 2x^3y^2 - (x^2 + y^2)^(3/2)x^2; y!=0`

Collect like terms:

`(x^2 + y^2)^(2)y^2 = (x^2 + y^2)^(3/2)(thetay^2 - x^2) - 2x^3y^2; y!=0`

The substitution for `theta` breaks the above into 3 equations:

`(x^2 + y^2)^(2)y^2 = (x^2 + y^2)^(3/2)(tan^-1(y/x)y^2 - x^2) - 2x^3y^2; y > 0 and x > 0`

`(x^2 + y^2)^(2)y^2 = (x^2 + y^2)^(3/2)(tan^-1(y/x) + pi)y^2 - x^2) - 2x^3y^2; y != 0 and x < 0`

`(x^2 + y^2)^(2)y^2 = (x^2 + y^2)^(3/2)((tan^-1(y/x) + 2pi)y^2 - x^2) - 2x^3y^2; y < 0 and x > 0`