Question

What is the Cartesian form of `r^2+theta = -sin^2theta+4cot^3theta`?

Answer 1

Please see the explanation, because this breaks into 3 equations.

Explanation:

Multiply both sides by `r^5sin^3(theta)`:

`r^7sin^3(theta) + (theta)r^5sin^3(theta) = 4r^5cos^3(theta) - r^5sin^5(theta)`

Do some regrouping:

`r^4(rsin(theta))^3 + (theta)r^2(rsin(theta))^3 = 4r^2(rcos(theta))^3 - (rsin(theta))^5`

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

`r^4y^3 + (theta)r^2y^3 = 4r^2x^3 - y^5`

Substitute `(x^2 + y^2)` for `r^2`

`(x^2 + y^2)^2y^3 + (theta)(x^2 + y^2)y^3 = 4(x^2 + y^2)x^3 - y^5`

We cannot merely substitute `tan^-1(y/x)` for `theta`; we must specify that `x!=0 and y!=0` and then handle special cases for `x < 0`, `x > 0 and y>=0`, and `x > 0 and y < 0`

For `x > 0 and y>0`:

`(x^2 + y^2)^2y^3 + tan^-1(y/x)(x^2 + y^2)y^3 = 4(x^2 + y^2)x^3 - y^5`

For `x < 0`:

`(x^2 + y^2)^2y^3 + (tan^-1(y/x) + pi)(x^2 + y^2)y^3 = 4(x^2 + y^2)x^3 - y^5`

For `x > 0 and y < 0`

`(x^2 + y^2)^2y^3 + (tan^-1(y/x) + 2pi)(x^2 + y^2)y^3 = 4(x^2 + y^2)x^3 - y^5`