Question

What is the Cartesian form of `r-5theta = sin^3theta+sectheta`?

Answer 1

`sqrt(x^2+y^2)-5sin^-1(y/sqrt(x^2+y^2))=(y/sqrt(x^2+y^2))^3+sqrt(x^2+y^2)/x`

Explanation:

To transform an expression in polar form to Cartesian (or rectangular) form, we need the formulas that relate `r` and `theta` to `x` and `y`.

Step 1. Recall the relationships between polar and Cartesian.

`x^2+y^2=r^2`

`x=rcos(theta)` or `cos(theta)=x/r` or `theta=cos^-1(x/r)`

`y=rsin(theta)` or `sin(theta)=y/r` or `theta=sin^-1(y/r)`

`tan(theta)=y/x`

These relationships can be determined from drawing a point in one of two ways, as radius and angle or as `x`-coordinate and `y`-coordinate as seen below.

Step 2. Convert each term from polar to Cartesian.
We are given `r-5theta=sin^3(theta)+sec(theta)`. Let's take each part and put it terms of `x` and `y`.

`r=sqrt(x^2+y^2)`

`5theta=5sin^-1(y/r)`

Plugging `r=sqrt(x^2+y^2)` into the previous gives

`5theta=5sin^-1(y/sqrt(x^2+y^2))`

`sin^3(theta)=(y/r)^3=(y/sqrt(x^2+y^2))^3`

`sec(theta)=1/cos(theta)=r/x=sqrt(x^2+y^2)/x`

Step 3. Plug these back into the original.

`r-5theta=sin^3(theta)+sec(theta)` becomes

`sqrt(x^2+y^2)-5sin^-1(y/sqrt(x^2+y^2))=(y/sqrt(x^2+y^2))^3+sqrt(x^2+y^2)/x`

You could spend a lot of time trying to simplify and trying to find a function `y` in terms of the independent variable `x`, but this is not what your question asked. So this is a great place to stop!