Question

What is the Cartesian form of `(r+3)^2 = sin^3theta-sec^2theta`?

Answer 1

`(r+3)^2=sin^3theta-sec^2theta` when expressed in catesian form becomes

`(3+sqrt(x^2+y^2))^2=(y/sqrt(x^2+y^2))^3-(1+(y/x)^2)`
which can be simplified as required

Explanation:

`x=rcostheta`

`y=rsintheta`

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

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

`sec^2theta=1+tan^2theta`
`tan^2theta=(tantheta)^2`
`theta=tan^-1(y/x)`
`tantheta=y/x`
`tan^2theta=(y/x)^2`
`sec^2theta=1+(y/x)^2`
Now, we have
`(r+3)^2=(3+sqrt(x^2+y^2))^2`
`sin^3theta=(y/sqrt(x^2+y^2))^3`
`sec^2theta=1+(y/x)^2`

Thus,
`(r+3)^2=sin^3theta-sec^2theta` when expressed in catesian form becomes

`(3+sqrt(x^2+y^2))^2=(y/sqrt(x^2+y^2))^3-(1+(y/x)^2)`
which can be simplified as required