Question

Convert to a rectangular equation? `r + rsintheta = 1`

Answer is `x^2 + 2y = 1`

Thanks in advance

Answer 1

`r + r sin theta = 1`

becomes

`x^2 + 2y = 1`

Explanation:

We know

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

`x = r cos theta`

`y = r sin theta`

so

`r + r sin theta = 1`

becomes

`\sqrt{x^2 + y^2 }+ y = 1`

`\sqrt{x^2 + y^2} = 1-y`

`x^2 + y^2 = 1 - 2y + y^2`

`x^2 + 2y = 1`

The only iffy step is the squaring of the square root. Usually for polar equations we allow negative `r`, and if so the squaring doesn't introduce a new part.

Answer 2

Procedure in explanation.

Explanation:

To convert from polar to rectangular, we may use the following substitutions: `x=rcosθ`
`y=rsinθ`
`r^2 =x^2+y^2`
`tanθ=y/x`
Using 1 and 3,
`sqrt(x^2 + y^2) + y = 1`
Square the equation. Using the expansion of `(a + b)^2`
`x^2 + y^2 +y^2 +2ysqrt(x^2 + y^2) = 1`
`implies x^2 + 2y^2 + 2ysqrt(x^2 + y^2) = 1`
`implies x^2 + 2y(y + sqrt(x^2 + y^2)) = 1`

Notice that the coefficient of 2y is 1.(See the first equation I wrote using 1 and 3)
So `x^2 + 2y = 1`

Hope this helps!

Answer 3

`x^2 - 2y = 1`

Explanation:

`r + rsintheta = 1`

We need to convert from polar to rectangular form.

We know that:
`x = rcostheta`

`y = rsintheta`

and

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

`------------------`

We can substitute in these values for `color(red)r` and `color(red)(rsintheta)`:
`color(red)(sqrt(x^2 + y^2) + y) = 1`

Subtract `color(red)y` from both sides of the equation:
`sqrt(x^2 + y^2) +y quadcolor(red)(-quady)= 1 quadcolor(red)(-quady)`

`sqrt(x^2 + y^2) = 1-y`

Square both sides of the equation:
`(sqrt(x^2 + y^2))^color(red)(2) = (1-y)^color(red)(2)`

`x^2 + y^2 = 1 - 2y + y^2`

Subtract `color(red)(y^2)` from both sides of the equation so they cancel:

`x^2 + cancel(y^2 quadcolor(red)(-quady^2)) = 1 - 2y + cancel(y^2 quadcolor(red)(-quady^2))`

`x^2 = 1 - 2y`

Add `color(red)(2y)` to both sides of the equation to get the final answer in rectangular form:
`x^2 - 2y = 1`

Hope this helps!