Convert the polar equation to rectangular form.

[Lebombo]

Homework Statement

r = 3sin\theta

since

x= rcos\theta

x = 3sin\thetacos\theta

and since:

y = rsin\theta

y = 3sin^2\theta


Then I'm sort of stuck..

 

[LCKurtz]

Homework Statement

r = 3sin\theta

since

x= rcos\theta

x = 3sin\thetacos\theta

and since:

y = rsin\theta

y = 3sin^2\theta


Then I'm sort of stuck..

You are making it way too difficult. Put $\sin\theta =\frac y r$ in your first equation then see if you can get an xy equation from that.

 

[Dick]

Homework Statement

r = 3sin\theta

since

x= rcos\theta

x = 3sin\thetacos\theta

and since:

y = rsin\theta

y = 3sin^2\theta


Then I'm sort of stuck..

Try putting sinθ=y/r.

 

[Lebombo]

You are making it way too difficult. Put $\sin\theta =\frac y r$ in your first equation then see if you can get an xy equation from that.

Like so:

r = 3sinθ becomes (r=3\frac{y}{r})

= (r =\sqrt{3y}) ?


or solving for y, (y = \frac{r^2}{3})



or do you mean something like this:



(\frac{y}{r} = sinθ) becomes (\frac{y}{3sinθ} = sinθ)


= (y = 3sin^{2}θ)

 

[Dick]

Like so:

r = 3sinθ becomes (r=3\frac{y}{r})

= (r =\sqrt{3y}) ?or solving for y, (y = \frac{r^2}{3})
or do you mean something like this:
(\frac{y}{r} = sinθ) becomes (\frac{y}{3sinθ} = sinθ)= (y = 3sin^{2}θ)

You also want to use $r^2=x^2+y^2$. You need to express both $sin(\theta)$ and r in terms of x and y. Try once more.