Convert the polar equation to rectangular form.

[Lebombo]

Homework Statement

r = 3sin$\theta$

since

x= rcos$\theta$

x = 3sin$\theta$cos$\theta$

and since:

y = rsin$\theta$

y = 3$sin^2\theta$


Then I'm sort of stuck..

 

[LCKurtz]

Homework Statement

r = 3sin$\theta$

since

x= rcos$\theta$

x = 3sin$\theta$cos$\theta$

and since:

y = rsin$\theta$

y = 3$sin^2\theta$


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$\theta$cos$\theta$

and since:

y = rsin$\theta$

y = 3$sin^2\theta$


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.