Convert the polar equation to rectangular form.

[Lebombo]

Homework Statement

r = 3sin[itex]\theta[/itex]

since

x= rcos[itex]\theta[/itex]

x = 3sin[itex]\theta[/itex]cos[itex]\theta[/itex]

and since:

y = rsin[itex]\theta[/itex]

y = 3[itex]sin^2\theta[/itex]


Then I'm sort of stuck..

 

[LCKurtz]

Homework Statement

r = 3sin[itex]\theta[/itex]

since

x= rcos[itex]\theta[/itex]

x = 3sin[itex]\theta[/itex]cos[itex]\theta[/itex]

and since:

y = rsin[itex]\theta[/itex]

y = 3[itex]sin^2\theta[/itex]


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[itex]\theta[/itex]

since

x= rcos[itex]\theta[/itex]

x = 3sin[itex]\theta[/itex]cos[itex]\theta[/itex]

and since:

y = rsin[itex]\theta[/itex]

y = 3[itex]sin^2\theta[/itex]


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[itex]\frac{y}{r}[/itex])

= (r =[itex]\sqrt{3y}[/itex]) ?


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



or do you mean something like this:



([itex]\frac{y}{r} = sinθ [/itex]) becomes ([itex]\frac{y}{3sinθ} = sinθ[/itex])


= (y = 3sin[itex]^{2}θ[/itex])

 

[Dick]

Like so:

r = 3sinθ becomes (r=3[itex]\frac{y}{r}[/itex])

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

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.