Convert the polar equation to rectangular form.

  • Thread starter Lebombo
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  • #1
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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..
 

Answers and Replies

  • #2
LCKurtz
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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.
 
  • #3
Dick
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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.
 
  • #4
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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])
 
  • #5
Dick
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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.
 

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