Graph set of points satisfying in polar coordinates

[Damoo]

r = 1 + sin(theta^2)

Conditions: 0<=theta<=square root of 6pi

How do I make this into rectangular coordinates?

 

[Deleted member 4993]

r = 1 + sin(theta^2)

Conditions: 0<=theta<=square root of 6pi

How do I make this into rectangular coordinates? ..............................Why?

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If the problem states "graph",

why do you have to convert to cartesian co-ordinates?

Anyway, if you so desire -

for cartesian co-ordinates,

r = \(\displaystyle \sqrt{x^2+y^2}\)

and

\(\displaystyle \theta \ = \ tan^{-1}\left({\frac{y}{x}}\right )\)

Please share your work with us indicating exactly where you are stuck - so that we may know where to begin to help you.

 

[HallsofIvy]

I would suggest that you start by multiplying the equation by r: \(\displaystyle r^2= r+ rsin(\theta)\). Then use \(\displaystyle r= \sqrt{x^2+ y^2}\) and \(\displaystyle r sin(\theta)= y\).

 

[Deleted member 4993]

I would suggest that you start by multiplying the equation by r: \(\displaystyle r^2= r+ rsin(\theta)\). Then use \(\displaystyle r= \sqrt{x^2+ y^2}\) and \(\displaystyle r sin(\theta)= y\).

As posted, the function to be plotted is:

\(\displaystyle r = 1 + rsin(\theta^2)\)

I cannot see an easy substitution for Θ².