Find a Cartesian equation for the curve and identify it.

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Homework Help Overview

The discussion revolves around finding a Cartesian equation for the curve defined by the polar equation r = 3sin(θ). Participants are exploring the conversion from polar to Cartesian coordinates and identifying the resulting curve.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the transformation of the polar equation into Cartesian form, noting specific steps such as the use of r^2 and the relationship between x, y, and θ. Questions arise regarding the origin of certain terms in the final equation, particularly the (3/2) value in the context of completing the square.

Discussion Status

The conversation is active, with participants seeking clarification on the steps involved in the transformation process. Some guidance is provided regarding the completion of the square, although there is no explicit consensus on the clarity of the instructional material referenced.

Contextual Notes

Participants express confusion over the presentation of the solution in a video format, indicating a potential gap in assumed knowledge regarding completing the square. This highlights the importance of clear communication in instructional resources.

tangibleLime
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Homework Statement



Find a Cartesian equation for the curve and identify it.

r = 3sin(θ)

Homework Equations


The Attempt at a Solution



I know that the answer is as follows...

r^2=3(r*sin(\theta))
x^2+y^2 = 3y
x^2 + (y-\frac{3}{2})^2 = (\frac{3}{2})^2

I understand everything but the last part. Where the hell is (3/2) coming from? I know if I want to solve the equation, x can be (3/2) and y can be (3/2) or... but x is still in here and so is y.
 
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tangibleLime said:

Homework Statement



Find a Cartesian equation for the curve and identify it.

r = 3sin(θ)


Homework Equations





The Attempt at a Solution



I know that the answer is as follows...

r^2=3(r*sin(\theta))
x^2+y^2 = 3y
x^2 + (y-\frac{3}{2})^2 = (\frac{3}{2})^2

I understand everything but the last part. Where the hell is (3/2) coming from? I know if I want to solve the equation, x can be (3/2) and y can be (3/2) or... but x is still in here and so is y.
Whoever wrote the answer is completing the square in the y terms.
x2 + y2 - (3/2)y = 0
To complete the square, (3/2)2 must be added to both sides.
 
Oooh thanks. It would have been cool if he said that instead of just doing it (it was a video).
 
Whoever made the video is assuming that you are familiar with the operation of completing the square.
 

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