converting polar to rectangular

[vmbs]

Please help a math-challenged girl who does want to understand this!!

First of all, how do I convert the following to rectangular form

r = 15 / 2-sin(theta)

Then, how do I graph it? Is it an ellipse?

THANK YOU!

 

[galactus]

There are some equivalencies you can use.

$\displaystyle r^{2}=x^{2}+y^{2}$

$\displaystyle y=rsin({\theta})$

$\displaystyle x=rcos({\theta})$


\(\displaystyle \L\\r=\frac{15}{2-sin({\theta})}\)

\(\displaystyle \L\\2r-rsin({\theta})=15\)

\(\displaystyle \L\\2r-y=15\)

Square both sides:

\(\displaystyle \L\\4r^{2}=(15+y)^{2}\)

\(\displaystyle \L\\4(x^{2}+y^{2})=(15+y)^{2}\)

\(\displaystyle \L\\4x^{2}+4y^{2}=y^{2}+30y+225\)

Now, can you finish up and see what you have?.

Check the discriminant.

As for graphing, does your calculator graph in polar. I bet it does.

Set the graph mode form rectangular to polar.

 

[vmbs]

but wouldnt I end up with 4x^2/225 on one side? how can one graph an ellipse like that?

 

[vmbs]

and if I divide 225 by 4, I get 56.25, and that would mean the minor axis is twice the square root of 56.25?

Sorry, I'm just trying to understand this

 

[galactus]

Here's the ellipse. That's how you graph it.

Let's group terms and complete squares:

\(\displaystyle \L\\4x^{2}+3y^{2}-30y=225\)

\(\displaystyle \L\\4x^{2}+3(y^{2}-10y)=225\)

\(\displaystyle \L\\4x^{2}+3(y^{2}-10y+25)=300\)<-----I added 75 to compensate for the 3*25 added on the left side. See?.

\(\displaystyle \L\\4x^{2}+3(y-5)^{2}=300\)

Divide through by 300:

\(\displaystyle \L\\\frac{x^{2}}{75}+\frac{(y-5)^{2}}{100}=1\)

There's the ellipse. Revealed at last. Can you see what the major and minor axes are now?.

 

[vmbs]

Yes!!

Thank you so much, I understand it now!