Help with polar equation please :-)

G

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Guest
Hi everyone
Thanks for this great service. :)

I am having trouble trying to figure out how to convert:

r=-8cos(theta) into Cartesian coordinates?

There is a note saying to do this: r^2=-8rcos(theta)

but I have no idea where that comes from unless they are using:
x^2 + y^2 = r^2 but that doesn't make sense, so I'm not sure.

Thanks for all your help.
Take care,
Beckie
 
You should have memorized these basic formulas:

. . . . .x = rcos(@)

. . . . .y = rsin(@)

If x^2 + y^2 = r^2 and r^2 = -8rcos(@) = -8(x), then you have been given that:

. . . . .x^2 + y^2 = -8x

. . . . .x^2 + 8x + y^2 = 0

Convert to the center-vertex form for circles.

Eliz.
 
If x^2 + y^2 = r^2 and r^2 = -8rcos(@) = -8(x), then you have been given that:


Thanks for this but I still don't get it. All I see is this:

Convert r = -8cos@ into Cartesian equation?

I don't see how r^2=-8rcos@

Where is that step where r^2 can equal -8rcos@?? Is there an equation that I'm missing?

Thanks
Take care,
Beckie
 
Beckie said:
All I see is this: "Convert r = -8cos@ into Cartesian equation." I don't see how r^2=-8rcos@.
Click to expand...
. . . . .r = -8cos(@)

. . . . .r(r) = r(-8cos(@))

. . . . .r^2 = -8rcos(@)

Eliz.
 
Hi Stapel
Thank you so much for all of your help. I really appreciate it.

I still don't understand where the extra r comes from on the right side of the equation. I think the r on the left side of the equation comes from the x^2 + y^2 = r^2 but I'm not sure, but the left side doesn't make sense to me.

r(r) = r(-8cos(@))

Can I just pull an r out of nowhere and multiply it to both sides of my equation???

Thanks again for all of your help.

Take care,
Beckie
 
Ahhh, but you do that all the time. If I told you
.5x = 3
wouldn't you "just pull a 2 out of nowhere and multiply it to both sides of my equation???"
 
Thank you thank you thank you thank you!!

I get it now!!

I started writing an explanation to the .5x = 3 and then when I was doing it I saw what you were saying and it all came clear to me. I've just never done it with a variable before.

This is my conclusion: I can do this because I want to get r^2 as a result on the left side, so I make it happen by doing it to both sides of the equation? Is this correct?

Thanks again for everyone's help.

Take care,
Beckie
 
You've got it. The only difference with a variable is that you have to remember you did it and it should NOT take on a value of zero after that.
 
Thanks again Gene. I have made a note of that so I won't forget. Wow I wonder when I missed this rule. I must have missed it in Algebra or something. I hate it when that happens. I found out just the other day that I missed a whole entire way of substitution for integrals. I had been doing my substitutions a certain way and it had been working but then all of a sudden it wasn't working and I found out that I missed a totally unique way of substitution from Calculus I, luckily it happened on a homework question not on a test. :) lol lol

Thanks!!!
Take care,
Beckie
 
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