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Converting from Polar Coordinates to Rectangular - Problem 1

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Teacher/Instructor Norm Prokup
Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

I'm converting polar equations to rectangular form. This time I also want to identify the resulting curve. So let’s start with the equation r equals 10 cosine theta minus 24 sine theta.

Now the first thing I want to do here, is multiply through by r. The reason I want to do this is, that I know that r cosine theta equals x, and r sine theta equals y. So I want to get our cosine and our sine theta. And also r² is x² plus y², so let me start replacing r’s with x’s and y’s.

So I have x² plus y² on the left, then I have 10 times r cosine theta 10x, minus 24 r sine theta that’s y, times 24y. And then let me pull this stuff to the left side. Now technically, this is already a rectangular equation, but I want to identify the curves. So I need to do a little more work here.

Y² plus 24y equals 0. Now I get the sense that this might end up being the equation of a circle, because I have an x², multiple of x, y², multiple of y. I just need to complete the square on both of these in order to figure out what the circle is, where the center is, or what the radius is.

So I need to add half of this quantity squared, -5² or 25. Need to add that to both sides. And I have to add half of this quantity squared, so 12² 144 to both sides. And that’s going to give me this quantity which is now a perfect square and can be written (x minus 5) squared, plus this trinomial can be written as y plus 12 quantity squared, equals 25 plus 144, 169.

So now I know that this is a circle. It’s centered at 5,-12, and its radius is 13. This is 13². So there it is. This original equation was a circle centered at 5,-12 radius 13.

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