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Converting from Polar Coordinates to Rectangular - Problem 1
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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