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Find the equation in the xy-plane whose graph includes x = 8 sin t and y = 8 cos t.
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I got \[y=\sqrt{1-x^2}\]
You are close! If you calculate x² + y², you get: 64sin²t + 64cos²t=64(sin²t + cos²t)=64*1=64. So: \( x^2 + y^2 = 64\) and therefore: \(y^2 = 64 - x^2\), which means: \(y=\pm\sqrt{64-x^2}\).
If you make a drawing, \(y=\sqrt{64-x^2}\) is the upper half of a circle with radius 8 and \(y=-\sqrt{64-x^2}\) is the lower half.
oh, I see. I think I tried to get one equation equal to another to plug in, but then my 8s cancelled and I was left with 1.
Yeah, the 8 blows it up eightfold...
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Thanks!
YW!
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