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How do I find the trace of the curve (e^(-t) cos t , e^(-t) sin t))? I know that if it was simply (cos t, sin t), that it would be a circle of equation x^2 + y^2 = 1, but I don't know how to account for the deformation the e^(-t) is going to cause.
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i think its like a circle but the radius is decreasing
|dw:1337863065897:dw|
Thank you!
do you understand why ?
I am going to make a graph, I think it's because of the 1/e^t, e^t gets greater and so 1/e^t gets smaller and smaller.
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