This isn't really a HW question, it's just something that's been confusing me in my Calc class.(adsbygoogle = window.adsbygoogle || []).push({});

We recently went over how to find curvatures of curves in 3D space. In lecture, the professor went over a simple example: a circle of radius 3 at any given point.

Maybe it's because I don't remember my Calc II class that well, but he parameterized the equation of a circle,

x^{2}+ y^{2}= 9

to become r(t) = 3cos(t)i + 3sin(t)j .

So from:

x^{2}+ y^{2}= r^{2}

If x = cos(t) and y = sin(t), the identity cos^{2}(t) + sin^{2}(t) = 1 still holds true.

If the radius is 3,

x^{2}+ y^{2}= 3^{2}

3(x^{2}+ y^{2}) = 9(1^{2})

9cos^{2}(t) + 9sin^{2}(t) = 9(1^{2})

9(cos^{2}(t) + sin^{2}(t)) = 9(1^{2})

is how I understand it. If there's an easier way to picture it, please let me know!

So then r(t) = 3cos(t)i + 3sin(t)j describes the circle when "traced."

My question: why does the "x" part and the "y" part of the parameterization have to be the respective cos, sin trig functions? In class, one student said that x was sin; but the prof said that no, it's cos. Why does x have to be cos, rather than sin? Does it have to do something with the direction the curve "goes in?" How can I determine the "direction" the curve goes in? Or did I just hear my professor wrong...?

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# Homework Help: Parameterization of a Circle Question

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