Parameterization of a Circle Question

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This isn't really a HW question, it's just something that's been confusing me in my Calc class.

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,

x2 + y2 = 9

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

So from:

x2 + y2 = r2

If x = cos(t) and y = sin(t), the identity cos2(t) + sin2(t) = 1 still holds true.

If the radius is 3,

x2 + y2 = 32

3(x2 + y2) = 9(12)

9cos2(t) + 9sin2(t) = 9(12)

9(cos2(t) + sin2(t)) = 9(12)

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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Try drawing a picture. Do you remember your unit circle? For a right-handed coordinate system, x = rcos[itex]\theta[/itex] and y the sine function. Positive theta corresponds to counter-clockwise motion from the positive x-axis.
Now when you parametrize the equation [itex]\theta[/itex] becomes [itex]\dot{\theta}[/itex]t, where [itex]\dot{\theta}[/itex] is how quickly you are going around the circle. In your case this is simply one.
Strictly speaking, x doesn't HAVE to be the cosine. It's just convention to use a right handed coordinate system. If you measured [itex]\theta[/itex] from a different axis or you switched positive x being in the vertical, your sines and cosines may change.
 
royblaze said:
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...?

Yes, it has to do with the orientation of the curve and where you want it to be when t = 0. Parameterizations are in general not unique. If the curve is specified to start at (1,0) and go counterclockwise then R(t) = < cos(t), sin(t) > is a natural choice. If it went clockwise you might use <cos(t), -sin(t)>. If nothing is given about the orientation of the curve then you could use <sin(t), cos(t)> or other choices; they are all equally correct.
 
royblaze said:
This isn't really a HW question, it's just something that's been confusing me in my Calc class.

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,

x2 + y2 = 9

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

So from:

x2 + y2 = r2

If x = cos(t) and y = sin(t), the identity cos2(t) + sin2(t) = 1 still holds true.

If the radius is 3,

x2 + y2 = 32

3(x2 + y2) = 9(12)

9cos2(t) + 9sin2(t) = 9(12)

9(cos2(t) + sin2(t)) = 9(12)

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...?
They don't have to be. There exist an infinite number of parametric equations describing any curve. [itex]x= 3sin(t)[/itex], [itex]y= 3cos(t)[/itex] are perfectly good parametric equations and, in fact, have the same "direction" as [itex]x= 3cos(t)[/itex] [itex]y= 3sin(t)[/itex]: with the first, as t goes from 0 to [itex]\pi/2[/itex], (x, y) goes from (0, 3) to (3, 0), counter-clockwise around the circle and the second goes from (3, 0) to (0, 3), also counterclockwise.

But before you go telling your teacher that we say he/she is wrong, note that they do have different points for specific values of t. The parametric equations [itex]x= cos(t)[/itex], [itex]y= sin(t)[/itex] "start" (t= 0) at (1, 0) and then go around the circle. Perhaps for some reason, that is important.
 
Wow, thank you all for the absolutely fantastic replies. It REALLY cleared it up for me, big time.

Thanks again!