How Do I Convert Circular Parametric Equations Using Arc Length as a Parameter?

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For an assignment, I am supposed to find the parametric equation for the circle:
x^2+y^2=a^2,
using as a parameter the arc length, s, measured counterclockwise from the point (a,0) to the point (x,y).

I understand that the parametric equation for a circle is x=a*cos(t) and y=a*sin(t), but I'm not sure what they are asking me to do in this problem.

Would anyone be able to get me started on this problem?

Thanks
 
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How is t (in radians?) related to the arc length, s?
 
Nevermind, I was able to figure it out. They are looking for x=a*cos(\Theta)=a*cos(s/a) and y=a*sin(\Theta)=a*sin(s/a).

Thanks, though.
 
Yes.

You're welcome.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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