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Find the parametric equation for the curve.

$$x^{2}+y^{2}=10$$

I haven't learned parametric equations fully yet, so I wanted to check with you guys and see if you can confirm if I'm doing this correctly and possibly go more in depth on the problem if you can?

Because it's centered at (0,0) the origin, and it has a radius of sqrt(10) then the answer is this, right?

$$(x(t),y(t)) = (\sqrt{10}\cos t\,,\, \sqrt{10}\sin t)$$

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    $\begingroup$ Yes, it is...and you should probably remark that $\;0\le t\le 2\pi\;$ $\endgroup$ – DonAntonio May 4 '14 at 21:15
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taking $x=a\sin t,y=a\cos t$ from $x^2+y^2=10$ we get $$a^2\sin^2t+a^2\cos^2t=a^2(\sin^2t+\cos^2t)=a^2=10$$ from above $a=\sqrt{10}$

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You are correct. Here's an easy check: note that the position vector at any point is given by $\langle\sqrt{10}\cos(t), \sqrt{10}\sin(t) \rangle$. If we take the magnitude of this vector, we get:

$$\sqrt{10\cos^2{t} + 10\sin^2{t}} = \sqrt{10}$$

And therefore the curve is a constant $\sqrt{10}$ distance from the origin, so it must be at least a part of a circle centered at the origin with radius $\sqrt{10}$. To make the argument that it's a complete circle, you can use a bit of calculus to verify that the curve never turns back on itself. To do this, take a derivative of the position function to yield the velocity function, and note that it is never $0$.

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