How do you determine the parametric equations of the path of a particle that travels the circle: $(x−3)^2+(y−1)^2=9$ on a time interval of $0<=t<=2$ ?

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Answer 1 · 2016-07-05T22:36:26

$x =3+ 3 cos pi t$
$y =1 + 3 sin pi t$

$(x−3)^2+(y−1)^2=9$

which is a circle that we can simplify as follows:
$u^2+v^2=3^3$

where $u = x-3, v = y -1$

so

$u = 3 cos psi, v = 3 sin psi implies u^2 + v^2 = 3^3$

$x-3 = 3 cos psi, y-1 = 3 sin psi$

$implies x =3+ 3 cos psi, y =1 + 3 sin psi$

in terms of periodicity, we can say that $psi = omega t$ where $omega = (2 pi) /T$

if the period T is $t = 2$ , then $omega = (2 pi) /2 = pi$

So
$x =3+ 3 cos pi t$
$y =1 + 3 sin pi t$

How do you determine the parametric equations of the path of a particle that travels the circle: (x−3)^2+(y−1)^2=9(x−3)^2+(y−1)^2=9 on a time interval of 0<=t<=20<=t<=2?