Find the parametric equations of a circle with radius of $5$ where you start at point $(5,0)$ at $v=0$ and you travel clockwise with a period of $3$.
So, I know that I require to have a $x(v)$ and $y(v)$ answer.
So for $x(v)$, Since it starts at $5$, I figured the answer would be $x(v) = 5 + 5\cos()$ and for $y(v)$, since it starts at $0$, its simply $y(v) = 0 + 5 \sin()$. However, I am not too sure about what to enter in the parentheses.( I know that we would need $bv$, where $b$ is the coefficient to $v$). Since it says $3$ period, is it found like this?
$$3b = 2 \pi, \quad b = \frac{2}{3} \pi$$
And so, the answer for $x$ would be $x(v) = 5 + 5\cos(2/3\pi v)$ and for $y; y(v) = 5 \sin(2/3\pi v)$. My answers are wrong, could you explain where I went wrong. Thanks!
calculus
You have the period under control, so we will not deal with that. If we are travelling around the circle with centre the origin, and radius $r$, in mathematicians' favourite direction, counterclockwise, starting at $(r,0)$, then the parametric equation is of the shape $x=r\cos( kt)$, $y=r\sin(kt)$.
For travelling clockwise, replace $t$ by $-t$. But $\cos(-kt)=\cos(kt)$ and $\sin(-kt)=-\sin(kt)$, so the parametric equation can be written as $x=r\cos( kt)$, $y=-r\sin(kt)$.
If at time $0$ we are at some other point $P$ on the circle, find an angle $\theta$ such that $P=(r\cos\theta,r\sin\theta)$. Then the parametric equation for counterclockwise travel is $x=r\cos(kt+\theta)$, $y=r\sin(kt+\theta)$. For clockwise, a manipulation similar to the one above gives $$x=r\cos( kt-\theta), \qquad y=-r\sin(kt-\theta).$$ In your particular case, we have $\theta=0$.
One can generalize the above to travelling around a circle with centre $(a,b)$.