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
Best Answer
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 $(a,b)$.