Finding the position of a moving point

Question

A point is moving on a given curve. For example, curve equation is:

$$x^2 + y^2 - 10y = 0,$$

which is a circle with $5$ meter radius.

If point is on $(0,0)$ at $t = 0$ and is moving on the curve with speed $0.5$ m/s clockwise, how to calculate the position of point at $t$ seconds (suppose $t = 5$)?

Answer 1

Let $\tau = 2\pi$ denote the ratio of the circle's circumference $C$ to its radius $r$: $$\tau = \frac{C}{r} \implies C = \tau r = 5 \tau \mbox{ m}$$ Let $d$ be the displacement: $$v = \frac{d}{t} \implies d = vt = (0.5 \mbox{ m/s}) \cdot (5 \mbox{ s}) = 2.5 \mbox{ m}$$ Let $P$ be the proportion moved around the circle: $$P = \frac{2.5 \mbox{ m}}{5 \tau \mbox{ m}} = \frac{1}{2 \tau}$$ Let $\theta$ be the angle moved through: $$\theta = P \cdot \tau = \frac12$$

Then form an isosceles triangle with the base as the line between the two points:

The problem is now: we have two $5 \mbox{ m}$ lines whose ends join at a point at an angle of $1/2 \mbox{ rad}$ to each other. Calculate the end of the second line given the equation of the first line.

Answer 2

In what follows, angles are measured in radians.

Move the circle downward by $5$ to standard position. After the move, our starting point is $(0,-5)$.

We are travelling at $0.5$ meters per second. So in time $t$ we travel a distance $(0.5)t$. Let us find the angle we have travelled through. A complete revolution is $(2\pi)(5)$, so in time $t$ we have travelled through $\frac{0.5t}{(2\pi)(5)}$ of a complete revolution.

It follows that we have travelled through an angle $(0.5 t)/5$, that is, $0.1 t$.

If we were travelling counterclockwise starting at $(5,0)$, we would after time $t$ be at the point $(5\cos(0.1t), 5\sin(0.1t))$. But we are going clockwise. So if we started at $(5,0)$, after time $t$ we would be at $(5\cos(-0.1t), 5\sin(-0.1t))$. Rewrite this as $(5\cos(0.1t), -5\sin(0.1t))$.

But we started at $(0,-5)$. Thus after time $t$, we are at $(5\cos(0.1t+\pi/2), -5\sin(0.1t+\pi/2))$. This is more simply written as $$(-5\sin(0.1t), -5\cos(0.1t)).$$

Finally, let us push everything up by $5$, to solve the problem for the circle we were given. We get that at time $t$ our position is given by $$x=-5\sin(0.1t),\qquad y= 5-5\cos(0.1t).$$

Remark: We used a technique sometimes called Transform, Solve, Transform Back. One can use the reasoning above to develop a general formula, with arbitrary circle, and arbitrary initial position on that circle.