Problem on parameterization and speed

An ant is on a merry-go-ground that is rotating clockwise at $\omega$ radians per second. Initially, the ant is at $(R,0)$. From the ant's perspective, it walks toward the center with speed $v$. Several snapshots in time are as follows:

Find the parameterization of the path taken by the ant (relative to the ground)
Compute the speed of the ant as a function of $t$. When is it largest?
Set up, but do not evaluate, an integral for the arc length of the path taken by the ant between $t=0$ and when the ant reaches the origin

Solution
Part a
Recall that
Parameterization of a circle

A point with distance to the origin $r$ and angle $\theta$ with respect to the $x$-axis has coordinates $r\langle \cos \theta, \sin \theta\rangle$.
Because the ant is always walking to the center with speed $v$, it's distance to the origin at time $t$ is $r(t) = R - v t$.
Because the ant is always walking directly toward the center from its perspective, it stays on the same ray emanating from the origin. Hence, it rotates around the origin at the same angular speed as the merry-go-round. Hence $\theta(t) = - \omega t$. Note the negative sign because the merry-go-round is rotating clockwise.
We conclude that the position of the at is given by \begin{align}\bfx(t) &= (R - vt) \langle \cos(-\omega t), \sin(-\omega t) \rangle\\ &= (R - vt) \langle \cos \omega t, -\sin \omega t \rangle.\end{align}
Part b
Recall that
Speed of a parameterization

If $\bfx(t)$ is the parameterization of a curve, the speed at $t$ is given by the magnitude of the velocity: $| \bfv | = \left |\frac{d \bfx}{d t} \right|.$
Velocity of a parameterization

We now compute the velocity vector by the product rule \begin{align}
\frac{d \bfx}{dt}(t) &= -v \langle \cos \omega t, - \sin \omega t \rangle + (R - vt) \langle -\omega \sin \omega t, - \omega \cos \omega t \rangle.
\end{align}
Now we compute the length of this vector. Recall that
Dot product and vector length

$|\bfv|^2 = \bfv \cdot \bfv$
We can compute that \begin{align}
\left |\frac{d\bfx}{dt} \right|^2 = &v^2 \langle \cos \omega t, - \sin \omega t \rangle \cdot \langle \cos \omega t, - \sin \omega t \rangle \\ &- 2 v \omega (R - vt) \langle \cos \omega t, - \sin \omega t \rangle\cdot \langle - \sin \omega t, - \cos \omega t \rangle \\ & + \omega^2 (R-vt)^2 \langle -\sin \omega t, - \cos \omega t \rangle \cdot \langle -\sin \omega t, - \cos \omega t \rangle.
\end{align}
Direct computation of dot product

We note that \begin{align}
\langle \cos(\omega t), - \sin(\omega t) \rangle \cdot \langle \cos(\omega t), - \sin(\omega t) \rangle &= 1, \\
\langle \cos(\omega t), - \sin(\omega t) \rangle\cdot \langle - \sin (\omega t), - \cos(\omega t) \rangle &= 0, \\
\langle -\sin (\omega t), - \cos(\omega t) \rangle \cdot \langle -\sin (\omega t), - \cos(\omega t) \rangle &=1.
\end{align}
And we get that \begin{align}
&\left |\frac{d\bfx}{dt} \right|^2 = v^2 + \omega^2(R-vt)^2, \\
&\left |\frac{d\bfx}{dt} \right| = \sqrt{v^2 + \omega^2(R-vt)^2}. \\
\end{align}
This expression is largest when $t=0$.
Part c
Recall that
Arc length of a parameterization

The length of the curve $\bfx(t)$ traced out between $t=a$ and $t=b$ is given by $$s = \int_a^b \left | \frac{d \bfx(t)}{dt} \right| dt.$$
We identify that $a=0$ and $b$ is the time at which the ant reaches the origin. Because the ant has a distance $R$ to travel at speed $v$, it takes $R/v$ time to reach the origin. Hence, we identify that $b = R/v$.
The arc length of the curve traced by the ant is $$s = \int_0^{R/v} \sqrt{v^2 + \omega^2(R-vt)^2} dt.$$

Problem on parameterization and speed

Solution

Part a

Part b

Part c

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