Question

a bug is located at the point (4, 0) at time t = 0 and crawls at the rate of 6 units/minute in the counterclockwise direction along the circle centered at the origin of radius 4. find the coordinates (x, y) which give the location of the bug after 11 minutes. after how many minutes will the bug first return to the location (4, 0)? give the coordinates (x(t), y(t)) at an arbitrary time t ≥ 0. x = 3.5 y = 0

Explanation:

Step1: Recall the parametric equations for a circle

The parametric equations for a circle centered at the origin with radius $r$ moving counter - clockwise are $x = r\cos(\theta)$ and $y = r\sin(\theta)$. The angle $\theta$ changes with time. If the bug crawls at a rate of $\omega$ radians per minute, and $\theta=\omega t$. Here $r = 4$ and $\omega=6$. So $x(t)=4\cos(6t)$ and $y(t)=4\sin(6t)$.

Step2: Find the coordinates at $t = 11$ minutes

Substitute $t = 11$ into the parametric equations.
$x(11)=4\cos(6\times11)=4\cos(66)$
$y(11)=4\sin(66)$
Using a calculator, $\cos(66)\approx - 0.4067$ and $\sin(66)\approx0.9136$
$x(11)=4\times(- 0.4067)=-1.6268\approx - 1.63$
$y(11)=4\times0.9136 = 3.6544\approx3.65$

Step3: Find the time when the bug returns to $(4,0)$

The bug returns to $(4,0)$ when $\cos(6t)=1$ and $\sin(6t)=0$. We know that $\cos\theta = 1$ when $\theta = 2k\pi$ for $k\in\mathbb{Z}$. So $6t=2k\pi$. We want the first non - negative $t>0$, when $k = 1$, $6t = 2\pi$, then $t=\frac{\pi}{3}\approx1.05$ minutes.

Answer:

$x(11)\approx - 1.63$, $y(11)\approx3.65$, time to return to $(4,0)$ is $t\approx1.05$ minutes