Question

a particle moves with its position given by

$x = \cos(2t)$ and $y = \sin(t)$ with $t \geq 0$

where the position is given in feet from the origin and time $t$ is in seconds.

a. find the speed of the particle at time $t$ seconds. (include units.)

speed: $v=$

b. find the first time value at which the particle comes to a stop. (include units.)

first stop: $t=$

c. if $n$ is any odd integer, write a formula (in terms of $n$) for all times at which the particle comes to a stop. (include units.)

general formula for odd $n$: $t=$

Explanation:

Step1: Find derivatives of \(x\) and \(y\)

Given \(x = \cos(2t)\), using the chain - rule \((\cos(u))^\prime=-\sin(u)\cdot u^\prime\) with \(u = 2t\) and \(u^\prime=2\), we get \(x^\prime=-\sin(2t)\cdot2=-2\sin(2t)\).
Given \(y=\sin(t)\), using the formula \((\sin(u))^\prime=\cos(u)\cdot u^\prime\) with \(u = t\) and \(u^\prime = 1\), we get \(y^\prime=\cos(t)\).

Step2: Use the speed formula \(v=\sqrt{(x^\prime)^2+(y^\prime)^2}\)

Substitute \(x^\prime=-2\sin(2t)\) and \(y^\prime=\cos(t)\) into the formula \(v=\sqrt{(x^\prime)^2+(y^\prime)^2}\).
Since \(\sin(2t)=2\sin(t)\cos(t)\), then \((x^\prime)^2 = 4\sin^{2}(2t)=16\sin^{2}(t)\cos^{2}(t)\) and \((y^\prime)^2=\cos^{2}(t)\).
\(v=\sqrt{4\sin^{2}(2t)+\cos^{2}(t)}=\sqrt{16\sin^{2}(t)\cos^{2}(t)+\cos^{2}(t)}=\cos(t)\sqrt{16\sin^{2}(t) + 1}\) feet per second.

Step3: Find when the particle stops (\(v = 0\))

Set \(v = 0\). Since \(v=\cos(t)\sqrt{16\sin^{2}(t)+1}\), and \(\sqrt{16\sin^{2}(t)+1}\gt0\) for all real \(t\), then \(\cos(t)=0\).
The general solution of \(\cos(t)=0\) is \(t=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}\).
For \(t\geq0\), the first value when \(k = 0\) is \(t=\frac{\pi}{2}\) seconds.

Step4: Find the general formula for odd \(n\)

Let \(n = 2k + 1,k\in\mathbb{Z}\). If \(t=\frac{\pi}{2}+k\pi\), when \(n\) is odd (\(k=\frac{n - 1}{2}\)), \(t=\frac{n\pi}{2}\) seconds.

Answer:

a. \(v=\cos(t)\sqrt{16\sin^{2}(t)+1}\) feet per second
b. \(t=\frac{\pi}{2}\) seconds
c. \(t=\frac{n\pi}{2}\) seconds