Consider a particle following the parametric equations

\(\begin{align*} x &= 1 + \sin (\pi t),\\ y &= 3 \sin (\pi t). \end{align*}\)

Find the first four positive times t when this particle visits the point (3/2, 3/2) and list them in increasing order.

Please, help is really appreciated

Guest Aug 10, 2022

#1**0 **

So I was looking through it, and I found that CPhill already answered this question, but it is wrong. Can someone please help

Thank you in advance.

Guest Aug 10, 2022

#2**+1 **

Hello Guest!

When the particle visits the point \((\frac{3}{2},\frac{3}{2})\) we get the following equations:

\(\dfrac{3}{2}=1+sin(\pi t) \iff sin(\pi t)=\dfrac{1}{2} \space \space \space \space \space \space[Eq.\space 1] \\ \space \\ \dfrac{3}{2}=3sin(\pi t) \iff sin(\pi t) = \frac{1}{2} \space \space \space \space \space \space [Eq. \space 2]\)

They are the same equation, so we need to solve:

\(sin(\pi t) =\dfrac{1}{2}\), and we want the first four positive solutions of t.

Lets solve for t:

\(t=\dfrac{1}{\pi}sin^{-1}(\dfrac{1}{2})\)

Now, \(sin^{-1}(\dfrac{1}{2})=30,150,390,510,750,870,...\) (degrees) and so on. (30 and 150 in first and second quadrant in the unit circle and then we keep adding 360 on each of them to get infinite solutions). However, we only need the first four, and since the question is in radians so we have to use radians.

Therefore,

\(t_1=\dfrac{1}{\pi}(\dfrac{\pi}{6})=\dfrac{1}{6}\)

\(t_2=\dfrac{1}{\pi}(\dfrac{5\pi}{6})=\dfrac{5}{6}\)

\(t_3=\dfrac{1}{\pi}(\dfrac{13\pi}{6})=\dfrac{13}{6}\)

\(t_4=\dfrac{1}{\pi}(\dfrac{17\pi}{6})=\dfrac{17}{6}\)

I hope this helps!

Guest Aug 10, 2022