Problem 40 - Thomas Calculus in SI Units

Question

The positions of two particles on the \(s\) -axis are \(s_{1}=\sin t\) and \(s_{2}=\sin (t+\pi / 3),\) with \(s_{1}\) and \(s_{2}\) in meters and \(t\) in seconds. a. At what time(s) in the interval \(0 \leq t \leq 2 \pi\) do the particles meet? b. What is the farthest apart that the particles ever get? c. When in the interval \(0 \leq t \leq 2 \pi\) is the distance between the particles changing the fastest?

Step-by-Step Solution

Answer

Particles meet at \( \frac{\pi}{3} \) and \( \frac{4\pi}{3} \); max distance is \( \sqrt{3} \); fastest change at \( \frac{5\pi}{12} \) and \( \frac{17\pi}{12} \).

Step 1: Equate the Positions

To find the times when the particles meet, we set their positions equal: \[ s_1 = s_2 \] This gives us the equation: \[ \sin t = \sin(t + \frac{\pi}{3}) \] Using the property of the sine function, we have two cases: either \[ t = t + \frac{\pi}{3} + 2k\pi \] for integers \(k\), or \[ t = \pi - (t + \frac{\pi}{3}) + 2k\pi. \] Solving these will give us values of \(t\) where the particles meet.

Step 2: Solve the Meeting Times

On simplifying, the first case \( t = t + \frac{\pi}{3} + 2k\pi \) is not valid as it implies \(\frac{\pi}{3} = 0\) which is incorrect.The second equation simplifies to: \[ 2t = \pi - \frac{\pi}{3} + 2k\pi \] \[ 2t = \frac{2\pi}{3} + 2k\pi \] \[ t = \frac{\pi}{3} + k\pi. \]Checking for \( k = 0 \): \( t = \frac{\pi}{3} \) is within the interval Checking for \( k = 1 \): \( t = \frac{4\pi}{3} \) is within the interval. So, the particles meet at \( t = \frac{\pi}{3} \) and \( t = \frac{4\pi}{3} \).

Step 3: Find the Maximum Distance

The distance between the particles is given by \[ d(t) = |\sin t - \sin(t + \frac{\pi}{3})| \] Using the sine subtraction formula, \[ \sin a - \sin b = 2 \cos\left(\frac{a+b}{2}\right) \sin\left(\frac{a-b}{2}\right), \] we get \[ d(t) = 2 \left| \cos\left(t + \frac{\pi}{6}\right) \sin\left( -\frac{\pi}{6} \right) \right| = \sqrt{3} \left| \cos\left(t + \frac{\pi}{6}\right) \right|. \] To find the maximum distance, find where \( |\cos\left(t + \frac{\pi}{6}\right)| \) is maximized (equal to 1): At \( t + \frac{\pi}{6} = n\pi \) for integer \(n\), \( t = n\pi - \frac{\pi}{6} \). Within the interval, these values are \( t = \frac{5\pi}{6}, \frac{11\pi}{6} \), giving distance \( \sqrt{3}. \)

Step 4: Find When Distance is Changing Fastest

First, find the derivative of the distance function with respect to time to get the rate of change: \[ d'(t) = \frac{d}{dt} \left( \sqrt{3} \left| \cos\left(t + \frac{\pi}{6}\right) \right| \right). \] Instead, study \[ g(t) = \cos\left(t + \frac{\pi}{6}\right) \] which has critical points when \[ g'(t) = -\sin\left(t + \frac{\pi}{6}\right) = 0, \] i.e., \[ t + \frac{\pi}{6} = n\pi \] giving extrema at \( t = n\pi - \frac{\pi}{6}. \) The distance changes fastest between the critical points of \( d(t) \). The midpoint, where acceleration changes, is at \( t + \frac{\pi}{6} = \frac{(2n-1)\pi}{2} \). This corresponds to \( t = \frac{5\pi}{12}, \frac{17\pi}{12} \).

Step 5: Conclusion

The particles meet at \( t = \frac{\pi}{3} \) and \( t = \frac{4\pi}{3} \). The maximum distance between them is \( \sqrt{3} \), occurring at \( t = \frac{5\pi}{6} \) and \( t = \frac{11\pi}{6} \). The distance between particles changes most rapidly at \( t = \frac{5\pi}{12} \) and \( t = \frac{17\pi}{12}. \)

Key Concepts

Sinusoidal Motion

When dealing with sinusoidal motion, we often investigate how physical quantities like position and velocity vary sinusoidally over time. Sinusoidal motion can typically be described by functions involving sine or cosine. In the given exercise, the positions of two particles are represented by \(s_1 = \sin t\) and \(s_2 = \sin(t + \frac{\pi}{3})\). These are sine functions that describe periodic motion along the \(s\)-axis.

The sine function exhibits a wave-like behavior, characterized by its amplitude, which is the maximum value it can reach, and its period, which is the time it takes to complete one full cycle. For a standard sine function \(\sin t\), the amplitude is 1, and the period is \(2\pi\). A phase shift, noted by an additional angle in the function, indicates how much the function is horizontally shifted on the graph.

Understanding such waves can be integral to predicting the motion of particles, as sinusoidal equations help model the repetitive motion found in nature, mechanics, and electrical currents.

Maximum Distance

Determining the maximum distance between two particles involves analyzing how far apart they can get from each other given their respective positions as functions of time. In the exercise, the distance between particles at any time \(t\) is expressed as

• \(d(t) = \left| \sin t - \sin(t + \frac{\pi}{3}) \right|\).

Utilizing trigonometric identities makes it possible to rewrite this expression. The sine difference formula is essential here:

• \(\sin a - \sin b = 2\cos\left(\frac{a + b}{2}\right)\sin\left(\frac{a - b}{2}\right)\).

Substituting appropriately, we find that

• \(d(t) = \sqrt{3} \left| \cos\left(t + \frac{\pi}{6}\right) \right|\).

The maximum distance occurs when \( \left| \cos\left(t + \frac{\pi}{6}\right) \right| = 1\), indicating that the cosine function is at its peak. Given the periodic nature of the cosine function, you can calculate that this condition is satisfied at particular time points such as \(t = \frac{5\pi}{6}\) and \(t = \frac{11\pi}{6}\), where the maximum distance is calculated to be \(\sqrt{3}\).

This analysis is critical, especially in applications where timing and spacing are crucial, such as synchronized systems or wave interference phenomena.

Rate of Change

The rate of change of a function in mathematics refers to how quickly a variable quantity changes in relation to another. Here, we are interested in how quickly the distance between the two particles changes over time. We address this by examining the derivative of the distance function \(d(t)\).

In the exercise, the approach to finding when the distance changes most rapidly involves looking at the critical points of the function \(\cos\left(t + \frac{\pi}{6}\right)\). The derivative \(g'(t) = -\sin\left(t + \frac{\pi}{6}\right)\) assists in determining these points. These occur when \(-\sin\left(t + \frac{\pi}{6}\right) = 0\), or when

• \(t + \frac{\pi}{6} = n\pi\) is satisfied.

At these key intervals, the function transitions from increasing to decreasing or vice versa and, thus, indicates key instances where the rate of change is at significant turning points.

For the given particle's motion, the distance is changing at its fastest between these points, best noted for times such as \(t = \frac{5\pi}{12}\) and \(t = \frac{17\pi}{12}\). Recognizing where the rates of change peak helps in applications where the timing of interactions or reactions is critical, such as in mechanical or wave dynamics. Understanding these principles allows us to monitor and predict behaviors in systems characterized by changing motions.