Problem 79 - Precalculus Essentials

Question

A ball attached to a spring is raised 2 feet and released with an initial vertical velocity of 3 feet per second. The distance of the ball from its rest position after \(t\) seconds is given by \(d=2 \cos t+3 \sin t,\) Show that A ball attached to a spring is raised 2 feet and released with an initial vertical velocity of 3 feet per second. The distance of the ball from its rest position after \(t\) seconds is given by \(d=2 \cos t+3 \sin t .\) Show that $$2 \cos t+3 \sin t=\sqrt{13} \cos (t-\theta)$$ where \(\theta\) lies in quadrant I and \(\tan \theta=\frac{3}{2} .\) Use the identity to find the amplitude and the period of the ball's motion.

Step-by-Step Solution

Answer

The amplitude of the ball's motion is \(\sqrt{13}\) feet and the period of the motion is \(2\pi\) seconds.

Step 1: Recognize and Rewrite

Recognize the function \(d=2 \cos t+3 \sin t \) as a phase-shifted sinusoidal function of the form \(r \cos(t-\theta)\). The goal is to find the value of \(r\), which represents the amplitude of the motion and \(\theta\), the phase shift. Using the identity \(\cos(t-\theta) = \cos t \cos \theta + \sin t \sin \theta \), rewrite the function in the form \(2 \cos t+3 \sin t = r(\cos t \cos \theta + \sin t \sin \theta) = r \cos (t-\theta)\)

Step 2: Identify r and θ

For the equation to hold, \(r \cos \theta = 2\) and \(r \sin \theta = 3\). Solve these two equations simultaneously to find \(r\) and \(\theta\). Using the Pythagorean identity \(\cos^2 \theta + \sin^2 \theta = 1\), find \(r = \sqrt{(r \cos \theta)^2+(r \sin \theta)^2} = \sqrt{(2)^2+(3)^2} = \sqrt{13}\), which is the amplitude of the motion. Next, use \(\tan \theta = \frac{\sin \theta}{\cos \theta}=\frac{3}{2}\) to find \(\theta\).

Step 3: Calculate the period of motion

The period of a cosine function in general form \(y=r \cos (bt-\theta)\) is \(T=\frac{2\pi}{b}\). Here, as \(b = 1\), the period of the function is \(T=\frac{2\pi}{1}=2\pi\) seconds. This is the time it takes for one complete oscillation.

Key Concepts

Amplitude of Motion

In the context of sinusoidal functions, the amplitude represents the maximum displacement from the rest position. It is an indicator of how far the oscillating object moves above and below its equilibrium position. For a function of the form d = Asin(t) + Bcos(t), where A and B are constants, the amplitude can often be found by formulating the function as a single sinusoidal function, d = Rcos(t - \theta).The amplitude R can be determined using the Pythagorean identity, which involves squaring the coefficients A and B, summing them, and then taking the square root. In the given exercise, we establish that the amplitude of the ball's motion is \(\sqrt{13}\) by computing \(\sqrt{2^2 + 3^2}\). This amplitude tells us that the ball reaches 2 feet above and 3 feet below the equilibrium line, culminating in a total displacement range of \(\sqrt{13}\) feet during its motion.

Phase Shift

The concept of a phase shift is related to the horizontal translation of a sinusoidal function on a graph. It shows how much the function is shifted left or right from its standard position. We describe this shifting using the variable \(\theta\) in a sinusoidal function like d = Rcos(t - \theta) or d = Rsin(t - \theta). The value of \(\theta\) can determine the initial state of the motion at t = 0.

In the given problem, identifying the phase shift involves solving trigonometric equations based on the initial function d = 2cos(t) + 3sin(t). After expressing the function in the form of Rcos(t - \theta), we find that \(\tan \theta = \frac{3}{2}\) by dividing \(r \sin \theta\) by \(r \cos \theta\). Calculating the arctangent of \(\frac{3}{2}\) yields the phase shift \(\theta\), which tells us by how much the motion has shifted from its usual cosine start point.

Pythagorean Identity

The Pythagorean identity is an essential tool in trigonometry that is rooted in the Pythagorean Theorem. It is expressed as \(\cos^2 \theta + \sin^2 \theta = 1\). For sinusoidal functions, this identity helps us relate the coefficients of sin and cos when the function is presented in a certain format.

In our exercise, to find the amplitude R, we use the identity to combine the cosine and sine elements into one sinusoidal function. It assists us in deriving a single value for a spatial parameter that defines the motion's extent. Concretely, we square the coefficients 2 and 3, add the squares to get 13, and then take the square root, using the Pythagorean identity to confirm that \(2^2 + 3^2\) does indeed equal \(\sqrt{13}^2\). This process completes the proof that the combined motion described by 2cos(t) + 3sin(t) can be expressed with an amplitude \(\sqrt{13}\), fundamental for understanding the full scope of the ball's movement.