Problem 79 - Algebra and Trigonometry

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 $$ 2 \cos t+3 \sin t=\sqrt{13} \cos (t-\theta) $$ 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 given trigonometric identity is verified and the amplitude of the ball's motion is \(\sqrt{13}\) feet, while the period is \(2\pi\) seconds.

Step 1: Simplify the given expression

Simplify \(2 \cos t+3 \sin t=\sqrt{13} \cos (t-\theta)\) to the form \(a \cos t + b \sin t = r \cos (t - \theta)\). Here, \(a = 2\), \(b = 3\), and \(r = \sqrt{13}\).

Step 2: Calculate theta

Using the given identity \(\tan \theta = \frac{b}{a} = \frac{3}{2}\), we can find the value of \(\theta\) in Quadrant I by applying an inverse tangent function. Thus \(\theta = \arctan(\frac{3}{2})\).

Step 3: Verify identity

Validate the given trigonometric identity by substituting \(\theta\) into \(2 \cos t+3 \sin t = \sqrt{13} \cos (t - \theta)\) and simplify each side. If they are equal, then the identity is verified.

Step 4: Determine amplitude and period

The amplitude of the sinusoidal function is the absolute value of the coefficient of the cosine function, hence the amplitude is \(\sqrt{13}\). The period is obtained from the coefficient of \(t\) in the cosine function. The coefficient is \(1\), therefore the period is \(2\pi / 1 = 2\pi\).