Problem 82 - Precalculus: Functions and Graphs

Question

If \(I(t)\) is the current (in amperes) in an alternating current circuit at time \(t\) (in seconds), find the smallest exact value of \(t\) for which \(I(t)=k\). $$I(t)=40 \sin (100 \pi t-4 \pi) ; \quad k=20$$

Step-by-Step Solution

Answer

The smallest exact value of \(t\) is \(\frac{1}{24}\) seconds.

Step 1: Set Up the Equation

We start with the given equation for the current: \[ I(t) = 40 \sin(100\pi t - 4\pi) \]and equate it to \(k\), which is 20:\[ 40\sin(100\pi t - 4\pi) = 20 \]

Step 2: Solve for the Sine Function

Divide both sides of the equation by 40 to isolate the sine term:\[ \sin(100\pi t - 4\pi) = \frac{20}{40} = \frac{1}{2} \]

Step 3: Determine the Angle for Sine

Recall that \(\sin(\theta) = \frac{1}{2}\) for angles (in radians) such as \(\frac{\pi}{6}\), \(\frac{5\pi}{6}\), and others given by the periodicity of the sine function. We use \(\theta = \frac{\pi}{6}\) for the smallest value. This gives:\[ 100\pi t - 4\pi = \frac{\pi}{6} \]

Step 4: Solve for Time t

Solve the equation for \(t\):Add \(4\pi\) to both sides:\[ 100\pi t = \frac{\pi}{6} + 4\pi \]Combine the terms on the right:\[ 100\pi t = \frac{\pi}{6} + \frac{24\pi}{6} = \frac{25\pi}{6} \]Divide both sides by \(100\pi\):\[ t = \frac{25\pi}{6 \cdot 100\pi} = \frac{25}{600} = \frac{1}{24} \]

Step 5: Review Additional Solutions

Consider further solutions by adding the period of \(2\pi\) to \(\theta\), but since we need the smallest positive \(t\), \(t = \frac{1}{24}\) remains the smallest value.

Key Concepts

Amplitude

In the context of alternating current (AC) circuits, the term "amplitude" refers to the peak value of the current or voltage waveform. It is the highest value reached by the waveform, measured from the baseline (typically zero) to the peak.
In our problem, the amplitude is denoted by the number 40 in the equation \( I(t) = 40 \sin(100\pi t - 4\pi) \). This means that the maximum current reached is 40 amperes.
Amplitude gives an idea of the strength or intensity of the current or voltage in the circuit.

• The higher the amplitude, the stronger the current.
• The amplitude is always positive, regardless of its position in the waveform.

Understanding amplitude is crucial for predicting how much electrical power is delivered by the AC circuit.

Phase Shift

Phase shift in a wave equation indicates a horizontal shift along the time axis. It shows how much the current or voltage waveform is shifted from the usual starting point.
In the equation \( I(t) = 40 \sin(100\pi t - 4\pi) \), the term \(-4\pi\) represents the phase shift. It alters the starting point of the sine wave, effectively delaying or advancing the waveform.
This concept is important when you need to synchronize the phase of different waveforms in an electrical circuit. Phase shift can be visualized as:

• A negative phase shift \(-kx\) shifts the wave to the right.
• A positive phase shift \(+kx\) shifts the wave to the left.

Such adjustments are vital in applications like radio transmissions and signal processing, where wave synchronization ensures optimized performance.

Frequency

Frequency in AC circuits refers to the number of times a waveform repeats its cycle in one second. It's crucial in determining how many complete oscillations the wave completes per second.
Expressed in hertz (Hz), frequency is a fundamental concept that relates to the speed of the waveform.
In the function \( 40 \sin(100\pi t - 4\pi) \), the coefficient of \(t\), namely \(100\pi\), directly relates to the frequency. The standard form \(a\sin(bt - c)\) has the component \(b\) encoding the angular frequency, defined as \(2\pi f\), where \(f\) is the frequency. Thus, the frequency \(f\) can be calculated from \(b/2\pi\).

• For our problem, \( b = 100\pi \), so \( f = \frac{100\pi}{2\pi} = 50 \) Hz.

This tells us that the current oscillates 50 times every second. Frequency is key in designing circuits for specific applications, influencing timing and synchronization across various devices.

Trigonometric Equations

Trigonometric equations involve finding angles (or time instances in waveform cases) that satisfy specific trigonometric identities.
The problem requires solving the trigonometric equation \(40 \sin(100\pi t - 4\pi) = 20\), which is simplified to \(\sin(100\pi t - 4\pi) = \frac{1}{2}\).
To solve it, we recognize standard sine values. The angles \(\theta\) for which \(\sin(\theta) = \frac{1}{2}\) include \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\), among others.

• Because sine is periodic, the general solution involves adding multiples of \(2\pi\cdot n\), where \(n\) is an integer to these angles.
• In our problem, we start by setting up the equation \(100\pi t - 4\pi = \frac{\pi}{6}\) for the smallest value of \(t\).

Solving trigonometric equations accurately is essential for predicting exact time moments in applications such as signal processing and electrical engineering.