Problem 81 - 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 \(I(t)=k\). $$I(t)=20 \sin (60 \pi t-6 \pi) ; \quad k=-10$$

Step-by-Step Solution

Answer

The smallest exact value of \(t\) is \(\frac{43}{360}\) seconds.

Step 1: Set Equation Equal to k

We start by setting the given equation for current equal to the specified value of \(k\). So, \(I(t) = 20 \sin(60\pi t - 6\pi) = -10\).

Step 2: Isolate the Sine Function

Divide both sides of the equation by 20 to isolate the \(\sin\) function. This gives us \(\sin(60\pi t - 6\pi) = -\frac{1}{2}\).

Step 3: Identify the Angle for Sine Value

We know that \(\sin(\theta) = -\frac{1}{2}\) at \(\theta = -\frac{\pi}{6} + 2n\pi\) or \(\theta = \frac{7\pi}{6} + 2n\pi\), where \(n\) is an integer. These are the general solutions for the sine function equal to \(-\frac{1}{2}\).

Step 4: Calculate Specific Values for t

We identify the specific angle equivalent to \(60\pi t - 6\pi\). Let's use \(\theta = \frac{7\pi}{6}\). So, we set \(60\pi t - 6\pi = \frac{7\pi}{6}\).Solving for \(t\), we have:\[ 60\pi t - 6\pi = \frac{7\pi}{6} \]\[ 60\pi t = \frac{7\pi}{6} + 6\pi \]\[ 60\pi t = \frac{7\pi + 36\pi}{6} \]\[ 60\pi t = \frac{43\pi}{6} \]\[ t = \frac{43}{360} \] This is the smallest positive value for \(t\).

Step 5: Verify the Smallest Positive t

Verify if there are smaller values by checking \(\theta = -\frac{\pi}{6} + 2n\pi\) to find the first positive \(t\) that satisfies the equation. For \(\theta = -\frac{\pi}{6} + 2\pi\), the solution follows:\[ 60\pi t - 6\pi = -\frac{\pi}{6} + 2\pi \]\[ 60\pi t - 6\pi = \frac{11\pi}{6} \]\[ 60\pi t = \frac{11\pi}{6} + 6\pi \]\[ 60\pi t = \frac{11\pi + 36\pi}{6} \]\[ 60\pi t = \frac{47\pi}{6} \]\[ t = \frac{47}{360} \] This value is the next solution but is slightly larger than \(\frac{43}{360}\). Thus, \(\frac{43}{360}\) remains the smallest.

Key Concepts

Alternating Current

Alternating current (AC) is an electric current that periodically reverses direction. Unlike direct current (DC) which flows only in one direction, AC circuits are commonly used in power supplies because they can easily be transformed to different voltages and are more efficient over long distances.
In the context of this exercise, the current in an AC circuit is modeled by a sine function, which describes the oscillation over time.

• A key property of AC is its frequency, measured in hertz (Hz), which indicates how many cycles occur per second.
• The maximum current (amplitude) and the phase shift (horizontal shift of the wave) can be adjusted, which change the behavior of the current.

Understanding how variables like time and angle affect alternating current is crucial for electrical engineering applications.

Sine Function

The sine function is a periodic function that is fundamental in trigonometry. It describes a smooth wave that is symmetric about the origin, typically used to model oscillatory processes like sound waves, light waves, and AC currents.
The sine function can be written as \( \sin(x) \), where \( x \) is the angle in radians. Key properties of this function include:

• It has a range between -1 and 1.
• The period is \( 2\pi \), which means the function repeats every \( 2\pi \) radians.
• At different points in its cycle, the sine function has defined values (e.g., \( \sin(0) = 0 \), \( \sin(\pi/2) = 1 \)).

Sine functions in AC circuits represent how the current alternates over time, with parameters adjusted to reflect the circuit's specific characteristics.

General Solutions for Sine

General solutions for trigonometric equations provide a way to express all possible angles that satisfy the equation. For a sine function, if \( \sin(\theta) = a \), there are infinitely many solutions that take the form:
\( \theta = \sin^{-1}(a) + 2n\pi \) and \( \theta = (\pi - \sin^{-1}(a)) + 2n\pi \), where \( n \) is any integer.
This arises because the sine function repeats every \( 2\pi \) radians.
In the exercise, to find angles where \( \sin(\theta) = -\frac{1}{2} \), we use:

• \( \theta = -\frac{\pi}{6} + 2n\pi \)
• \( \theta = \frac{7\pi}{6} + 2n\pi \)

Using these general solutions allows us to express all possible instances where the sine function equals a specific value within its range.

Angle Calculation

Angle calculation in trigonometry often involves solving for an angle \( \theta \) that satisfies a given trigonometric equation. This is vital in modelling contexts like alternating current, where phase shifts represent delays or advances in the wave.
In problems involving a sine function, isolating the angle involves steps such as:

• Setting the trigonometric function equal to a known value.
• Dividing by coefficients to find the angle in a simplified form.

Once the angle is isolated, substitute possible general solutions to solve for practical application times or angles.
In the original exercise, calculating angle involves solving equations like \( 60\pi t - 6\pi = \frac{43\pi}{6} \) that allow us to find the smallest possible \( t \) satisfying conditions of the AC function, corresponding to a specific current.