Problem 29 - Algebra 2 and Trigonometry

Question

The voltage \(E\) (in volts) in an electrical circuit is given by the function $$ E=20 \cos (\pi t) $$ where \(t\) is time in seconds. a. Graph the voltage \(E\) in the interval \(0 \leq t \leq 2\) . b. What is the voltage of the electrical circuit when \(t=1 ?\) c. How many times does the voltage equal 12 volts in the first two seconds? d. Find, to the nearest hundredth of a second, the times in the first two seconds when the voltage is equal to 12 volts. (1) Let \(\theta=\pi t .\) Solve the equation \(20 \cos \theta=12\) in the interval \(0 \leq \theta<2 \pi\) (2) Use the formula \(\theta=\pi t\) and your answers to part \((1)\) to find \(t\) when \(0 \leq \theta<2 \pi\) and the voltage is equal to 12 volts.

Step-by-Step Solution

Answer

The voltage at \( t=1 \) is -20 volts. Voltage equals 12 volts twice: at approximately \( 0.29 \) seconds and \( 1.71 \) seconds.

Step 1: Understanding the Voltage Equation

The voltage is given by \( E = 20 \cos(\pi t) \), where \( t \) is the time in seconds. This is a cosine function with a period of 2 seconds since \(\cos(\pi t)\) completes a full cycle when \( \pi t \) goes from 0 to \( 2\pi \). This means that within our interval \( 0 \leq t \leq 2\), the graph will show one full cosine wave.

Step 2: Graphing the Voltage

The graph of \( E = 20 \cos(\pi t) \) in the interval \( 0 \leq t \leq 2 \) starts at 20 volts when \( t = 0 \), drops to -20 volts when \( t = 1 \), and returns to 20 volts at \( t = 2 \). This is because the standard period of \( \cos(x) \) is \( 2\pi \). Here, \( \cos(\pi t) \) compresses this period to \( 2 \) seconds, creating a full wave in that interval.

Step 3: Calculating Voltage at Specific Time (t=1)

To find the voltage at \( t = 1 \), substitute \( t = 1 \) into the voltage equation: \( E = 20 \cos(\pi \times 1) = 20 \cos(\pi) = 20 \times -1 = -20 \) volts.

Step 4: Determining When Voltage Equals 12 Volts

Set \( 20 \cos(\pi t) = 12 \). Solve for \( \cos(\pi t) \): \( \cos(\pi t) = 0.6 \). Now solve \( \theta = \pi t \) using the formula for angle cosine: in \( 0 \leq \theta < 2\pi \), \( \theta = \cos^{-1}(0.6) \) and \( \theta = -\cos^{-1}(0.6) + 2\pi \).

Step 5: Solving for Time Values

From Step 4, find \( \theta = \cos^{-1}(0.6) \approx 0.927 \) radians and \( \theta = 2\pi - \cos^{-1}(0.6) \approx 5.356 \) radians. Now solve \( \pi t = \theta \): For the first angle: \( \pi t = 0.927 \) gives \( t = \frac{0.927}{\pi} \approx 0.295 \) seconds. For the second angle: \( \pi t = 2\pi - 0.927 \) gives \( t = \frac{5.356}{\pi} \approx 1.705 \) seconds.

Key Concepts

Voltage Function

Understanding the voltage function can help us determine how the voltage in an electrical circuit changes over time. In our case, the voltage, denoted by \( E \), is given by the function \( E = 20 \cos(\pi t) \). Here, \( t \) is the time in seconds, and the number 20 represents the amplitude of the cosine wave. This function is particularly interesting because it describes how voltage fluctuates, and it uses a cosine trigonometric function to do so. The amplitude indicates the maximum voltage, which is 20 volts here. The term \( \pi t \) modifies the standard cosine function, compressing its period to 2 seconds. This is because one full cycle of \( \cos(\pi t) \) is completed when \( \pi t \) progresses from 0 to \( 2\pi \). Thus, by analyzing this function, we can predict the voltage at any specific time \( t \), track its periodic nature, and solve for problems related to voltage fluctuations.

Trigonometric Equations

Trigonometric equations arise when you need to solve for an unknown angle or variable within a trigonometric function, like the cosine function in our voltage equation. In our exercise, we set the equation \( 20 \cos(\pi t) = 12 \) to find when the voltage equals 12 volts. To solve this, we first rearrange the equation to isolate the cosine term: \( \cos(\pi t) = 0.6 \). Thus, our task is to find the values of \( \pi t \) that satisfy this equation. This often involves using the inverse cosine function, \( \cos^{-1} \), to find the principal angle that corresponds to \( \cos^{-1}(0.6) \approx 0.927 \) radians. However, cosine functions have symmetry properties that produce additional solutions. For a full range of angles, \( \theta = \pi t = 2\pi - \cos^{-1}(0.6) \approx 5.356 \) also satisfies the equation. This gives us two possible angles within the interval \( 0 \leq \theta < 2\pi \). These provide us values for \( t \) that correspond to \( 12 \) volts, critical in solving trigonometric equations.

Graphing Trigonometric Functions

Graphing trigonometric functions like the cosine function is an essential skill for visualizing how a voltage function behaves over time. For the function \( E = 20 \cos(\pi t) \), the graph helps us understand its periodic nature and amplitude. To graph \( E = 20 \cos(\pi t) \) in the interval \( 0 \leq t \leq 2 \):

• Start at \( t = 0 \) with the voltage at its maximum, 20 volts.
• As \( t \) increases to 1, the voltage decreases to its minimum, -20 volts, illustrating the downward peak of the wave.
• Then, as \( t \) approaches 2, the voltage returns to 20 volts, completing one full cycle of the wave.

This pattern repeats due to the periodicity of the cosine function. Understanding graphing helps confirm solutions to trigonometric equations, such as when we calculate specific time values for a target voltage. By visualizing the graph, such solutions become intuitive, especially in understanding voltages at particular times.