Problem 65 - Precalculus : Building Concepts and Connections

Question

The voltage in an electrical circuit is given by the function $$V(t)=\sin \left(3 t-\frac{\pi}{2}\right)$$ What is the smallest non-negative value of \(t\) at which the voltage is equal to \(0 ?\)

Step-by-Step Solution

Answer

The smallest non-negative value of \(t\) at which the voltage is equal to \(0\) is \(t = \frac{\pi}{2}\).

Step 1: Set Up Equation

Since we are asked to find the time \(t\) when the voltage \(V(t)\) is equal to zero, we can set the function \(V(t) = \sin \left(3t - \frac{\pi}{2}\right)\) equal to zero. This results in the equation: \(\sin \left(3t - \frac{\pi}{2}\right) = 0\).

Step 2: Solve for angle argument

The sine function is zero at \(\pi/2 + k \pi\) where \(k\) is any integer. Equate \(\pi/2 + k \pi\) to angle argument of sine function \(3t - \frac{\pi}{2}\). Therefore, \(3t - \frac{\pi}{2} = \frac{\pi}{2} + k \pi\), where \(k\) is any integer.

Step 3: Solve for \(t\)

From the above equation, we solve for \(t\) by following algebraic steps. Add \(\frac{\pi}{2}\) to both sides and then divide by 3 to get \(t\): \(t = \frac{\pi + 2k \pi}{6}\). Now, for \(t\) to be non-negative, select the smallest integer \(k\) such that \(t \geq 0\). As \(k=0\), \(t\) is negative, and with \(k=1\) gives \(t= \frac{\pi}{2}\). Thus, the smallest non-negative \(t\) is \(\frac{\pi}{2}\).

Step 4: Check Result

We can check our result by substituting \(t = \frac{\pi}{2}\) back into the original equation: \(\sin \left(3(\frac{\pi}{2}) - \frac{\pi}{2}\right) = \sin \left(\frac{3\pi}{2} - \frac{\pi}{2}\right) = \sin (\pi) = 0\). As it satisfies the equation, our result is verified.

Key Concepts

Understanding the Sine Function

The sine function, denoted as \( \sin(\theta) \), is one of the fundamental trigonometric functions used in mathematics. This function is periodic, meaning it repeats its values in regular intervals or periods, specifically every \(2\pi\) radians. Such periodicity makes sine invaluable in modeling repetitive phenomena such as sound waves and electrical signals. The role of the sine function is to associate every angle \(\theta\) with a value between -1 and 1:

• At \(\theta = 0 \), \( \sin(\theta) = 0\).
• At \(\theta = \pi/2 \), \( \sin(\theta) = 1\).
• At \(\theta = \pi \), \( \sin(\theta) = 0\) again, signaling a zero crossing.

Recognizing these specific points helps solve expressions where the sine function equates to zero, as its zero values are integral to problems involving periodic functions like voltage in circuits.

Navigating an Angle Argument

An angle argument in trigonometry is the expression inside the function, like the entire expression \(3t - \frac{\pi}{2}\) inside the sine function in this exercise. The goal is finding the values of \(t\) that make this argument result in zero for the function value of \(\sin(\theta) = 0\). In the sine function, this happens at multiples of \(\pi\), namely \(\theta = \pi k\) where \(k\) is any integer.
For the voltage problem:

• The expression \(3t - \frac{\pi}{2}\) is set equal to \(\pi k\).
• This derives into the formula \(3t - \frac{\pi}{2} = \pi k\), which simplifies into the equation for solution through algebra.

Understanding how to manipulate and substitute into this expression is key to finding specific solutions, especially in timing circuits where precision is critical.

Deriving an Algebraic Solution

An algebraic solution involves manipulating an equation to isolate a variable—in our scenario, the variable is \(t\). From the earlier steps, we solve \(3t - \frac{\pi}{2} = \pi k\) to find \(t\). Follow these steps to derive the solution:

• Add \(\frac{\pi}{2}\) to both sides of the equation: \(3t = \pi k + \frac{\pi}{2}\).
• Divide every term by 3 to solve for \(t\): \(t = \frac{\pi k + \pi/2}{3}\).

The goal is to find the smallest non-negative \(t\). By testing integers for \(k\), starting at \(k=1\), produces \(t = \pi/2\) which is non-negative and satisfies the equation when substituted back. This checks out as correct, demonstrating that each algebraic transformation remains consistent and valid through problem-solving.