Question

An alternating voltage is given by: \(v=100 \sin 200 t\) volts, where \(t\) is the time in seconds. Calculate the rate of change of voltage when (a) \(t=0.005 \mathrm{~s}\) and (b) \(t=0.01 \mathrm{~s}\)

Short answer

The rate at \( t = 0.005 \text{s} \) is approximately 10806 V/s, and at \( t = 0.01 \text{s} \) is -8322 V/s.

Step-by-step solution

Understand the Given Equation

The voltage is described by the equation \( v = 100 \sin(200t) \) volts. Here, \( v \) represents the voltage, and \( t \) is time in seconds. Our task is to find how fast the voltage is changing with time at specific moments.

Find the Derivative with Respect to Time

To find the rate of change of voltage, we need the derivative of \( v \) with respect to \( t \). Differentiate \( v = 100 \sin(200t) \) to get \( \frac{dv}{dt} = 100 \cdot 200 \cos(200t) = 20000 \cos(200t) \). This derivative gives us the rate of change of voltage at any time \( t \).

Calculate the Rate at \( t = 0.005 \text{s} \)

Substitute \( t = 0.005 \text{s} \) into the derivative \( \frac{dv}{dt} = 20000 \cos(200t) \). This gives: \( \frac{dv}{dt} = 20000 \cos(200 \times 0.005) = 20000 \cos(1) \). Use a calculator to find \( \cos(1) \) (in radians), which is approximately 0.5403. Thus, \( \frac{dv}{dt} \approx 20000 \times 0.5403 = 10806 \text{ volts per second} \).

Calculate the Rate at \( t = 0.01 \text{s} \)

Substitute \( t = 0.01 \text{s} \) into the derivative \( \frac{dv}{dt} = 20000 \cos(200t) \). This gives: \( \frac{dv}{dt} = 20000 \cos(200 \times 0.01) = 20000 \cos(2) \). Use a calculator to find \( \cos(2) \) (in radians), which is approximately -0.4161. Thus, \( \frac{dv}{dt} \approx 20000 \times (-0.4161) = -8322 \text{ volts per second} \).

Key concepts

Rate of Change

When we talk about the "Rate of Change," we refer to how a quantity changes with respect to time or another variable. It's like measuring speed - you want to know how fast or slow something is happening. For the alternating voltage problem, this means figuring out how quickly the voltage, given by the equation \( v = 100 \sin(200t) \), changes over time.
To do this, we calculate something called the derivative, which tells us the rate at which one quantity changes with respect to another. Here, it helps us find how fast the voltage changes as time goes by.

• The rate at \( t = 0.005 \text{s} \) was around 10806 volts per second, telling us the voltage was increasing at that moment.
• At \( t = 0.01 \text{s} \), the rate was around -8322 volts per second, indicating the voltage was decreasing then.

These numbers provide us an insight into the behavior of the voltage over time, showing whether it's ramping up or down at any given point.

Differentiation

Differentiation is a crucial concept in calculus, mainly focused on finding derivatives. It allows us to understand how a function changes at any given point and plays a significant role in solving the exercise about alternating voltage.
In our scenario, we differentiated \( v = 100 \sin(200t) \) to get the derivative \( \frac{dv}{dt} = 20000 \cos(200t) \).
Here’s why differentiation is valuable:

• It provides the exact rate of change at any particular time \( t \).
• By calculating \( \cos(200t) \) at specific moments like \( t = 0.005 \text{s} \) and \( t = 0.01 \text{s} \), we can see how quickly the voltage switches direction (increases or decreases).

This precise computation allows engineers to predict and manage electrical circuits' behavior effectively.

Trigonometric Functions

Trigonometric functions like sine and cosine are essential in various scientific and engineering fields, including analyzing wave functions like alternating voltage. In the given problem, the voltage equation \( v = 100 \sin(200t) \) is based on the sine function, which is fundamental in describing waves.
The cosine function, another trigonometric entity, appears in the derivative \( \frac{dv}{dt} = 20000 \cos(200t) \).
Trigonometric functions have unique properties:

• The sine function helps describe the alternating nature of the voltage, depicting how it rises and falls over time.
• The cosine function, derived from the sine derivative, helps determine the rate at which the voltage change happens.

Understanding these trigonometric functions is crucial for analyzing periodic phenomena like alternating currents in electrical engineering.