Problem 56 - Advanced Engineering Mathematics

Question

The voltage across a conductor is increasing at a rate of 2 volts/min and the resistance is decreasing at a rate of 1 ohm/min. Use \(I=E / R\) and the Chain Rule to find the rate at which the current passing through the conductar is changing when \(R=50\) ohms and \(E=60\) volts.

Step-by-Step Solution

Answer

The current changes at a rate of \( \frac{8}{125} \) amperes per minute.

Step 1: Understand the given data

We are given that the rate of change of voltage, \( \frac{dE}{dt} = 2 \) volts/min, and the rate of change of resistance, \( \frac{dR}{dt} = -1 \) ohm/min. We need to find the rate of change of current, \( \frac{dI}{dt} \), when \( R = 50 \) ohms and \( E = 60 \) volts.

Step 2: Express current formula and differentiate

The current \( I \) is given by \( I = \frac{E}{R} \). To find \( \frac{dI}{dt} \), we differentiate \( I \) with respect to \( t \) using the Quotient Rule: \( \frac{dI}{dt} = \frac{R \cdot \frac{dE}{dt} - E \cdot \frac{dR}{dt}}{R^2} \).

Step 3: Substitute known values

Substitute \( R = 50 \), \( E = 60 \), \( \frac{dE}{dt} = 2 \), and \( \frac{dR}{dt} = -1 \) into the differentiated formula: \[ \frac{dI}{dt} = \frac{50 \cdot 2 - 60 \cdot (-1)}{50^2}. \]

Step 4: Simplify the expression

Calculate the expression: \[ \frac{dI}{dt} = \frac{100 + 60}{2500} = \frac{160}{2500}. \] Simplify this to: \[ \frac{dI}{dt} = \frac{16}{250} = \frac{8}{125}. \]

Step 5: Conclusion

The rate at which the current is changing is \( \frac{8}{125} \) amperes per minute.

Key Concepts

Rate of Change

In calculus, the concept of "rate of change" refers to how quickly a variable quantity is changing over time. It's akin to measuring how fast something is speeding up or slowing down. This is often used in real-world applications such as physics, economics, and biology. For this exercise, we are specifically looking at how the electrical current in a circuit changes over time. The problem states that voltage across a conductor increases by 2 volts per minute (\( \frac{dE}{dt} = 2 \) V/min) and resistance decreases by 1 ohm per minute (\( \frac{dR}{dt} = -1 \) Ω/min). Our task is to calculate the change in current over time, \( \frac{dI}{dt} \).

• Understand the change: Voltage and resistance affect current; changes in them mean changes in current.
• Application: Knowing electric current's rate of change helps in designing efficient circuits and preventing failures.

This exercise involves using the Chain Rule and Quotient Rule in calculus to figure out this change, giving us insight into how interconnected quantities affect each other.

Quotient Rule

The Quotient Rule in calculus is essential when you're dealing with divisions of functions, such as electrical formulas like \( I = \frac{E}{R} \). In our problem, the current \( I \) is given by this formula, where \( E \) is the voltage and \( R \) is the resistance. Both are changing over time, requiring us to find \( \frac{dI}{dt} \).To apply the Quotient Rule, remember:\[\frac{d}{dt}\left( \frac{E}{R} \right) = \frac{R \cdot \frac{dE}{dt} - E \cdot \frac{dR}{dt}}{R^2}\]This rule helps us differentiate without errors, providing a structured approach to handle division of two functions:

• Top function derivative: Multiply the bottom by the derivative of the top.
• Bottom function derivative: Multiply the top by the derivative of the bottom.
• Subtract and divide: Subtract and place over the square of the bottom function.

Using the Quotient Rule helps accurately calculate complex rates of change when variables are expressed as fractions, making it invaluable in scenarios like this one.

Electrical Current and Circuits

Electrical current is the flow of electric charge, and its behavior in circuits is fundamental to understanding how electrical systems work. The formula \( I = \frac{E}{R} \) tells us that current depends on both voltage (\( E \)) and resistance (\( R \)). With changing E and R, the current I also changes in response.In this problem, we examined how increases in voltage and decreases in resistance affect the current across a conductor.

• Increasing voltage: More voltage with the same resistance increases the current.
• Decreasing resistance: Lower resistance with the same voltage increases the current.

By calculating the rate of change of the current, engineers can predict and manage how an electrical system will perform as conditions change, such as component aging or varying supply. Current stability is crucial in preventing malfunctions and optimizing system efficiency. Understanding these principles helps design safer and more efficient circuits, critical in everything from household electronics to complex industrial machinery.