Problem 9 - Active Calculus Multivariable

Question

In a simple electric circuit, Ohm's law states that \(V=I R,\) where \(\mathrm{V}\) is the voltage in volts, I is the current in amperes, and \(\mathrm{R}\) is the resistance in ohms. Assume that, as the battery wears out, the voltage decreases at 0.03 volts per second and, as the resistor heats up, the resistance is increasing at 0.02 ohms per second. When the resistance is 100 ohms and the current is 0.02 amperes, at what rate is the current changing? ___________ amperes per second .

Step-by-Step Solution

Answer

The rate at which the current is changing when the resistance is 100 ohms and the current is 0.02 amperes is approximately \(-0.000304 \, \text{amperes/sec}\).

Step 1: Write down the given rate of changes

We are given the following rates of change: - Voltage (V): -0.03 volts/second - Resistance (R): 0.02 ohms/second

Step 2: Write down the given values for when we need to find the rate

We are asked to find the rate at which the current (I) is changing when: - Resistance (R): 100 ohms - Current (I): 0.02 amperes

Step 3: Differentiate the Ohm's law equation with respect to time t

Differentiate the equation \(V = IR\) with respect to time (t) to obtain the rate of change of current: \[\frac{dV}{dt} = \frac{d(IR)}{dt}\] Keep in mind that \(I, R, \) and \(V\) are all functions of time, so we'll use the Chain Rule to differentiate the product \(IR\): \[\frac{dV}{dt} = R \frac{dI}{dt} + I \frac{dR}{dt}\]

Step 4: Plug in the given values and rates of change

Now, we can substitute the given values and rates of change into the equation: \[-0.03 = 100 \cdot \frac{dI}{dt} + 0.02 \cdot 0.02\]

Step 5: Solve for dI/dt

Next, solve for the rate of the change of the current, \(\frac{dI}{dt}\): \[-0.03 = 100 \cdot \frac{dI}{dt} + 0.0004\] \[-0.0304 = 100 \cdot \frac{dI}{dt}\] \[\frac{dI}{dt} = \frac{-0.0304}{100}\] \[\frac{dI}{dt} = -0.000304\]

Step 6: Write the final answer

Thus, the rate at which the current is changing when the resistance is 100 ohms and the current is 0.02 amperes is: \[-0.000304 \, \text{amperes/sec}\]

Key Concepts

Differential Calculus

Differential calculus is a subfield of calculus concerned with the study of how things change. It allows us to calculate the rate of change of quantities. In the exercise, we applied differential calculus to determine how quickly the current was changing in an electric circuit. By taking the derivative of Ohm's Law \(V = IR\) with respect to time, we are able to find this rate of change, known as \(\frac{dI}{dt}\).

This involves applying the Chain Rule, a fundamental technique in calculus used to differentiate a function of a function, because both current \(I\) and resistance \(R\) are variables that change over time. The Chain Rule is expressed as \(\frac{df(g(t))}{dt} = f'(g(t)) \cdot g'(t)\), and in the context of the given problem, it means taking the derivative of each term independently while considering the other as a constant, and then summing the results.

Electric Circuits

Electric circuits are pathways through which electricity flows. Ohm’s Law, a fundamental principle in the study of electric circuits, illustrates the relationship between voltage \(V\), current \(I\), and resistance \(R\) in a simple circuit. According to this law, \(V = IR\), meaning that the voltage (the force driving the current) is equal to the product of the current (the flow rate of charge) and the resistance (which opposes the flow).

In practical terms, if you alter the resistance or the voltage in a circuit, the current will change accordingly. This change is what students were asked to find in the exercise. Understanding how any of these variables affect one another is crucial for students who wish to work with electric circuits, whether it be in designing them or troubleshooting existing systems.

Multivariable Calculus

Multivariable calculus extends the principles of calculus to functions of more than one variable. In the context of the provided exercise, voltage, current, and resistance are all functions of time, representing a multivariable situation because the rate of change of one is affected by the rates of change of the others.

When applying multivariable calculus, we often work with partial derivatives, sine each variable's rate of change is considered with the understanding that other variables are also changing. However, in this problem, we are using the time derivative of Ohm’s Law because we are examining how the current changes over time due to changes in both voltage and resistance. This is an example of how multivariable calculus principles can be narrowed down to handle real-world problems involving several changing variables.