Problem 17 - Active Calculus Multivariable

Question

The voltage \(V\) (in volts) across a circuit is given by Ohm's Law: \(V=I R\), where \(I\) is the current (in amps) in the circuit and \(R\) is the resistance (in ohms). Suppose we connect two resistors with resistances \(R_{1}\) and \(R_{2}\) in parallel as shown in Figure \(10.5 .5 .\) The total resistance \(R\) in the circuit is then given by $$\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$$ a. Assume that the current, \(I\), and the resistances, \(R_{1}\) and \(R_{2}\), are changing over time, \(t\). Use the Chain Rule to write a formula for \(\frac{d V}{d t}\). b. Suppose that, at some particular point in time, we measure the current to be 3 amps and that the current is increasing at \(\frac{1}{10}\) amps per second, while resistance \(R_{1}\) is 2 ohms and decreasing at the rate of 0.2 ohms per second and \(R_{2}\) is 1 ohm and increasing at the rate of 0.5 ohms per second. At what rate is the voltage changing at this point in time?

Step-by-Step Solution

Answer

The voltage across the circuit is changing at a rate of 2 volts per second at this point in time.

Step 1: We're given the equation for the total resistance in the circuit with resistors in parallel: \(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\) We are also given Ohm's Law which relates voltage, current, and resistance: \(V=IR\) #Step 2: Differentiate both sides of the total resistance equation with respect to time#

Since all the variables change with time, we need to find the derivative of each term with respect to time. \(\frac{d}{dt}\left(\frac{1}{R}\right)=\frac{d}{dt}\left(\frac{1}{R_{1}}\right)+\frac{d}{dt}\left(\frac{1}{R_{2}}\right)\) #Step 3: Use Chain Rule to calculate the derivatives#

Step 2: Apply the Chain Rule to each term, and we get: \(-\frac{1}{R^{2}}\frac{dR}{dt}=-\frac{1}{R_{1}^{2}}\frac{dR_{1}}{dt}-\frac{1}{R_{2}^{2}}\frac{dR_{2}}{dt}\) #Step 4: Find the formula for the rate of change of voltage with respect to time#

Differentiate Ohm's law equation with respect to time (using Chain Rule and Product Rule) as follows: \(\frac{dV}{dt} = \frac{d(IR)}{dt} = I\frac{dR}{dt}+R\frac{dI}{dt}\) Now we can find the relationship between the rate of change of resistance \(\frac{dR}{dt}\) and other given variables from the equation derived in Step 3: \(\frac{dR}{dt} = R^{2}\left(\frac{1}{R_{1}^{2}}\frac{dR_{1}}{dt}+\frac{1}{R_{2}^{2}}\frac{dR_{2}}{dt}\right)\) Substitute this expression into the formula for \(\frac{dV}{dt}\), we get: \(\frac{dV}{dt} = IR^{2}\left(\frac{1}{R_{1}^{2}}\frac{dR_{1}}{dt}+\frac{1}{R_{2}^{2}}\frac{dR_{2}}{dt}\right) + R\frac{dI}{dt}\) #Step 5: Use the given values to find the rate of change of voltage#

Step 3: We are given the following values at a particular point in time: \(I = 3 A\), \(R_1 = 2\Omega\), \(R_2 = 1\Omega\), \(\frac{dI}{dt}=0.1 \frac{A}{s}\), \(\frac{dR_1}{dt}=-0.2\frac{\Omega}{s}\), and \(\frac{dR_2}{dt}=0.5\frac{\Omega}{s}\). First, we need to find the total resistance at this point in time using the provided equation: \(\frac{1}{R} = \frac{1}{2}+\frac{1}{1}\), which gives \(R = \frac{2}{3}\Omega\). Now we can plug the given values and calculated total resistance into the formula for the rate of change of voltage: \(\frac{dV}{dt} = 3\left(\frac{2}{3}\right)^{2}\left(\frac{1}{2^{2}}\left(-0.2\right)+\frac{1}{1^{2}}\left(0.5\right)\right) + \frac{2}{3}\left(0.1\right)\) #Step 6: Calculate the rate of change of voltage#

Perform the calculation to find the rate of change of voltage at this point in time: \(\frac{dV}{dt} = 3\left(\frac{2}{3}\right)^{2}\left(\frac{-0.2}{4}+\frac{0.5}{1}\right) + \frac{2}{3}\left(0.1\right)\) \(\frac{dV}{dt} = 3\left(\frac{4}{9}\right)\left(-0.05+0.5\right) + \frac{1}{5}\) \(\frac{dV}{dt} = \frac{12}{9}(0.45) + \frac{1}{5}\) \(\frac{dV}{dt} = 1.8 + 0.2 = 2\frac{V}{s}\) Therefore, the voltage across the circuit is changing at a rate of 2 volts per second at this point in time.

Key Concepts

Chain Rule differentiation

Understanding the Chain Rule in differentiation is crucial when dealing with functions that are composed of other functions. Essentially, the Chain Rule helps us differentiate composite functions, and it states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. A classic example of applying the Chain Rule emerges in calculus problems involving circuits, like the one described in the exercise.

In the context of a changing resistance and current in a circuit, we apply the Chain Rule to differentiate complex expressions of resistance and current in terms of time. This is necessary to find the rate at which these quantities are changing. For instance, differentiating \(\frac{1}{R}\) with respect to time requires us to handle the derivative of the reciprocal function \(\frac{1}{R(t)}\), which involves applying the Chain Rule. By mastering the Chain Rule, we gain the ability to track dynamic changes in circuits and many other systems.

Resistors in parallel

When it comes to circuits, resistors can be configured in various ways, with 'parallel' being one of the most common. In a parallel configuration, resistors are connected across the same two points, creating multiple paths for the current to flow. According to Ohm's Law, the voltage across resistors in parallel is the same, but the total resistance offered by the parallel combination differs from that of any single resistor.

The formula for calculating the total resistance \(R\) of parallel resistors, \(\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}}\), is counterintuitive because it shows that adding more resistors in a parallel configuration actually decreases the total resistance. This principle is vital when trying to manage current flow and voltage distribution in electrical systems and plays a significant role in the calculus problem presented, where we were required to find the relationship between the time rate of change in resistance and voltage.

Rate of change of voltage

Voltage changes over time are central to understanding electrical dynamics and are particularly important in applying calculus to electricity. The rate of change of voltage \(\frac{dV}{dt}\) tells us how quickly the voltage is increasing or decreasing at a given moment. It's influenced by variations in current \(I\) and resistance \(R\), as described by Ohm's Law \(V = IR\).

The significance of calculating \(\frac{dV}{dt}\) is underscored when we consider activities like charging a battery, where we are concerned with how quickly the voltage builds up, or in monitoring a circuit's response to changing conditions. In the given problem, this concept is explored by finding the voltage's rate of change as the current and resistances of the parallel resistors vary over time. The complex relationship uncovered by utilizing Ohm's Law and calculus allows engineers and physicists to predict and control the behavior of electronic circuits effectively.