Problem 43 - Calculus Single Variable

Question

If \(R_{1}\) and \(R_{2}\) are parallel variable resistances, then the resulting resistance \(R\) satisfies \(1 / R=1 / R_{1}+1 / R_{2} .\) If \(R_{1}\) increases at the rate of \(0.6 \Omega / \mathrm{s}\) when \(R_{1}=40 \Omega,\) and \(R_{2}=20 \Omega,\) then at what rate must \(R_{2}\) decrease if \(R\) remains constant? (The unit by which resistance in an electric circuit is usually measured, the \(o h m,\) is denoted by \(\Omega .)\)

Step-by-Step Solution

Answer

\(R_2\) must decrease at a rate of \(0.15 \Omega/s\).

Step 1: Understand the Problem

We need to find the rate at which resistance \(R_2\) must decrease when \(R_1\) is increasing, such that the total resistance \(R\) remains constant.

Step 2: Write the Given Information

We have the following: \(\frac{dR_1}{dt} = 0.6 \Omega/s\), \(R_1 = 40 \Omega\), \(R_2 = 20 \Omega\). We need to find \(\frac{dR_2}{dt}\) when \(R\) remains constant.

Step 3: Formula for Parallel Resistance

The formula for the total resistance \(R\) for two parallel resistors \(R_1\) and \(R_2\) is given by: \[\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}\]

Step 4: Differentiate the Formula

Differentiate both sides of the equation with respect to time \(t\), remembering that \(R\) is constant: \[0 = \frac{-1}{R_1^2} \cdot \frac{dR_1}{dt} + \frac{-1}{R_2^2} \cdot \frac{dR_2}{dt}\]

Step 5: Substitute Known Values

Substitute the known values \(R_1 = 40 \Omega\), \(\frac{dR_1}{dt} = 0.6 \Omega/s\), and \(R_2 = 20 \Omega\) into the differentiated equation:\[0 = \frac{-1}{(40)^2} \times 0.6 + \frac{-1}{(20)^2} \cdot \frac{dR_2}{dt}\]

Step 6: Solve for \(\frac{dR_2}{dt}\)

Rearrange the equation to solve for \(\frac{dR_2}{dt}\):\[\frac{dR_2}{dt} = \frac{0.6 \times (20)^2}{(40)^2}\]Calculate the result to find \(\frac{dR_2}{dt}\).

Step 7: Calculate the Value

Perform the calculation:\[\frac{dR_2}{dt} = \frac{0.6 \times 400}{1600} = 0.15 \Omega/s\]Thus, \(R_2\) must decrease at a rate of \(0.15 \Omega/s\).

Key Concepts

Parallel Resistance

The concept of parallel resistance is crucial in understanding circuits where multiple paths are available for current to flow. When resistors are placed in parallel, the total or equivalent resistance is determined by combining the reciprocal of each resistance. In mathematical terms, the formula is given by:

• \( \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \)

Here, \( R \) is the effective resistance, and \( R_1 \) and \( R_2 \) are the individual resistances. Each path contributes inversely to the total resistance, meaning the path with the smallest resistance can dominate the overall current flow.
When resistors are added in parallel, the total resistance decreases. This is contrary to adding resistors in a series, where resistances sum up and the total value increases. Understanding how resistances combine in parallel is vital especially in designing circuits that require specific current and resistance values.

Rate of Change

In calculus, the rate of change is a crucial concept used to describe how one quantity changes relative to another. In the context of the given exercise, it tells us how the resistance \( R_2 \) must change over time to keep the total resistance \( R \) constant as \( R_1 \) changes. The rate of change is represented by the derivative, denoted as \( \frac{dR_2}{dt} \), indicating the change in \( R_2 \) per unit time.

• \( \frac{dR_1}{dt} = 0.6 \Omega/s \) – this is the rate at which \( R_1 \) is increasing.
• We need to find \( \frac{dR_2}{dt} \) to maintain constant \( R \).

Differentiating the equation of parallel resistance with respect to time allows us to set up a relationship between the rates of change of \( R_1 \) and \( R_2 \). By substituting known values, we calculate the required rate of \( R_2 \) to change. This example illustrates the power of using calculus to solve real-world problems by understanding and manipulating rates of change.

Calculus in Physics

Calculus is an indispensable tool in physics, as it allows for the mathematical modeling of dynamic systems and changes over time. In the realm of electricity and circuits, calculus helps in understanding how varying one parameter can influence others. This exercise demonstrates how differentiating the formula for parallel resistance aids in analyzing how changes in one resistor impact another. Many physics concepts inherently rely on calculus, including:

• Motion and its related concepts such as velocity and acceleration.
• Force and energy calculations that depend on rates of change.
• Circuit analysis, particularly in analyzing alternating currents and impedance.

In this scenario, by employing calculus we can determine how to adjust component values dynamically, ensuring circuits operate as intended despite changes. This is just one of many examples where calculus serves as the backbone for problem-solving in physics, providing insight into the nature and behavior of physical systems.