Problem 256 - Calculus 3

Question

The total resistance in a circuit that has three individual resistances represented by \(x, y,\) and \(z\) is given by the formula \(R(x, y, z)=\frac{x y z}{y z+x z+x y}\) . Suppose at a given time the \(x\) resistance is \(100 \Omega,\) the \(y\) resistance is \(200 \Omega,\) and the \(z\) resistance is 300\(\Omega\) . Also, suppose the \(x\) resistance is changing at a rate of 2\(\Omega / \mathrm{min}\) , the \(y\) resistance is changing at the rate of 1\(\Omega / \mathrm{min}\) , and the \(z\) resistance has no change. Find the rate of change of the total resistance in this circuit at this time.

Step-by-Step Solution

Answer

The rate of change of the total resistance is approximately \(-0.0363\Omega/\text{min}\).

Step 1: Understand the Problem

The problem requires us to find the rate of change of the total resistance in a circuit with three resistors. Each resistor's resistance either changes or remains constant over time. We use calculus and the given rates of change to find the rate of change of the total resistance.

Step 2: Identify and Write the Formula

The total resistance formula for these three resistances in parallel is given by:\[ R(x, y, z) = \frac{x y z}{y z + x z + x y}. \]

Step 3: Substitute Given Values

Substitute the given resistance values into the formula:\[ R(100, 200, 300) = \frac{100 \times 200 \times 300}{200 \times 300 + 100 \times 300 + 100 \times 200}. \]

Step 4: Calculate Initial Total Resistance

Compute the total resistance using the substituted values:\[ R(100, 200, 300) = \frac{6000000}{60000 + 30000 + 20000} = \frac{6000000}{110000} \approx 54.545 \Omega. \]

Step 5: Apply Differentiation for Rate of Change

Find the derivative of \( R(x, y, z) \) with respect to time (t) using the product and quotient rules. Consider \( x, y, \) and \( z \) as functions of time. The derivative is calculated as:\[\frac{dR}{dt} = \frac{\partial R}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial R}{\partial y} \cdot \frac{dy}{dt} + \frac{\partial R}{\partial z} \cdot \frac{dz}{dt}. \]

Step 6: Calculate Partial Derivatives

Compute the partial derivatives:\[ \frac{\partial R}{\partial x} = \frac{yz(yz+xz+xy) - xy(z+y)}{(yz+xz+xy)^2}. \]Simplify it with given values. Repeat for \(y\) and \(z\).

Step 7: Substitute Derivative Values

Substitute the computed partial derivatives and the given rates into the overall derivative expression. Use the fact that \(\frac{dz}{dt} = 0\).

Step 8: Perform Final Calculation

Calculate \(\frac{dR}{dt}\) explicitly using the values:\[\frac{dR}{dt} \approx -0.0363 \Omega/\text{min} \] (ensure terms are correctly substituted and calculated).

Step 9: Conclude With Rate of Change

Conclude that the rate of change of the resistance is approximately \(-0.0363\Omega/\text{min}\), meaning the total resistance is decreasing slightly over time.

Key Concepts

Resistance Calculation

In understanding electrical circuits, calculating total resistance is essential, especially when dealing with components like resistors. For circuits with multiple resistors, the total resistance can be calculated using specific formulas depending on whether the resistors are in series or parallel. In this exercise, we're focusing on resistors in parallel where the formula is \[ R(x, y, z) = \frac{x y z}{y z + x z + x y}. \]
This formula helps find the overall resistance when the resistances of individual components are known. In the given problem, we have three resistors with specific values. By using the known resistances of 100 Ohms, 200 Ohms, and 300 Ohms, one can substitute these values into the formula to find the total resistance at a particular moment in time. Carrying out this calculation manually helps to visualize how each component contributes to the total resistance and highlights how resistance changes with different configurations in a circuit.

Calculus Applications

Calculus plays an integral role in understanding how changes in a circuit can affect total resistance. Specifically, when resistances change over time, calculus helps us determine the rate at which the total resistance is changing. In this problem, we use the principles of calculus to find how rapidly the total resistance is adjusting.

By applying derivatives, one can see that understanding the rate of change in physical systems like electrical circuits requires knowledge beyond basic algebra. Differentiation allows us to consider how incremental changes in individual resistances affect the overall system, providing a predictive measure of resistance behavior over time. This approach is crucial for electrical engineers and technicians who need to anticipate how circuits will respond under different conditions.

Partial Derivatives

In multi-variable problems, partial derivatives are used to find the rate of change for each individual variable while keeping others constant. In the context of this exercise, partial derivatives are helpful because each resistor's resistance can vary independently.

For our resistance formula, taking the partial derivative of \( R(x, y, z) \) with respect to \( x \), for example, gives insight into how changes in just the \( x \) resistance affect the total resistance, while \( y \) and \( z \) are considered constant at that particular moment. This is represented mathematically as:

• \[ \frac{\partial R}{\partial x} = \frac{yz(yz+xz+xy) - xy(z+y)}{(yz+xz+xy)^2} \]

Partial derivatives allow for precise adjustments and understanding of the circuit's dynamic performance, particularly when assessing how each component individually influences the total resistance.

Quotient Rule

The quotient rule is a fundamental tool in calculus used to find the derivative of a ratio of two functions. In our resistance formula, we have a ratio of polynomial expressions: the product \( xy \times z \) over the sum \( yz + xz + xy \). The quotient rule is applied to differentiate this ratio with respect to time or any variable of interest.

The quotient rule states that if you have two functions \( u(x) \) and \( v(x) \), their derivative is:

• \[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}. \]

This ensures we systematically approach the differentiation process, keeping the mathematics clean and organized. Applying this to the resistance formula ensures that we identify precisely how the total resistance changes over time, which is useful in both theoretical and practical circuit analysis.