Problem 6 - Thomas Calculus

Question

Electrical power The power \(P\) (watts) of an electric circuit is related to the circuit's resistance \(R\) (ohms) and current \(I\) (amperes) by the equation \(P=R I^{2}\) . a. How are \(d P / d t, d R / d t,\) and \(d I / d t\) related if none of \(P, R,\) and \(I\) are constant? b. How is \(d R / d t\) related to \(d I / d t\) if \(P\) is constant?

Step-by-Step Solution

Answer

a. \(\frac{dP}{dt} = I^2 \frac{dR}{dt} + 2RI \frac{dI}{dt}\). b. \(\frac{dR}{dt} = -\frac{2R}{I} \frac{dI}{dt}\) if \(P\) is constant.

Step 1: Differentiate the Power Equation

The given equation for electrical power is \( P = R I^2 \). Since none of \( P, R, \) and \( I \) are constant, we need to differentiate with respect to time \( t \) to find the relationship between \( \frac{dP}{dt}, \frac{dR}{dt}, \) and \( \frac{dI}{dt} \). Use the product rule and chain rule for differentiation: \[\frac{dP}{dt} = \frac{d}{dt}(R I^2) = \frac{dR}{dt} \cdot I^2 + R \cdot \frac{d}{dt}(I^2)\] Apply the chain rule on \( I^2 \), which gives \( 2I \frac{dI}{dt} \). Therefore, \[\frac{dP}{dt} = \frac{dR}{dt} \cdot I^2 + R \cdot 2I \frac{dI}{dt} = I^2 \frac{dR}{dt} + 2RI \frac{dI}{dt}\]

Step 2: Solve for Constant Power

For part b, we assume \( P \) is constant, so \( \frac{dP}{dt} = 0 \). Using the equation derived in Step 1, set \( \frac{dP}{dt} = 0 \) and solve for the relationship between \( \frac{dR}{dt} \) and \( \frac{dI}{dt} \): \[0 = I^2 \frac{dR}{dt} + 2RI \frac{dI}{dt}\] Rearrange the terms to express \( \frac{dR}{dt} \) in terms of \( \frac{dI}{dt} \): \[I^2 \frac{dR}{dt} = -2RI \frac{dI}{dt}\] Divide both sides by \( I^2 \) (assuming \( I eq 0 \)): \[\frac{dR}{dt} = -\frac{2R}{I} \frac{dI}{dt}\]

Key Concepts

Electrical Circuits

Electrical circuits are fundamental components in many devices. They consist of elements like resistors, capacitors, and power sources, which allow the flow of electrical current. In physics, understanding circuits involves knowing how different quantities like power, resistance, and current relate and affect each other.
One of the core relationships in circuits is given by the power equation: \( P = R I^2 \). This equation links power \(P\) in watts, resistance \(R\) in ohms, and current \(I\) in amperes.

• **Power** represents the rate at which energy is consumed or transformed in a circuit.
• **Resistance** is a measure of how much a component opposes the flow of current.
• **Current** refers to the flow of electric charge.

Understanding these relationships helps in designing circuits that are efficient and safe for use.

Product Rule

The product rule is a key concept in calculus, especially useful for differentiation when you have a product of two functions. In the case of the power equation \( P = R I^2 \), \( R \) and \( I^2 \) are multiplied.
When differentiating the product of these two functions with respect to time \( t \), the product rule comes into play. The product rule states that if you have two functions \( u(t) \) and \( v(t) \), the derivative of their product is given by: \[(uv)' = u'v + uv'\] Here, \( u = R \) and \( v = I^2 \). Applying the product rule to the power equation, we have:\[\frac{d}{dt}(RI^2) = \frac{dR}{dt} \cdot I^2 + R \cdot \frac{d}{dt}(I^2)\] This differentiation is crucial to understand how changes in resistance and current affect power.

Chain Rule

The chain rule is another essential calculus tool for differentiation, particularly when dealing with composite functions. In our context, we need the chain rule to differentiate \( I^2 \) with respect to time.
The chain rule states that if a variable \( y \) is a function of \( u \), which is itself a function of \( t \), then the derivative of \( y \) with respect to \( t \) is:\[\frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt}\] For \( I^2 \), we recognize \( y = I^2 \) and \( u = I \), so:\[\frac{d}{dt}(I^2) = 2I \frac{dI}{dt}\] This differentiation reveals the rate at which the square of the current changes with respect to time, and it is crucial for solving the exercise.

Time Rate of Change

Time rate of change is a concept that helps us understand how a quantity changes as time progresses. In our exercise, we are interested in seeing how power \( P \), resistance \( R \), and current \( I \) vary over time.

• \(\frac{dP}{dt}\) is the rate at which power changes.
• \(\frac{dR}{dt}\) is the rate at which resistance changes.
• \(\frac{dI}{dt}\) is the rate at which current changes.

Knowing these rates is important when analyzing circuits, especially in dynamic situations where circuit parameters are not constant.
For example, in the given task, we find that when power is constant, any increase in resistance is compensated by a decrease in current. This understanding is vital in managing and controlling electrical circuits to maintain desired performance.