Problem 59 - Calculus - AP Edition

Question

The electrical resistance \(R\) of a certain wire is given by \(R=k / r^{2},\) where \(k\) is a constant and \(r\) is the radius of the wire. Assuming that the radius \(r\) has a possible error of \(\pm 5 \%,\) use differentials to estimate the percentage error in \(R .\) (Assume \(k\) is exact.)

Step-by-Step Solution

Answer

The percentage error in resistance \( R \) is approximately \( \pm 10\% \).

Step 1: Differentiate the Formula for Resistance

The electrical resistance is given by the formula \( R = \frac{k}{r^2} \). To find how errors in \( r \) affect \( R \), differentiate \( R \) with respect to \( r \). This gives: \( \frac{dR}{dr} = -\frac{2k}{r^3} \).

Step 2: Use Differentials for Error Estimation

The differential \( dR \) represents the approximate change in resistance due to a change in radius \( dr \). Using the formula for differentials, we have \( dR = \frac{dR}{dr} \cdot dr = -\frac{2k}{r^3} \cdot dr \).

Step 3: Relate Differentials to Percentage Errors

To find the percentage error in \( R \), divide \( dR \) by \( R \) and multiply by 100%. So, the expression is \[ \frac{dR}{R} = \frac{-\frac{2k}{r^3} \cdot dr}{\frac{k}{r^2}} = -2 \frac{dr}{r} \]. Multiply by 100% to get: \( \frac{dR}{R} \times 100\% = -2 \times \frac{dr}{r} \times 100\% \).

Step 4: Substitute the Given Error in Radius

The problem states there's a \( \pm 5\% \) error in \( r \). Express this as \( dr = \pm 0.05r \). Substitute into the percentage error formula: \(-2 \times \frac{dr}{r} \times 100\% = -2 \times \pm 0.05 \times 100\% = \pm 10\% \).

Key Concepts

Electrical Resistance

Electrical resistance is a fundamental concept in physics and electronics, describing how strongly a material opposes the flow of electric current. It is like a traffic cop slowing down vehicles in traffic, only it slows down electrons in a wire.
The formula for electrical resistance used here is given by \( R = \frac{k}{r^2} \), where \( R \) is the resistance, \( k \) is a constant, and \( r \) is the radius of the wire. This equation tells us that resistance decreases as the radius of the wire increases.
This inverse relationship makes intuitive sense: a wider wire allows more room for electric current to flow through, thereby reducing resistance.

• Resistance \( R \) is directly proportional to the constant \( k \).
• Resistance \( R \) is inversely proportional to the square of the radius \( r \).

Understanding electrical resistance and its dependence on wire dimensions is crucial for designing efficient electrical circuits.

Percentage Error

Percentage error is a way to express how much an approximate or measured value deviates from the exact or true value. It allows us to understand the accuracy of measurements in a readily accessible way.
In the context of this problem, percentage error analysis helps us estimate how variations in the radius of a wire affect the resistance. The basic formula for percentage error is:

• Find the difference between the measured and true values.
• Divide the difference by the true value.
• Multiply the result by 100 to convert it to a percentage.

Percentage error = \( \left( \frac{{\text{{approximate value}} - \text{{true value}}}}{{\text{{true value}}}} \right) \times 100 \%\).
The exercise simplifies this process using differentials to approximate how an error in radius translates to an error in resistance, ultimately giving a \( \pm 10\% \) error in resistance due to a \( \pm 5\% \) error in radius.

Radius Error Estimation

Radius error estimation is a method used to predict the impact of any deviation in the radius of a wire on its resistance. This process is crucial in scenarios where precise measurements are necessary, such as in electrical installations or component manufacturing.
In the given exercise, error estimation begins with recognizing that a possible error in the radius can affect the wire's resistance. A radius error expressed as \( dr = \pm 0.05r \) indicates a \( \pm 5\% \) possible error in the radius measurement.
By utilizing calculus, specifically differentials, it's possible to see how small changes in the radius \( dr \) translate into changes in resistance \( dR \). The relation \( \frac{dR}{R} = -2 \frac{dr}{r} \) shows that a \( \pm 5\% \) error in the radius translates into a \( \pm 10\% \) error in resistance, due to the \( -2 \) multiplier in the formula. This approach underscores the sensitivity of our measurements and the importance of precision.