Problem 27 - The Calculus with Analytic Geometry

Question

The measure of the electrical resistance of a wire is proportional to the measure of its length and inversely proportional to the square of the measure of its diameter. Suppose the resistance of a wire of given length is computed from a measurement of the diameter with a possible \(2 \%\) error. Find the possible percent error in the computed value of the resistance.

Step-by-Step Solution

Answer

The possible percent error in the computed value of the resistance is 4%.

Step 1 - Understand the formula for resistance

The resistance, R, of a wire is given by the formula: \[ R = k \frac{L}{D^2} \] where k is a constant of proportionality, L is the length of the wire, and D is the diameter of the wire.

Step 2 - Set up the error relationship

Given that the diameter has an error of 2%, express this as a differential change in D: \[ dD = 0.02D \] The goal is to find the corresponding change in resistance, dR/R.

Step 3 - Differentiate the resistance formula

Calculate the derivative of the resistance formula with respect to D: \[ R = k \frac{L}{D^2} dR = -2k \frac{L}{D^3}dD \] Using the original formula for R, rewrite this as \[ dR = -2 \frac{R}{D} dD \]

Step 4 - Express dR/R in terms of dD/D

Rewrite the expression using the known relationship: \[ \frac{dR}{R} = -2 \frac{dD}{D} \] By substituting \( dD = 0.02D \), this becomes: \[ \frac{dR}{R} = -2 \times 0.02 = -0.04 \]

Step 5 - Determine the absolute percent error

The absolute value of the calculated percent error in R is 4%, disregarding the negative sign which indicates direction. Thus, the possible percent error in the computed value of the resistance is: \( 4\% \)

Key Concepts

Proportional Relationships

Understanding proportional relationships is key to grasping electrical resistance calculations. In this exercise, resistance (\( R \)) is proportional to the length (\( L \)) of the wire and inversely proportional to the square of its diameter (\( D \)). Specifically, the formula is given by: \[ R = k \frac{L}{D^2} \] \ The constant (\( k \)) accounts for material properties and other fixed parameters. When you see that resistance is 'proportional' to length, it means if you double the length, the resistance doubles too. On the other hand, 'inversely proportional' to the square of the diameter means if you double the diameter, the resistance becomes one-fourth. This inverse square relationship is crucial for understanding how small changes in diameter can significantly affect resistance. So, always watch if changes in one variable will linearly increase/decrease (proportional) or significantly alter another (inversely proportional).

Error Analysis

Error analysis helps us understand how measurement inaccuracies affect calculated outcomes. In this case, a 2% error in diameter (\bigskip\( dD = 0.02D \bigskip\)) will impact the resistance calculation. To analyze this, we differentiate the resistance formula concerning diameter to see how errors propagate. \ First, we have: \( R = k \frac{L}{D^2} \bigskip\ \bigskip\) \Taking the differential: \( dR = -2k \frac{L}{D^3}dD \bigskip\) \Rewritten using the original formula for \bigskip\ R:\bigskip\ \( dR = -2 \frac{R}{D} dD \bigskip\) The goal is to express \bigskip\( dR/R \bigskip\) relative to \( dD/D \): \bigskip\ \( \frac{dR}{R} = -2 \frac{dD}{D} \bigskip\) \Substituting, we find \bigskip\ \( \frac{dR}{R} = -2 \times 0.02 = -0.04 \bigskip \)+. The negative indicates direction, but the magnitude (absolute percent error) is 4%. Hence, a small error in diameter measurement leads to a laser in a 4% error in resistance computation. Always perform error analysis to know how accurate your measurements need to be.

Differentiation

Differentiation in calculus helps us find how a function changes as its input changes. Here, it's used to determine how errors in diameter (\bigskip\( D \bigskip\)) affect resistance (\bigskip\( R \bigskip\)). Starting with the formula \ \(\bigskip\( R = k \frac{L}{D^2} \ \bigskip\)\bigskip\), we calculate the derivative with respect to \bigskip\( D \bigskip\):\ The derivative \bigskip\( dR \bigskip\) tells us how a small change in diameter affects resistance: \bigskip\  \( dR = -2k \frac{L}{D^3}dD  \bigskip\) Simplifying, we get \bigskip\: \( dR = -2 \frac{R}{D} dD \bigskip \) . This is useful because it allows you to relate relative changes: \bigskip\( \frac{dR}{R} = -2 \frac{dD}{D} \bigskip\) This simplifies real-world calculations. For example, substituting \bigskip\( dD = 0.02D \bigskip\), we find that a 2% error in diameter translates to a 4% error in resistance. Differentiation is a powerful tool that lets you see how changes in one part of your system (like diameter) significantly influence another part (like resistance). So, by mastering differentiation, you can better predict and manage such relationships.