Problem 55 - Calculus - AP Edition

Question

The side of a square is measured to be \(10 \mathrm{ft}\), with a possible error of \(\pm 0.1 \mathrm{ft}\). (a) Use differentials to estimate the error in the calculated area. (b) Estimate the percentage errors in the side and the area.

Step-by-Step Solution

Answer

(a) The error in the area is \(\pm 2\) sq ft. (b) Side error is 1%, area error is 2%.

Step 1: Understand the Problem

We have a square with a side measured as \(s = 10\) ft. There's a potential error in the measurement of the side, \( \Delta s = \pm 0.1 \) ft. We need to find the error in the area of the square using differentials, and the percentage errors in both the side and the area.

Step 2: Use Differentials for Area Error

The area \(A\) of a square with side \(s\) is given by \(A = s^2\). The differential \(dA\) represents a small change in area, so we can express it using the formula \(dA = 2s \cdot ds\). Substitute \(s = 10\) ft and \(ds = \pm 0.1\) ft to find \(dA\).

Step 3: Calculate the Differential

Substituting into the formula: \[dA = 2 \cdot 10 \cdot 0.1 = 2 \cdot 1 = 2 \]Thus, the estimated error in the area is \(\pm 2\) square feet.

Step 4: Calculate Percentage Error in Side

The percentage error in the side length is given by \[\left( \frac{\mid ds \mid}{s} \right) \times 100\% = \left( \frac{0.1}{10} \right) \times 100\% = 1\%.\]Thus, the percentage error in the side length is 1%.

Step 5: Calculate Percentage Error in Area

The percentage error in the area is given by \[\left( \frac{\mid dA \mid}{A} \right) \times 100\% = \left( \frac{2}{100} \right) \times 100\% = 2\%.\]This occurs because we previously found the area as \(A = 10^2 = 100\) square feet.

Key Concepts

Error Estimation

Error estimation is an important part of measurements and calculations, as it helps us understand how inaccuracies in measurements can propagate and affect final outcomes. In the problem of finding the error in the area of a square, differentials offer a straightforward method for estimation. The area of a square is expressed as \(A = s^2\), where \(s\) is the side of the square. A small change in the side \(ds\) causes a change \(dA\) in the area, which can be approximated using the derivative: \(dA = 2s \cdot ds\). By substituting for \(s = 10\) ft and \(ds = \pm 0.1\) ft, we find that the corresponding change in area, \(dA\), is \(\pm 2\) square feet. This technique is particularly useful because it simplifies calculations by linearly approximating changes through the use of calculus principles.

Percentage Error

When calculating measurements, it's helpful to know the percentage error, as it relates the error to the actual value, providing a sense of the relative size of the error.

To find the percentage error of the side of the square, use the formula:

• \(\left( \frac{|ds|}{s} \right) \times 100\%\)
• Substituting our values gives: \(\frac{0.1}{10} \times 100\% = 1\%\).

Thus, the side length has a percentage error of 1%.

For the area, the percentage error is calculated as follows:

• \(\left( \frac{|dA|}{A} \right) \times 100\%\)
• Substituting the area \(A = 100\) square feet and \(dA = 2\) square feet, gives: \(\frac{2}{100} \times 100\% = 2\%\).

The area has a higher percentage error compared to the side because the area is a squared quantity, inflating the effect of small errors in measurements.

Area Calculation

Calculating the area of a geometric shape like a square is a fundamental mathematical task, especially in applications involving construction, design, and numerous fields of engineering. For a square with side length \(s\), the area is given by \(A = s^2\).

In the context of this problem, where \(s = 10\) ft, the area computes to \(100\) square feet. If there is an uncertainty in the side measurement, it will directly affect the precision of the calculated area. Since any error is squared in the area calculation, it's important to account for possible deviations, which can be estimated using the concept of differentials, as we've elaborated. Understanding how these elements interplay helps in making more informed decisions when analyzing or designing with real-world objects.