Problem 65 - Calculus - AP Edition

Question

The area of a circle is to be computed from a measured value of its diameter. Estimate the maximum permissible percentage error in the measurement if the percentage error in the area must be kept within \(\pm 1 \%\).

Step-by-Step Solution

Answer

The maximum permissible percentage error in the diameter is \(\pm 0.5\%\).

Step 1: Understand the Relationship between Diameter and Area

The area of a circle is calculated using the formula \( A = \pi r^2 \), where \( r \) is the radius. Since the diameter \( D \) is twice the radius (\( D = 2r \)), we can express the radius as \( r = \frac{D}{2} \). Substitute this into the area formula to get \( A = \pi \left(\frac{D}{2}\right)^2 = \frac{\pi D^2}{4} \).

Step 2: Differentiate the Area with Respect to Diameter

To find the relationship between the error in the diameter and the area, differentiate the area with respect to the diameter. The derivative is \( \frac{dA}{dD} = \frac{\pi D}{2} \). This derivative represents how the area changes with a small change in the diameter.

Step 3: Relate Errors Using Differentials

The percentage change in area \( \frac{dA}{A} \) can be expressed using differentials: \( \frac{dA}{A} = \frac{\frac{\pi D}{2} \cdot dD}{\frac{\pi D^2}{4}} = \frac{2 \cdot dD}{D} \). This simplifies to \( \frac{dA}{A} = 2 \cdot \frac{dD}{D} \).

Step 4: Solve for Permissible Error in Diameter

We are given that \( \frac{dA}{A} \leq \pm 0.01 \) (since 1% is a 0.01 proportion). From the equation \( \frac{dA}{A} = 2 \cdot \frac{dD}{D} \), set \( 2 \cdot \frac{dD}{D} \leq \pm 0.01 \). Solving for \( \frac{dD}{D} \), we get \( \frac{dD}{D} \leq \pm 0.005 \).

Step 5: Interpret the Result

A \(0.5\%\) error in the diameter translates to a \(1\%\) error in the area. Thus, to keep the area error within \(\pm 1\%\), the diameter measurement must be kept within \(\pm 0.5\%\) error.

Key Concepts

Area of a Circle

Understanding how to calculate the area of a circle is foundational in math. The area represents the space contained within the circle's boundaries. It's vital for various applications in science, engineering, and even art. The formula to find the area is given by:\[ A = \pi r^2 \]where,

• \( A \) is the area,


• \( \pi \) is approximately 3.14159,


• \( r \) is the radius of the circle.

When we deal with measurements, such as diameter (denoted as \( D \)), it's useful to remember that the diameter is twice the radius. Therefore, the radius can be found using \( r = \frac{D}{2} \). By substituting \( r = \frac{D}{2} \) into the area formula, one can rewrite it in terms of the diameter:\[ A = \pi \left(\frac{D}{2}\right)^2 = \frac{\pi D^2}{4} \]This formula is crucial because measuring the diameter can often be more straightforward than measuring the radius directly. Knowing how to manipulate the formula helps to solve real-world problems involving the dimensions of circles.

Differentiation

Differentiation is a mathematical process used to determine the rate at which one quantity changes with respect to another. In the context of this problem, it helps us understand how changes in the diameter of a circle impact the area. By differentiating the area formula with respect to the diameter, we can identify the relationship between a small change in diameter (\( dD \)) and a consequential change in area (\( dA \)). After taking the derivative of the area formula \( A = \frac{\pi D^2}{4} \) with respect to the diameter, the result is:\[ \frac{dA}{dD} = \frac{\pi D}{2} \]This result is insightful because it shows how sensitive the area is to changes in the diameter. More practically, it lets us express the percentage change in the area (\( \frac{dA}{A} \)) in terms of the percentage change in the diameter (\( \frac{dD}{D} \)), which is crucial for estimating errors.

Percentage Error

Understanding percentage error is vital in measurement and calculations. It represents the discrepancy between an approximate value and a true value, expressed as a percentage. Essentially, it quantifies the "room for error" in measurements. In this context, the percentage error in the area is related to the error in the diameter measurement.Given that the desired percentage error in the area is limited to \( \pm 1\% \), we can set up the relationship using the formula derived for errors:\[ \frac{dA}{A} = 2 \times \frac{dD}{D} \]This equation tells us that changing the diameter by a certain percentage results in double the percentage change in the area. Thus, to keep the area error within \( \pm 1\% \), the error in the diameter must be smaller, specifically within \( \pm 0.5\% \). This is crucial in fields like engineering where precision is key. Understanding the link between errors ensures accurate and reliable measurements, maintaining the integrity of calculations and designs.