Problem 61 - Calculus - AP Edition

Question

The area of a right triangle with a hypotenuse of \(H\) is calculated using the formula \(A=\frac{1}{4} H^{2} \sin 2 \theta,\) where \(\theta\) is one of the acute angles. Use differentials to approximate the error in calculating \(A\) if \(H=4 \mathrm{cm}\) (exactly) and \(\theta\) is measured to be \(30^{\circ},\) with a possible error of \(\pm 15^{\prime}\).

Step-by-Step Solution

Answer

The error in the area is approximately \( \pm 0.01745 \) square centimeters.

Step 1: Define the Function and Variables

The area of the right triangle is given as \( A = \frac{1}{4} H^2 \sin 2\theta \). We know that \( H = 4 \) cm exactly, and \( \theta = 30^{\circ} \). The possible error in \( \theta \) is \( \pm 15' \) (minutes), which is equivalent to \( \pm \frac{1}{4} \) of a degree (since there are 60 minutes in a degree).

Step 2: Compute the Derivative

We need to find \( \frac{dA}{d\theta} \) to use differentials. Differentiate the area function with respect to \( \theta \):\[\frac{dA}{d\theta} = \frac{1}{4} H^2 \cdot 2 \cos 2\theta = \frac{1}{2} H^2 \cos 2\theta.\]

Step 3: Evaluate the Derivative at Given Values

Plug in \( H = 4 \) cm and \( \theta = 30^{\circ} \) into the derivative:\[\frac{dA}{d\theta} = \frac{1}{2} \times 4^2 \times \cos 60^{\circ} = \frac{1}{2} \times 16 \times \frac{1}{2} = 4.\]

Step 4: Use Differentials to Approximate the Error

The differential \( dA \approx \frac{dA}{d\theta} \cdot d\theta \). The error in \( \theta \) (\( d\theta \)) is \( \pm \frac{1}{4}^\circ \), which in radians is \( \pm \frac{\pi}{720} \) (since \( \frac{\pi}{180} \) radians is a degree). So,\[dA \approx 4 \times \frac{\pi}{720} \approx \pm \frac{\pi}{180}.\]

Step 5: Calculate the Error

Numerically evaluate \( \frac{\pi}{180} \) using \( \pi \approx 3.1416 \):\[dA \approx \pm \frac{3.1416}{180} \approx \pm 0.01745.\]Therefore, the approximate error in the area \( A \) is about \( \pm 0.01745 \) square centimeters.

Key Concepts

Calculating Area

When we talk about calculating the area of a right triangle, we are typically using a formula that includes known angles and side lengths. The formula given in the exercise is specific to a triangle with a hypotenuse. It can be written as:

• \(A = \frac{1}{4} H^2 \sin 2\theta\)

For this formula:

• \(H\) represents the hypotenuse, the longest side of the right triangle.
• \(\theta\) is one of the acute angles.
• The term \(\sin 2\theta\) accounts for the two acute angles, since right triangles have one \(90^\circ\) angle and each other angle is \(\theta\).

The purpose of this formula is to use readily measurable quantities—side lengths and angles—to determine the area quickly and efficiently. In this specific exercise, we know \(H = 4\) cm and \(\theta = 30^\circ\). To make precise calculations, it’s essential to be as accurate with these measurements as possible, especially when angles are involved, as small errors in angles can lead to noticeable discrepancies in area computations.

Derivative Evaluation

Evaluating derivatives is central to solving problems involving rate of change. In our problem, we need the derivative of the area formula with respect to \(\theta\) to approximate the change in area due to a small change in \(\theta\). The derivative is found as follows:

• Differentiate the area function: \( \frac{dA}{d\theta} = \frac{1}{4} H^2 \cdot 2 \cos 2\theta = \frac{1}{2} H^2 \cos 2\theta \).
• This derivative calculates how much the area changes as \(\theta\) changes.
• In this problem, we substitute the given values, \(H = 4\) and \(\theta = 30^\circ\), into the derivative formula.

Plugging these values, we evaluate \( \frac{dA}{d\theta} = 4 \). This tells us that for a small change in \(\theta\), the area changes at a rate of 4 square centimeters per degree of \(\theta\). This rate of change provides a tangible understanding of how sensitive the area is to changes in the angle, which is crucial for the next step: estimating the error.

Error Approximation

Error approximation involves estimating how much a result can change due to small errors in input measurements. In this case, we consider how a slight error in measuring \(\theta\) will affect the calculated area \(A\).To estimate this error:

• We begin by using the differential of \(A\): \(dA \approx \frac{dA}{d\theta} \cdot d\theta\).
• From previous steps, we know \(\frac{dA}{d\theta} \) is 4. We need \(d\theta\), which is the error in \(\theta\). The exercise gives a measurement error of \( \pm 15' \) which is \(\pm \frac{1}{4}\) degree.
• Convert this small angle error to radians, because derivatives in the context of trigonometric functions often require radians: \(d\theta = \pm \frac{\pi}{720}\).
• Calculate \(dA \approx 4 \times \frac{\pi}{720} \approx \pm \frac{\pi}{180} \approx \pm 0.01745\) sq cm.

This calculation shows that the error due to the inexact measurement of \(\theta\) could result in a small miscalculation of the area, up to about \(\pm 0.01745\) square cm. Understanding this error helps us ensure the reliability and accuracy of our measurements, especially when precise area calculations are vital.