Problem 58 - Calculus - AP Edition

Question

One side of a right triangle is known to be \(25 \mathrm{cm}\) exactly. The angle opposite to this side is measured to be \(60^{\circ},\) with a possible error of \(\pm 0.5^{\circ}\). (a) Use differentials to estimate the errors in the adjacent side and the hypotenuse. (b) Estimate the percentage errors in the adjacent side and hypotenuse.

Step-by-Step Solution

Answer

Errors: Adjacent \(\approx -0.145\ \text{cm}\), Hypotenuse \(\approx -0.0725\ \text{cm}\); Percentage Errors: \(-1.00\%\) and \(-0.25\%\).

Step 1: Understand the Relationships in the Triangle

In a right triangle, the sides and angles are related by trigonometric functions. For this triangle, given the angle \(\theta = 60^{\circ}\) and the opposite side \(O = 25\ \text{cm}\), we can use the formulas for tangent and sine:\[ \tan(\theta) = \frac{O}{A} \quad \text{and} \quad \sin(\theta) = \frac{O}{H} \]where \(A\) is the adjacent side and \(H\) is the hypotenuse.

Step 2: Differentiate the Trigonometric Functions

To find the error estimates using differentials, differentiate the formulas with respect to \(\theta\).For the adjacent side:\[A = \frac{O}{\tan(\theta)} \implies \frac{dA}{d\theta} = -\frac{O}{\sin^2(\theta)}\]For the hypotenuse:\[H = \frac{O}{\sin(\theta)} \implies \frac{dH}{d\theta} = -\frac{O \cos(\theta)}{\sin^2(\theta)}\]

Step 3: Calculate the Errors

Given \( \Delta \theta = 0.5^{\circ} = 0.0087\ \text{radians} \), the error in \(A\) and \(H\) can be calculated as:\[\Delta A = \frac{dA}{d\theta} \cdot \Delta \theta = -\frac{25}{\sin^2(60^{\circ})} \cdot 0.0087\]\[\Delta H = \frac{dH}{d\theta} \cdot \Delta \theta = -\frac{25 \cos(60^{\circ})}{\sin^2(60^{\circ})} \cdot 0.0087\]

Step 4: Plug in Values and Solve

Substitute \( \sin(60^{\circ}) = \frac{\sqrt{3}}{2} \) and \(\cos(60^{\circ}) = \frac{1}{2}\) into the error formulas:\[\Delta A = -\frac{25}{\left(\frac{\sqrt{3}}{2}\right)^2} \cdot 0.0087 = -16.67 \cdot 0.0087 \approx -0.145\]\[\Delta H = -\frac{25 \times \frac{1}{2}}{\left(\frac{\sqrt{3}}{2}\right)^2} \cdot 0.0087 = -8.335 \cdot 0.0087 \approx -0.0725\]

Step 5: Percentage Error Calculation

To find the percentage errors:For the adjacent side:\[\text{Percentage Error in } A \approx \left(\frac{\Delta A}{A}\right) \times 100 = \left(\frac{-0.145}{14.43}\right) \times 100 \approx -1.00\%\]where \(A = 25 \div \tan(60^{\circ}) = 14.43\ \text{cm}\).For the hypotenuse:\[\text{Percentage Error in } H \approx \left(\frac{\Delta H}{H}\right) \times 100 = \left(\frac{-0.0725}{28.87}\right) \times 100 \approx -0.25\%\]where \(H = \frac{25}{\sin(60^{\circ})} = 28.87\ \text{cm}\).

Key Concepts

Error Estimation

In differential calculus, estimating the error is a useful technique when dealing with small variations in a measurement or calculation. To understand error estimation, think about how small changes in input values can affect the output of a function. In our problem, this is about how a tiny angle error affects the lengths of sides in a right triangle.

When using differentials, we can analyze each trigonometric relationship in the triangle. For the adjacent side and hypotenuse, there's a potential error due to a small angle change. This concept is essential as it helps predict how precise or inaccurate a result might be given a change in input (like \(\Delta \theta\) in this exercise).

The differentiation process involves working with the trigonometric functions tangent and sine. We find the derivatives of these functions relative to the angle, which helps us understand how these changes play out. These equations capture how sensitive each calculation is to the angle's change, showing the importance of precision in angle measurement in calculated results.

Trigonometric Functions

Trigonometric functions are mathematical tools that relate the angles of a right triangle to its side lengths. They play a key role in this exercise by helping determine the changes in side lengths based on angle measurements.

In our problem, the tangent and sine functions are used. They are defined as follows:

• \( \tan(\theta) = \frac{O}{A} \), where \(O\) is the opposite side and \(A\) is the adjacent side.
• \( \sin(\theta) = \frac{O}{H} \), where \(O\) is the opposite side and \(H\) is the hypotenuse.

By working with these trigonometric equations, you can understand how different side lengths relate to the angles of the triangle.

This is important because any error or imprecision in measuring these trigonometric functions can lead to variation in the calculated side lengths. They provide an essential framework for solving triangle problems and understanding how each function behavior changes with angle variations.

Percentage Error Calculation

Percentage error is a valuable way to express the relative accuracy of a measurement. It gives us an idea of how far off a calculated value might be from its true value.

Calculating percentage error involves the error value divided by the actual (or true) value, multiplied by 100 to convert it into a percentage:
\[\text{Percentage Error} = \left( \frac{\Delta \text{value}}{\text{actual value}} \right) \times 100\]In the triangle exercise, percentage errors are calculated for both the adjacent side and hypotenuse. It's crucial because it contextualizes the estimated error relative to the true size of the side.

For instance, if the calculated side length is significantly larger than the error, the percentage error is small, indicating high measurement precision. This method helps clarify how small errors in input (angle measurement in this case) can influence the certainty of your calculated results. Understanding this allows better decisions related to measurement accuracy and precision in applications.