Problem 9 - Calculus - AP Edition

Question

Let \(\theta\) (in radians) be an acute angle in a right triangle, and let \(x\) and \(y,\) respectively, be the lengths of the sides adjacent to and opposite \(\theta .\) Suppose also that \(x\) and \(y\) vary with time. (a) How are \(d \theta / d t, d x / d t,\) and \(d y / d t\) related? (b) At a certain instant, \(x=2\) units and is increasing at 1 unit/s, while \(y=2\) units and is decreasing at \(\frac{1}{4}\) unit/s. How fast is \(\theta\) changing at that instant? Is \(\theta\) increasing or decreasing at that instant?

Step-by-Step Solution

Answer

a) The relationship is \( \sec^2(\theta) \frac{d\theta}{dt} = \frac{x \frac{dy}{dt} - y \frac{dx}{dt}}{x^2} \). b) \( \theta \) is decreasing at \(-\frac{5}{16}\) radians/s.

Step 1: Identify the relationship between x, y, and θ

In a right triangle, the tangent of angle \( \theta \) is equal to the ratio of the opposite side \( y \) to the adjacent side \( x \). Therefore, we have the equation: \[ \tan(\theta) = \frac{y}{x} \].

Step 2: Differentiate to find relationship between rates

To find how the rates \( \frac{d\theta}{dt} \,\), \(\frac{dx}{dt} \), and \(\frac{dy}{dt} \) are related, differentiate both sides of \( \tan(\theta) = \frac{y}{x} \) with respect to \( t \):\[ \frac{d}{dt}(\tan(\theta)) = \frac{d}{dt}\left(\frac{y}{x}\right) \].

Step 3: Apply differentiation rules

Using the derivative of the tangent function, we have:\[ \sec^2(\theta) \frac{d\theta}{dt} = \frac{x \frac{dy}{dt} - y \frac{dx}{dt}}{x^2} \].This equation relates the rates of change of \( \theta \), \( x \), and \( y \).

Step 4: Evaluate at the given instant values

We are given \( x = 2 \), \( y = 2 \), \( \frac{dx}{dt} = 1 \), and \( \frac{dy}{dt} = -\frac{1}{4} \). Since \( x = y = 2 \), \[ \tan(\theta) = 1 \rightarrow \theta = \frac{\pi}{4} \].Substitute these values into the differentiated equation:\[ \sec^2\left(\frac{\pi}{4}\right) \frac{d\theta}{dt} = \frac{2 \times \left(-\frac{1}{4}\right) - 2 \times 1}{2^2} \].

Step 5: Simplify and solve for dθ/dt

Calculate \( \sec^2\left(\frac{\pi}{4}\right) = 2 \). Substitute it into the equation:\[ 2 \frac{d\theta}{dt} = \frac{-0.5 - 2}{4} = \frac{-2.5}{4} = -\frac{5}{8} \].Then solve for \( \frac{d\theta}{dt} \):\[ \frac{d\theta}{dt} = -\frac{5}{16} \].

Step 6: Determine whether θ is increasing or decreasing

Since \( \frac{d\theta}{dt} \) is negative, \( \theta \) is decreasing at that instant.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that helps us understand how functions change over time or at specific points. In our problem, we are keen on finding how the angle \( \theta \) changes by differentiating the function that represents the tangent of this angle over time.

When we say differentiation in relation to time, we are using the term 'related rates.' This type of differentiation involves more than a single variable changing over time. In our example, both the lengths \(x\) and \(y\) of the sides of the right triangle and the angle \(\theta\) are changing.

To find how these changes relate, we differentiate the equation \( \tan(\theta) = \frac{y}{x} \) with respect to time \( t \). This gives us an equation involving the derivatives \( \frac{d\theta}{dt}, \frac{dx}{dt}, \text{and} \frac{dy}{dt} \). From this differentiation, we get:

• \( \sec^2(\theta) \frac{d\theta}{dt} = \frac{x \frac{dy}{dt} - y \frac{dx}{dt}}{x^2} \)

This equation is crucial as it connects how fast each component is changing in relation to time.

Tangent Function

The tangent function is a trigonometric function represented as \( \tan(\theta) \), which is the ratio of the opposite side to the adjacent side in a right triangle. In relation to our exercise, it provides the fundamental equation necessary to relate the changing rates of the sides of the triangle and the angle.

Key features of the tangent function include:

• It is periodic, with a period of \( \pi \).
• It is undefined at odd multiples of \( \frac{\pi}{2} \), where the cosine function equals zero.
• It takes values from \( -\infty \) to \( +\infty \) as \( \theta \) approaches its vertical asymptotes, located where the cosine is zero.

Differentiating the tangent function gives the derivative \( \sec^2(\theta) \), which simplifies our task in finding the rate at which \( \theta \) changes. By using this derivative, we relate changes in \(\theta\) to the known changes in \(x\) and \(y\). This makes the tangent function indispensable in problems involving right triangle related rates.

Right Triangle Trigonometry

Right triangle trigonometry is all about understanding the relationships between the angles and sides of right triangles. In our case, it provides the framework for the problem using trigonometric identities and relationships.

In a right triangle:

• There is one angle that is a right angle (90 degrees or \( \frac{\pi}{2} \) radians).
• The hypotenuse is the longest side opposite the right angle, but not explicitly involved here.
• The other two sides are known as the opposite and adjacent relative to the angle \( \theta \).

To solve for the rate of change of the angle \( \theta \), we employed the concept of tangent (opposite/adjacent) given by \( \tan(\theta) = \frac{y}{x} \). Using this relationship and the values given at a particular time (\( x = 2 \) units and \( y = 2 \) units), we derived that \( \theta \) was changing at a rate of \( -\frac{5}{16} \) radians per second.

This negative rate indicates that the angle \( \theta \) is decreasing as the side opposite is shrinking while the adjacent side is increasing. Understanding these trigonometric relationships is essential for solving this and similar related rate problems.