Problem 43 - Calculus Early Transcendentals

Question

One side of a triangle is increasing at a rate of 3 \(\mathrm{cm} / \mathrm{s}\) and a second side is decreasing at a rate of 2 \(\mathrm{cm} / \mathrm{s} .\) If the area of the triangle remains constant, at what rate does the angle between the sides change when the first side is 20 \(\mathrm{cm}\) long, the second side is \(30 \mathrm{cm},\) and the angle is \(\pi / 6 ?\)

Step-by-Step Solution

Answer

The angle changes at a rate of \( \frac{\sqrt{3}}{90} \) radians per second.

Step 1: Identify the Given Rates and Initial Values

We have side lengths given as \( a = 20 \text{ cm} \) and \( b = 30 \text{ cm} \). The angle \( \theta \) between them is \( \frac{\pi}{6} \). The rates are \( \frac{da}{dt} = 3 \text{ cm/s} \) and \( \frac{db}{dt} = -2 \text{ cm/s} \). The area \( A \) is constant.

Step 2: Use the Formula for the Area of a Triangle

The area \( A \) of a triangle with sides \( a, b \) and angle \( \theta \) is given by \[ A = \frac{1}{2}ab\sin\theta \] Since the area is constant, \( \frac{dA}{dt} = 0 \).

Step 3: Differentiate the Area Formula with Respect to Time

Differentiate both sides of the area formula with respect to time \( t \): \[ \frac{d}{dt}\left( \frac{1}{2}ab\sin\theta \right) = 0 \] Applying the product rule, we get: \[ \frac{1}{2} \left( \frac{da}{dt} b \sin\theta + a \frac{db}{dt} \sin\theta + ab \frac{d}{dt}(\sin\theta) \right) = 0 \] Simplify it: \[ \frac{1}{2} \left( 3 \cdot 30 \cdot \frac{1}{2} + 20 \cdot (-2) \cdot \frac{1}{2} + 20 \cdot 30 \cdot \cos\left(\frac{\pi}{6}\right)\frac{d\theta}{dt}\right) = 0 \]

Step 4: Simplify and Solve for the Rate of Change of the Angle

Simplify further: \[ \frac{1}{2}(45 - 20 \cdot 2 + 20 \cdot 30 \cdot \cos\left(\frac{\pi}{6}\right)\frac{d\theta}{dt}) = 0 \] \[ 15 - 20 + 20 \cdot 30 \cdot \frac{\sqrt{3}}{2} \frac{d\theta}{dt} = 0 \] \[ -5 + 300\frac{\sqrt{3}}{2} \frac{d\theta}{dt} = 0 \] Solve for \( \frac{d\theta}{dt} \): \[ 300\frac{\sqrt{3}}{2} \frac{d\theta}{dt} = 5 \] \[ 150\sqrt{3} \frac{d\theta}{dt} = 5 \] \[ \frac{d\theta}{dt} = \frac{5}{150\sqrt{3}} = \frac{1}{30\sqrt{3}} \] Simplify further: \[ \frac{d\theta}{dt} = \frac{\sqrt{3}}{90} \]

Step 5: Conclude the Rate of Change of the Angle

Thus, the rate at which the angle \( \theta \) changes is \( \frac{\sqrt{3}}{90} \) radians per second.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that allows us to determine how a function changes as its input changes. In this problem, we are looking at how the angle between two sides of a triangle changes as the lengths of the sides change. This is where differentiation comes into play.
To solve such problems, you often need to differentiate equations with respect to time. This is known as finding the derivative. Here, we differentiate the area formula of a triangle with respect to time to find how changes in side length affect the angle. We use the product rule of differentiation, which is applied when differentiating a function that is the product of two or more smaller functions. In our problem, since the area is constant, the derivative with respect to time must equal zero, allowing us to solve for the rate of change of the angle.

Trigonometric Functions

Trigonometric functions are essential in calculations involving angles and sides of triangles. In this problem, the sine function is particularly important. The area of a triangle with two sides and an angle between them can be written in terms of sine, making it useful for determining how the angle changes.
We use the formula for the area in terms of sine: \( A = \frac{1}{2}ab\sin{\theta} \). Since the sides are changing, and the area is constant, \( \sin{\theta} \) helps relate the side changes to the angle change. During differentiation, the derivative of \( \sin{\theta} \) with respect to time is handled using trigonometric identities and rules like the chain rule.

• The cosine function often accompanies sine in differentiation, as we see in our area derivative.
• Understanding these functions helps in predicting how triangular dimensions and angles shift as sides vary over time.

Calculus Problems

Calculus problems, particularly involving related rates, can be complex but are solvable by breaking them into smaller, manageable parts. In this problem, we handle several tasks:
Initially, identify given values and understand what needs to be determined, like the rate of change of an angle when the sides of a triangle change.
Next, establish the relationship through a formula. Here, the area of the triangle defines our relationship between angles and sides. Then, differentiate the equation to find the rate of change.

• Consider all known rates of change and apply them in differentiation to solve for the unknown rate, as shown in our step-by-step solution.
• It's crucial to ensure units are consistent; rates should reflect the dimension they affect, such as cm for length rates and radians for angular rates.

This approach is systematic and helps in solving many calculus-related problems involving dynamic changes.