Problem 45 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

On New Year's Eve, Zain is watching the descent of a lighted ball from atop a tall building that is 600 feet away. The ball is falling at the rate of 20 feet per minute. At what rate is the angle of elevation of Zain's line of sight changing with respect to time when the ball is 800 feet from the ground?

Step-by-Step Solution

Answer

\( \frac{d \theta}{dt} = -\frac{3}{250} \) radians per minute when the ball is 800 feet from the ground.

Step 1: Define the Variables

Let \( \theta \) be the angle of elevation from Zain's viewpoint to the ball. Let \( y \) be the height of the ball above the ground. Zain is 600 feet away from the base of the building.

Step 2: Set Up the Relationship

Use the tangent function to relate the angle of elevation \( \theta \) to the height \( y \): \[ \tan( \theta ) = \frac{y}{600} \]

Step 3: Differentiate with respect to Time

Differentiate both sides with respect to time \( t \): \[ \frac{d}{dt} \left( \tan( \theta ) \right) = \frac{d}{dt} \left( \frac{y}{600} \right) \] Using the chain rule, this becomes: \[ \sec^2( \theta ) \frac{d\theta}{dt} = \frac{1}{600} \frac{dy}{dt} \]

Step 4: Find the Values at the Given Instance

When the ball is 800 feet from the ground, use \( y = 800 \) and \( \frac{dy}{dt} = -20 \). Calculate \( \tan( \theta ) \): \[ \tan( \theta ) = \frac{800}{600} = \frac{4}{3} \] Then find \( \sec( \theta ) \): \[ \sec( \theta ) = \sqrt{1 + \tan^2( \theta )} = \sqrt{1 + \left( \frac{4}{3} \right)^2} = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3} \] Therefore, \[ \sec^2( \theta ) = \left( \frac{5}{3} \right)^2 = \frac{25}{9} \]

Step 5: Solve for the Rate of Change of the Angle

Use the differentiated equation and solve for \( \frac{d\theta}{dt} \): \[ \frac{25}{9} \frac{d\theta}{dt} = \frac{1}{600} \cdot (-20) \] \[ \frac{25}{9} \frac{d\theta}{dt} = -\frac{20}{600} = -\frac{1}{30} \] Therefore, \[ \frac{d\theta}{dt} = -\frac{1}{30} \cdot \frac{9}{25} = -\frac{3}{250} \]

Key Concepts

angle of elevation

When solving related rates problems, understanding the angle of elevation is crucial. The angle of elevation is the angle between the horizontal ground and an observer's line of sight to an object above the ground. In this problem, Zain observes the ball dropping from a tall building. His line of sight to the ball creates an angle of elevation with the ground. Identifying this angle correctly simplifies the use of trigonometric functions, such as the tangent function. By focusing on the height of the ball and the distance from the observer, students can relate these measurements to the angle of elevation.

differentiation with respect to time

Differentiation with respect to time is essential in related rates problems. Here, our goal is to determine how the angle of elevation changes over time as the ball falls. In mathematics, when quantities change over time, we use derivatives. In this problem, we differentiate both sides of the equation involving the tangent function with respect to time. This action allows us to find the rate at which the angle of elevation changes. By understanding the concept of instantaneous rates of change, students can translate physical movement into mathematical derivatives.

tangent function

The tangent function relates the angle of elevation to the height of the ball above the ground. For Zain, the key relationship is given by: \[\tan( \theta ) = \frac{y}{600}\] Here, \(\theta\) is the angle of elevation, and \(y\) is the ball's height above the ground. The tangent function connects these two variables, enabling us to differentiate the equation with respect to time. Understanding the properties of the tangent function, and how it varies as \(\theta\) changes, is fundamental to solving the problem correctly.

chain rule

The chain rule is a powerful tool in calculus, especially for differentiation with respect to time. Given the function \(\tan( \theta )\), we need to differentiate it with respect to time. The chain rule explains how changes in \(\theta\) over time affect \(\tan( \theta )\). Using the chain rule, we express the rate of change of \(\tan( \theta )\) as: \[\frac{d}{dt}( \tan( \theta ) ) = \sec^2( \theta ) \frac{d\theta}{dt}\] This equation separates the derivative into parts, making it easier to manage. By applying the chain rule, we can effectively solve for the desired rate of change, \(\frac{d\theta}{dt}\).