Problem 6 - Active Calculus

Question

A baseball diamond is a square with sides 90 feet long. Suppose a baseball player is advancing from second to third base at the rate of 24 feet per second, and an umpire is standing on home plate. Let \(\theta\) be the angle between the third baseline and the line of sight from the umpire to the runner. How fast is \(\theta\) changing when the runner is 30 feet from third base?

Step-by-Step Solution

Answer

0.24 radians per second

Step 1: Understand the Problem

A baseball diamond is a square, and the runner moves from second to third base. The player's speed is 24 feet per second, and the distance to the third base changes. We need to find the rate at which the angle \( \theta \) between the third baseline and the line of sight from the umpire to the runner is changing when the runner is 30 feet from third base.

Step 2: Set Up the Right Triangle

The distance from home plate to third base forms one leg of the right triangle, which is 90 feet. The distance from the runner to third base forms the other leg of the triangle. Let \( x \) be the distance from the runner to third base, and it changes over time.

Step 3: Use Related Rates

We need the relationship between \( \theta \), \( x \), and the known distances. Using the tangent function, \[ \tan(\theta) = \frac{90}{x} \]. Differentiate both sides with respect to time \( t \): \[ \frac{d}{dt} \left( \tan(\theta) \right) = \frac{d}{dt} \left( \frac{90}{x} \right) \].

Step 4: Differentiate and Simplify

Using the chain rule, \[ \sec^2(\theta) \cdot \frac{d\theta}{dt} = -\frac{90}{x^2} \cdot \frac{dx}{dt} \]. Substitute the values \( x = 30 \) and \( \frac{dx}{dt} = -24 \) feet per second (negative because the distance is decreasing), and solve for \( \frac{d\theta}{dt} \).

Step 5: Calculate \( \frac{dx}{dt} \)

Given \[ \frac{d\theta}{dt} = \frac{-90 \cdot (-24)}{30^2 \cdot \sec^2(\theta)} \]. Calculating \[ \sec^2(\theta) = 1 + \tan^2(\theta) \]. Using \( \tan(\theta) = \frac{90}{30} = 3 \), we find \[ \sec^2(\theta) = 1 + 9 = 10 \].

Step 6: Substitute and Solve

Finally, \[ \frac{d\theta}{dt} = \frac{90 \cdot 24}{30^2 \cdot 10} = \frac{2160}{9000} = 0.24 \]. Therefore, \( \frac{d\theta}{dt} \) is 0.24 radians per second.

Key Concepts

Related Rates

In calculus, 'related rates' problems deal with finding the rate at which one quantity changes in relation to another quantity. Essentially, it involves the rate of change of two or more related variables with respect to time. By understanding how one quantity varies, we can infer how another connected quantity changes simultaneously. This concept is central to the given problem, where we are trying to determine how fast the angle \( \theta \) changes as the runner moves towards third base.
Here's how to approach related rates problems: first, identify all given information and what you need to find. Then, create an equation that links the variables in question. Finally, differentiate the equation with respect to time (using implicit differentiation), and substitute the known values to solve for the unknown rate.

• Identify variables and rates of change.
• Set up an equation relating the variables.
• Differentiate with respect to time.
• Substitute known values and solve.

Differentiation

Differentiation is the core tool we use in the problem to find how the angle \( \theta \) changes over time. Differentiation allows us to find rates of change and is an essential concept in calculus. This involves applying derivative rules and sometimes combining multiple rules, as we did in our solution using the chain rule.
In the given problem, we used the chain rule for differentiation. The chain rule states that if you have a composite function, you differentiate the outer function and multiply it by the derivative of the inner function.
Here's what we did in steps:

• Differentiated both sides of the equation \( \tan(\theta) = \frac{90}{x} \) with respect to time \( t \).
• Applied the chain rule to find the derivative of \( \tan(\theta) \).
• Simplified and solved for \( \frac{d\theta}{dt} \).

This method allowed us to find how quickly the angle \( \theta \) changes as the runner approaches the third base.

Trigonometric Functions

Trigonometric functions, such as sine, cosine, and tangent, relate the angles of a triangle to the lengths of its sides. In this problem, we use the tangent function, because it helps in expressing the relationship between the angle \( \theta \), the distance from the umpire to the runner, and the baseline of the baseball field.
Here, we set up the relationship using the tangent function:
\( \tan(\theta) = \frac{90}{x} \)
Where \( \theta \) is the angle between the baseline and the line from the umpire to the runner, and \( x \) is the distance from the runner to third base. The derivative of the tangent function is given by \( \sec^2(\theta) \), which plays a crucial role in our differentiation step.
Using the trigonometric identity \( \sec^2(\theta) = 1 + \tan^2(\theta) \), we calculate the values accurately. Understanding trigonometric functions and their derivatives is crucial for solving related rates problems that involve angles.

Calculus

Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. It helps us understand changes and motion. The concepts within calculus, particularly related rates, differentiation, and trigonometric functions, are used to solve dynamic problems in real life, like the baseball runner problem.
In this specific problem:

• We established a relationship between the angle and the changing distances using tangent.
• Used differentiation to find the rate of change of this angle.
• Applied trigonometric identities to simplify our equations and find the derivatives.

By going through these steps, calculus allows us to break down complex situations into solvable problems and determine how one changing quantity impacts another. This problem illustrates the application of calculus in a practical and engaging way, showing how mathematical concepts are used to find precise results.