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The Pythagorean theorem is a powerful tool in geometry and calculus that helps us relate the sides of a right triangle. It's named after the ancient Greek mathematician Pythagoras. The theorem states that in a right triangle, the sum of the squares of the lengths of the two shorter sides is equal to the square of the length of the longest side (the hypotenuse). This relationship can be written as:\[ a^2 + b^2 = c^2 \]In the context of the exercise, the baseball diamond is viewed as a right triangle when the ball is moving along the third-base line. Here, the distance the ball has traveled (denoted as \(x\)) and the fixed distance from third base to home plate (90 feet) make up the two shorter sides of the triangle. The distance from the ball to first base (denoted as \(y\)) is the hypotenuse. By applying the Pythagorean theorem, we get:\[ x^2 + 90^2 = y^2 \]This equation is crucial for understanding how the distances are related and allows us to proceed with calculus to find the rates of change.

Differentiation is the process of finding the rate at which quantities change. In calculus, it's a fundamental concept used to understand how functions behave. When dealing with physical situations, such as the movement of a baseball, differentiation allows us to calculate how fast distances are changing over time.In the exercise, we differentiate the Pythagorean relation \(x^2 + 90^2 = y^2\) with respect to time \(t\). This is done because both \(x\) and \(y\) are functions of time. Differentiating both sides, we obtain:\[ 2x \frac{dx}{dt} = 2y \frac{dy}{dt} \]Here:

• \(\frac{dx}{dt}\) is the rate at which the ball is moving towards third base, given as \(-100\) ft/s (negative because it's moving closer).
• \(\frac{dy}{dt}\) is the rate at which the distance from the ball to first base is changing, which is what we want to find.

Differentiation helps us reveal these dynamic relationships and calculate rates of change, which are the core of related rates problems.

The concept of a rate of change refers to how one quantity changes in relation to another. It's particularly important in scenarios where time is a variable, such as moving objects. Rates of change can be positive or negative, indicating increasing or decreasing trends.In the problem, we explore the rate at which the distance from the ball to first base changes while it moves along the third base line. By rearranging the differentiated equation:\[ \frac{dy}{dt} = \frac{2x \frac{dx}{dt}}{2y} \]we find the formula for the rate of change of \(y\) with respect to time. Substituting the given \(\frac{dx}{dt} = -100\) ft/s, we get:\[ \frac{dy}{dt} = \frac{-100x}{y} \]This emphasizes how the geometry of the triangle and the movement of the ball influence the rate at which the distance changes.

Solving calculus problems involving related rates requires a logical step-by-step approach. The key is to understand how changes in one quantity affect another related quantity. Here's how you can systematically tackle such problems:

• Draw a diagram: Visualizing the problem with a diagram can help you understand the relationships between variables.
• Identify variables and rates: Determine which variables represent specific quantities and which rates of change are given or need to be found.
• Use known formulas (like the Pythagorean theorem): Apply these formulas to set up relationships between the variables.
• Differentiate with respect to time: This step reveals how the variables relate to each other dynamically.
• Solve for the unknown rate: Rearrange the differentiated equation to find the rate of change of interest.

By following these systematic steps, you can decode related rates problems and find solutions. In the baseball diamond exercise, this approach enabled us to track how the distance to first base changed as the ball moved.