/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 A softball diamond is a square w... [FREE SOLUTION] | 91Ó°ÊÓ

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A softball diamond is a square with each side measuring 60 feet. Suppose a player is running from second base to third base at a rate of 22 feet per second. a. At what rate is the distance between the runner and home plate changing when the runner is halfway to third base? b. How far is the runner from home plate at this time?

Short Answer

Expert verified
The rate of change of the distance from home plate is \( \frac{22\sqrt{5}}{5} \) ft/s, and the runner is \( 30\sqrt{5} \) feet away from home plate.

Step by step solution

01

Understanding the Problem

The softball diamond is a square of side 60 feet. The runner moves from second base to third, which means the runner is moving along one side of the square. We are tasked with finding how fast the distance between the runner and home plate is changing, as well as the current distance from home plate when the runner is exactly halfway between second and third base.
02

Defining Variables and Equations

Let the distance from second base to the runner be denoted by \( x \). The rate of change of \( x \), \( \frac{dx}{dt} \), is given as 22 ft/s. The distance from the runner to home plate can be calculated using the Pythagorean theorem: \( d^2 = x^2 + (60)^2 \), where \( d \) is the distance from the runner to home plate.
03

Applying the Pythagorean Theorem

When the runner is halfway from second to third base, \( x = 30 \) feet. Substituting \( x = 30 \) into the equation \( d^2 = x^2 + (60)^2 \) gives \( d^2 = 30^2 + 60^2 \). Simplifying, \( d^2 = 900 + 3600 \), so \( d^2 = 4500 \). Thus, \( d = \sqrt{4500} = 30\sqrt{5} \) feet.
04

Differentiating and Finding Rate of Change

We need to find \( \frac{dd}{dt} \), the rate at which the distance from the runner to home plate is changing. Differentiate both sides of \( d^2 = x^2 + 3600 \) with respect to time \( t \): \[ 2d \frac{dd}{dt} = 2x \frac{dx}{dt} \]. Simplify to find \( \frac{dd}{dt} = \frac{x \frac{dx}{dt}}{d} \). Substitute \( x = 30 \), \( \frac{dx}{dt} = 22 \), and \( d = 30\sqrt{5} \). Thus, \( \frac{dd}{dt} = \frac{30 \times 22}{30\sqrt{5}} = \frac{22}{\sqrt{5}} \). Simplifying gives \( \frac{dd}{dt} = \frac{22\sqrt{5}}{5} \) ft/s.
05

Finding the Distance from Home Plate

We've already calculated the distance from the runner to home plate when he is halfway to third base as \( d = 30\sqrt{5} \) feet.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pythagorean Theorem
The Pythagorean Theorem is a fundamental principle in geometry dealing with right triangles. It relates the lengths of the sides of a right triangle, stating that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. This can be mathematically expressed as:
  • a2 + b2 = c2
In this specific exercise, the softball diamond forms a right triangle where:
  • a is the distance the runner covers from second base,
  • b is the distance from second base to home plate,
  • and c or d denotes the distance from the runner to home plate (the hypotenuse).
By applying this theorem, you can compute the unknown distance, which is vital as the problem involves determining how this distance changes over time.
Differentiation
Differentiation is a key concept in calculus, used to determine how a function changes as its inputs change. It's particularly powerful for finding rates of change and can be seen as finding the slope of the function at any given point.In our exercise, after setting up an equation using the Pythagorean Theorem, we need to determine how different quantities change concerning time as the runner moves. To do this, we use differentiation.By differentiating the derived equation with respect to time, we gain insights into how fast the distance to home plate changes as the runner speeds up along the side of the square towards third base. This involves taking the derivative of each term in our equation with respect to time, introducing variables such as
  • \( rac{dx}{dt} \) - the rate at which the runner approaches third base,
  • \( rac{dd}{dt} \) - the rate of change of the distance from the runner to home plate.
This enables us to ascertain the dynamic relationship between these changing distances.
Distance
Distance is an essential concept in this problem and generally in physics and geometry. It describes the amount of space between two points. In our scenario, we are interested in various distances:
  • The distance from second base to the runner, denoted as \( x \),
  • The fixed distance from home plate to second base, which is 60 feet, and
  • The distance from the runner to home plate, denoted as \( d \).
As the runner moves from second to third base, it's crucial to track how these distances change. By applying the Pythagorean Theorem, we calculated
  • the initial distance: when the runner is halfway, \( x = 30 \), making the calculated distance from home plate \( d = 30\sqrt{5} \).
This reliable calculation assists in further exploring how quickly this measured distance changes, which leads us into understanding rates of change.
Rates of Change
Rates of change refer to how a quantity varies over time. This mathematical concept is crucial in understanding dynamic systems where aspects like speed, velocity, or in our case, distances are constantly evolving.In the softball exercise, the runner's speed along the base path is given as 22 feet per second. This known speed helps us explore how the distance from the runner to home plate changes as they run. Calculating this rate of change involves differentiating our equation formed by the Pythagorean Theorem. Upon differentiation, we solve to find
  • \( \frac{dd}{dt} \), which shows how fast the distance between the runner and home plate is changing at the halfway point.
This value (
  • \( \frac{22\sqrt{5}}{5} \) feet per second
) gives us a precise understanding of the runner's dynamic relationship with respect to home plate, illustrating the application of calculus in analyzing real-world sports situations.

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