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A baseball diamond is a square with sides of 90 feet. A player, after hitting a home run, loped around the diamond at 10 feet per second. Let \(s\) represent the player's distance from home plate after \(t\) seconds. (a) Express \(s\) as a function of \(t\) by means of a four-part formula. (b) Express \(s\) as a function of \(t\) by means of a three-part formula.

Short Answer

Expert verified
Part (a): 4-part formula segments the run per base path. Part (b): 3-part formula combines segments where the formula doesn't change.

Step by step solution

01

Understanding the Diamond Layout

The baseball diamond is a square with each side measuring 90 feet. Thus, the total distance around the bases is the perimeter of the square, which equals \(4 \times 90 = 360\) feet. The player needs to run this distance to return to home plate.
02

Determine the Travelled Distance Function

Knowing the player runs at a constant speed of 10 feet per second, we can express the distance \(s\) as a function of time \(t\). Since distance \( \text{(Distance)} = \text{(Speed)} \times \text{(Time)} \), we have: \(s(t) = 10t\).
03

Investigate Each Segment Separately (Four-Part Formula)

For each base path (90 feet each), we define \(s(t)\) based on \(t\) values:- 1st base path (0 to 90 feet): \(s(t) = 10t\) for \(0 \leq t < 9\).- 2nd base path (90 to 180 feet): \(s(t) = 10t\) remains valid, \(9 \leq t < 18\).- 3rd base path (180 to 270 feet): Maintain function, \(s(t) = 10t\); \(18 \leq t < 27\).- 4th base path (270 to 360 feet): Maintain function, \(s(t) = 10t\); \(27 \leq t < 36\).
04

Group Segments to a Three-Part Formula

Combine consecutive segments that do not require a change in distance formula:- From home to second base (0 to 180 feet): \(s(t) = 10t\); \(0 \leq t < 18\).- Round second, heading to third (180 to 270 feet): Continue with \(s(t) = 10t\); \(18 \leq t < 27\).- From third base to home (270 to 360 feet): Continue with \(s(t) = 10t\); \(27 \leq t < 36\).
05

Finalize the Formulas

For the four-part formula:\[ s(t) = \begin{cases} 10t & \text{if } 0 \leq t < 9, \ 10t & \text{if } 9 \leq t < 18, \ 10t & \text{if } 18 \leq t < 27, \ 10t & \text{if } 27 \leq t < 36.\end{cases} \]For the three-part formula:\[ s(t) = \begin{cases} 10t & \text{if } 0 \leq t < 18, \ 10t & \text{if } 18 \leq t < 27, \ 10t & \text{if } 27 \leq t < 36.\end{cases} \]

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Key Concepts

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

Piecewise Functions
Piecewise functions are a powerful tool in mathematics. They let us describe a function that is defined by different expressions over different intervals. In the case of the baseball player, the function representing the distance is expressed for different sections of time intervals. This helps simplify complex situations into manageable parts.
For example, the player's path from home plate back to home plate can be divided into intervals corresponding to each base path. In this way, we use functions that are valid only within certain ranges.
  • From 0 to 90 feet (0 to 9 seconds): The function is linear, described as \(s(t) = 10t\).
  • From 90 to 180 feet (9 to 18 seconds): The same linear function continues since the speed remains constant.
  • And so on, for the third and fourth base paths.
This approach is particularly helpful in scenarios where conditions change across defined intervals.
Parametrization
Parametrization involves expressing variables using parameters, which can simplify the representation of paths or curves. Here, the player's path around the bases is parametrized using time \(t\) as the parameter.
In this context, parametrization provides a clear way to calculate the player's position based on time.
  • Distance \(s\) is dependent on time \(t\).
  • It gives us a precise way to model motion since time is a direct measure of the player's movement.
Since the player's speed is constant, the expression \(s(t) = 10t\) gives a straightforward relationship between distance and time. Parametrization is particularly useful when dealing with motion or paths that have a time-dependent nature.
Applied Mathematics
In applied mathematics, we use math concepts to solve real-world problems. The exercise involving a baseball player's run around a diamond is a classic example. It showcases how mathematical models can represent practical scenarios.
Consider the baseball player running: using mathematical concepts like piecewise functions and parametrization simplifies analyzing their movement.
  • We transform real-world actions into mathematical expressions.
  • This approach allows us to solve for unknowns, predict outcomes, and gain insights into physical processes.
Such applications help us understand the underlying patterns and principles of everyday experiences, bridging the gap between theory and practice. They can be used in various fields such as engineering, economics, and natural sciences.

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