How Does the Mathematical Function of Time Affect River Depth?

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Discussion Overview

The discussion revolves around the mathematical function representing the depth of a river over time, specifically how the function D = 9 + 3cos[pi(t)/14] can be analyzed to determine the minimum and maximum depths, as well as the timing of these occurrences. The scope includes mathematical reasoning and problem-solving related to the function's behavior over a specified time period.

Discussion Character

  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • Post 1 introduces the function for river depth and asks for help with determining minimum and maximum depths and the timing of the minimum.
  • Post 2 explains that the minimum depth occurs when the cosine function is at its minimum and the maximum when it is at its maximum, suggesting a method to find these values.
  • Post 3 confirms the minimum depth as 6m and maximum as 12m but expresses uncertainty about how to find the time of the first minimum.
  • Post 4 questions the notation of pi in the function, suggesting it may be a typographical error, and proposes a method involving derivatives to find extrema and their timing.

Areas of Agreement / Disagreement

Participants generally agree on the method to find the minimum and maximum depths based on the cosine function, but there is uncertainty regarding the timing of the minimum depth and the interpretation of the function's notation.

Contextual Notes

There are unresolved assumptions regarding the interpretation of the function and the calculation of time differences, as well as the potential for multiple minima in the function's behavior.

bagpiper
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D = 9 + 3cos[pi(t)/14]
where D meters is the depth of the water above the river bed at time t hours after 12 noon on 7 March 2005, Monday. This means that the depth of water only depends on time.
a) Write down the minimum and maximum depths of the river
b) Find the day and time when the water first reaches its minimum.

please help me with these questions! thanks!
 
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You know that:
[tex]-1 \leq \cos \alpha \leq 1[/tex]
So when the D is minimum is when [tex]\cos \frac{\pi t}{14}[/tex] is minimum and vice versa, when the D is maximum is when [tex]\cos \frac{\pi t}{14}[/tex] is maximum.
Can you go from here?
Viet Dao,
 
Yup. i got there. So minimum is 6m, maximum is 12m. But i don't know how to manipulate it formula to solve the second part.
 
Why is pi written as a function of time? Isn't pi a constant? I'll assume it's a typographic error or it means pi * t, i.e. multiplication.

1. Let's say you have D = f(t). If you set f'(t) = 0 that'll give you the extrema (minima and maxima) solutions, say tmin and tmax. Then take each of these solutions and look at f''(t) at this point. If f''(t) < 0 then it's a maximum; if f''(t) > 0 then it's a minimum. If f''(t) = 0 then it's an inflection point, I guess.
2. Now having found all your minima, calculate the time difference between the first such minima occurring (assuming there are more than one, if there is only one minimum then just take its time of occurrence, tmin) and "today" (7 March 2005): tmin - t7Mar05.
 

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