Сюжетные математические задачи и логика · Dynamic Rate of Consumption
Визуализатор задачи Ньютона о пасущихся коровах
Визуализатор классической задачи Ньютона о скорости роста травы и выедании пастбища
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Окно математической демонстрации● 60 FPS
Параметры эксперимента
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Расчёты и показатели в реальном времени
Суточный прирост травы
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Исходный запас травы
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Дней до исчерпания
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Макс. устойчивое стадо
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Вывод формул в реальном времени
Основная формула:
--Подстановка значений:
--Вывод: --
Разбор типовых задач и анализ экзаменационных ошибок
Worked Problem
Условие задачи: Pasture feeds 27 cows for 6 weeks, or 23 cows for 9 weeks. How long for 21 cows? Max cows to graze forever?
Пошаговый вывод формул:
- Weekly growth = (23*9 - 27*6)/(9-6) = 15 units/wk. Initial grass = 27*6 - 15*6 = 72 units. 21 cows: 15 eat growth, 6 eat initial stock: 72/6 = 12 weeks. Max 15 cows graze forever.
Worked Problem
Условие задачи: Station ticket hall queue: 4 windows clears in 30 min, 6 windows in 12 min. How fast for 8 windows?
Пошаговый вывод формул:
- Arrival rate = (120 - 72)/18 = 8/3. Initial queue = 40. With 8 windows: net clearance rate = 8 - 8/3 = 16/3. Time = 40 / (16/3) = 7.5 min.
Рабочий лист для урока и домашние задания
Q1. Variant 1: Leaking boat: 10 people bail in 20 min; 15 people in 10 min. How many people to clear in 5 min?
Answer Key: Leak rate = (200 - 150)/10 = 5 units/min. Initial water = 100. For 5 min: total water = 100 + 5*5 = 125. People needed = 125 / 5 = 25 people.
Q2. Variant 2: Reservoir flood: 3 gates empty in 36h; 5 gates in 18h. How many gates to empty in 12h?
Answer Key: Inflow rate = (108 - 90)/18 = 1 gate. Initial water = 72. Required clearance rate = 72 / 12 = 6 gates. Total gates = 6 + 1 = 7 gates.
Q3. Variant 3: Pasture feeds 20 cows for 10d, 15 cows for 20d. If grass stops growing in autumn, how long for 25 cows?
Answer Key: Growth = 10 units/d. Initial grass = 100. Autumn stops growth (rate = 0): Days = 100 / 25 = 4 days.
Часто задаваемые вопросы и типичные заблуждения
What fundamentally sets Newton's grazing problem apart from standard work problems?
The base stock is not static; it grows continuously while being consumed! It is a dynamic equilibrium inflow-outflow problem.
What happens if cow count is strictly less than daily growth rate?
Grass accumulates indefinitely! Inflow exceeds consumption, ensuring infinite grazing sustainability.
How to instantly spot disguise variants of Newton's grazing problem?
Look for the signature triplet: fixed initial backlog + steady uniform inflow rate + parallel outflow consumer agents (e.g. queues, leaks, dams).
Why is this problem named after Sir Isaac Newton?
Newton formulated it in his 1707 treatise Arithmetica Universalis as a foundational algebra case study for dynamic equilibrium systems.