Problemas matemáticos y razonamiento · Dynamic Rate of Consumption
Visualizador del problema de los pastos de Newton
Visualizador del problema de crecimiento de pastos y vacas de Newton: equilibrio dinámico
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Ventana de demostración matemática● 60 FPS
Variables de control del experimento
animales
días
animales
días
animales
Derivaciones y métricas en tiempo real
Crecimiento diario de hierba
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Hierba inicial
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Días hasta agotarse
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Ganado máximo sostenible
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Derivación de fórmulas en tiempo real
Fórmula básica:
--Sustitución de valores:
--Conclusión: --
Ejemplos clásicos y análisis de errores en exámenes
Worked Problem
Problema: Pasture feeds 27 cows for 6 weeks, or 23 cows for 9 weeks. How long for 21 cows? Max cows to graze forever?
Derivación paso a paso:
- 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
Problema: Station ticket hall queue: 4 windows clears in 30 min, 6 windows in 12 min. How fast for 8 windows?
Derivación paso a paso:
- 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.
Hoja de indagación en clase y tareas
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.
Preguntas frecuentes y errores conceptuales
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.