Problemas matemáticos e raciocínio lógico · Dynamic Rate of Consumption
Visualizador do problema dos pastos de Newton
Visualizador do problema de Newton sobre crescimento de pasto e consumo de vacas
- No navegador
- Sem cadastro
- Dedução ao vivo
- Modo projetor
- Folha de exercícios A4
Janela de demonstração matemática● 60 FPS
Variáveis de controle do experimento
animais
dias
animais
dias
animais
Deduções e métricas em tempo real
Crescimento diário do capim
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Capim inicial
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Dias até acabar
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Rebanho máximo sustentável
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Dedução de fórmulas em tempo real
Fórmula fundamental:
--Substituição de valores:
--Conclusão: --
Exemplos clássicos e armadilhas de exame
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?
Derivação passo a passo:
- 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?
Derivação passo a passo:
- 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.
Folha de investigação em aula e exercícios
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.
Perguntas frequentes e erros comuns
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.