Mathematische Sachaufgaben & Denkmodelle · Dynamic Rate of Consumption
Newtons Rinderproblem & Weide-Visualisierer
Newtons Weideproblem: Mathematische Modellierung von Graswachstum und Rinderweide
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Experimentelle Steuerungsparameter
Tiere
Tage
Tiere
Tage
Tiere
Echtzeit-Herleitung & Kennzahlen
Tägliches Graswachstum
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Anfänglicher Grasbestand
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Tage bis leer
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Maximale nachhaltige Herde
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Echtzeit-Formelherleitung
Grundformel:
--Einsetzungsschritte:
--Ergebnis: --
Typische Beispielaufgaben & Prüfungsfallen
Worked Problem
Aufgabe: Pasture feeds 27 cows for 6 weeks, or 23 cows for 9 weeks. How long for 21 cows? Max cows to graze forever?
Schrittweise Herleitung:
- 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
Aufgabe: Station ticket hall queue: 4 windows clears in 30 min, 6 windows in 12 min. How fast for 8 windows?
Schrittweise Herleitung:
- 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.
Arbeitsblatt für den Unterricht & Hausaufgaben
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.
Häufige Fragen & typische Missverständnisse
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.