Mathematical Modeling

Grade 12 · Math · Study Guide

Learn how to translate real-world phenomena into mathematical structures including linear, exponential, logistic, and differential equation models.

Mathematical modeling is the process of representing real-world situations with mathematical equations to analyze, predict, and optimize outcomes. The modeling cycle includes: (1) identify the problem, (2) make assumptions, (3) build the model, (4) solve/analyze it, (5) interpret results, (6) validate against real data, and (7) refine if necessary.

Linear models y = mx + b describe situations with constant rates of change. Exponential models y = ab^t describe growth (b > 1) or decay (0 < b < 1) with a constant percentage rate — like population growth or radioactive decay. The exponential model can be written as y = y₀e^(kt), where k > 0 is growth and k < 0 is decay.

Logistic models capture growth with a carrying capacity K: P(t) = K / (1 + Ae^(-kt)). Population growth in limited environments follows this S-shaped curve, growing rapidly at first, then leveling off at K.

Differential equations model rates of change directly. The equation dy/dt = ky gives exponential behavior; its solution is y = y₀e^(kt). More complex systems involve systems of differential equations. Regression analysis fits a mathematical model to real data, and residual plots help assess model fit. R² (coefficient of determination) measures what proportion of variation in the data the model explains.

Key terms

Mathematical Model
A mathematical representation of a real-world system used to analyze behavior, make predictions, or optimize outcomes
Exponential Growth Model
y = y₀e^(kt) with k > 0; describes populations, investments, or quantities growing at a constant percentage rate
Logistic Model
P(t) = K/(1+Ae^(-kt)); models growth with a carrying capacity K, producing an S-shaped curve
Carrying Capacity
The maximum population size K that an environment can sustainably support in a logistic model
Differential Equation
An equation involving a function and its derivatives; dy/dt = ky models exponential growth or decay
R² (Coefficient of Determination)
A measure from 0 to 1 of how well a regression model fits the data; closer to 1 indicates a better fit

Flashcards

Modeling Cycle
Problem → Assumptions → Build model → Solve → Interpret → Validate → Refine
Exponential Growth
y = y₀e^(kt), k > 0; or y = y₀b^t, b > 1; constant percentage rate of increase
Exponential Decay
y = y₀e^(-kt), k > 0; or y = y₀b^t, 0 < b < 1; models radioactive decay, cooling, depreciation
Half-Life Formula
A(t) = A₀ · (1/2)^(t/h), where h is the half-life period
Logistic Growth Equation
P(t) = K / (1 + Ae^(-kt)), where K = carrying capacity, A = (K - P₀)/P₀
Separable Differential Equation
An equation that can be written as g(y)dy = f(t)dt; solved by integrating both sides
Linear Regression Model
ŷ = mx + b fitted to minimize sum of squared residuals; R² measures goodness of fit
Residual
Observed value minus predicted value: e = y - ŷ; residual plots reveal model inadequacy

Questions

  1. A bacterial colony starts with 500 cells and doubles every 3 hours. Which model describes the population after t hours?

    1. P(t) = 500 + 2t
    2. P(t) = 500 · 2^(t/3)
    3. P(t) = 500e^(2t)
    4. P(t) = 500 · 3^(t/2)
    Show answer

    Answer: P(t) = 500 · 2^(t/3)

    Doubling every 3 hours: P(t) = 500 · 2^(t/3). After 3 hours: 500·2¹=1000. After 6 hours: 500·2²=2000. This correctly models exponential growth with a 3-hour doubling time.

  2. A logistic growth model predicts that a population will grow without bound if given enough time.

    Show answer

    Answer: False

    Logistic growth has a carrying capacity K. As t → ∞, P(t) → K. The population levels off and does not grow without bound.

  3. Radioactive carbon-14 has a half-life of 5,730 years. If a sample begins with 80 grams, which gives the amount remaining after t years?

    1. A(t) = 80(0.5)^(t/5730)
    2. A(t) = 80e^(5730t)
    3. A(t) = 80 - 5730t
    4. A(t) = 80(2)^(t/5730)
    Show answer

    Answer: A(t) = 80(0.5)^(t/5730)

    Half-life model: A(t) = A₀ · (1/2)^(t/h) = 80 · (0.5)^(t/5730). After 5730 years: 80 · 0.5 = 40 grams. This correctly models radioactive decay.

  4. In the differential equation dP/dt = 0.04P, the constant 0.04 represents the ___ rate.

    Show answer

    Answer: growth

    In dP/dt = kP, the constant k is the per-unit growth rate. With k = 0.04, the population grows at 4% per unit time.

  5. A model has R² = 0.92. This means the model explains what percentage of the variation in the data?

    1. 8%
    2. 92%
    3. 0.92%
    4. 8.7%
    Show answer

    Answer: 92%

    R² = 0.92 means the model accounts for 92% of the variance in the data. The remaining 8% is unexplained variation (residuals).

Get a personalized course on Mathematical Modeling built around your child.