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Grade 12 · Differential growth models

Differential growth models: A growing unit initial condition

Differential growth models: investigate a growing unit initial condition with initial amount a = 1; rate k = 0.2.

Predict what will happen, press Play, then pause and explain what changed. Every control also works without playback.

Watch the relationship

Paused
Differential growth models: A growing unit initial condition. Time / input: 0. Current amount: 1. Instantaneous rate: 0.2. Growth multiplier: 1An exact proportional ODE solution02.23024amount ytime t → · labeled axes rescale to this model
The rate k is constant and all initial values are positive. This is an exact elementary ODE model with arbitrary units, not a numerical solver or a general population forecast. The linked lesson develops the inverse-exponential prerequisite.

Starts paused. Play once, pause anywhere, or use Step to inspect the mathematics. Playback stops when this panel leaves the screen.

Make it your experiment

Change one value. Notice what follows.

The controls adjust the model. Numbers below describe the current frame. Decimals are rounded.

Time / input
0
Current amount
1
Instantaneous rate
0.2
Growth multiplier
1

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Understand what you are seeing

The idea behind the motion.

A differential equation constrains an unknown function through its instantaneous rate. For proportional growth, the exponential solves both the rate relation and the initial condition. Checking only a curve's starting value is insufficient: the derivative must equal k times the amount at every time. This investigation starts with Initial amount a = 1; Rate k = 0.2. Predict the result before playing, then change one parameter while keeping the other fixed to test the reason for the change.

A relationship to keep

y′=ky; y(0)=a; y(t)=a e^(kt)

Read the symbols alongside the explanation. A diagram shows the relationship; the assumptions tell you when it applies.

  1. STEP 1

    Set up the mathematical model

    Use the initial condition to fix the multiplicative constant in the exponential family. The starting case is “A growing unit initial condition.”

  2. STEP 2

    Follow the changing quantity

    Follow the solution over four time units. Compare the tangent slope with k times the current height.

  3. STEP 3

    Explain and test the result

    Check the zero-rate case and a negative-rate case. A negative derivative describes decreasing positive amount, not a negative amount.

Your turn to explain

Make a prediction. Test your reasoning.

Keep Initial amount a = 1; Rate k = 0.2. Pause the timeline at 40%. Given time / input = 1.6, calculate current amount, instantaneous rate, growth multiplier. Show the substitution into the displayed formula.

Use the values specified in the question. Reset restores the initial values for this investigation.

Compare your explanation

Substitution gives y=1 exp((0.2)(1.6))=1.377128. The rate is k y=(0.2)(1.377128)=0.275426, rather than just the amount y. Results: Current amount: 1.377; Instantaneous rate: 0.275; Growth multiplier: 1.377. Decimal values are rounded; retain the original parameters when checking.

Work through a full lesson

Connect the animation to a worked example and practice questions.

Grade 12 · Algebra

Undo a shifted exponential and state the recovered domain

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