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Math · Differential Equations

Chapter 2: Separable Equations and Models

Growth, Decay and Cooling

One equation behind three familiar phenomena.

Lesson
2
Time
About 23 minutes
0 of 12 done

Step 1: Let's Learn

Read it, or press Listen and follow the words.

dy/dt = ky says the rate of change is proportional to the amount present. Its solution is y = y₀e^(kt).

The sign of k

A positive k gives growth and a negative one gives decay. Nothing else about the equation changes.

Half-life

The time to halve is fixed and does not depend on how much you started with. That is the signature of exponential decay.

Newton's law of cooling

dT/dt = −k(T − A). The rate depends on the gap between the object and the ambient temperature A.

Why it is the same equation

Substituting u = T − A turns cooling into du/dt = −ku. It is exponential decay of the temperature gap.

Where it stops

The object approaches the ambient temperature but never quite reaches it. A is the stable equilibrium.

One equation behind three phenomena

dy/dt = ky says the rate is proportional to the amount. It produces exponential growth for positive k and decay for negative, and it is the same equation in both cases.

Half-life is independent of the starting amount

The time to halve depends only on k, not on how much you began with. That is a distinctive signature of exponential decay and is what makes radiometric dating possible.

Newton's law of cooling

An object cools at a rate proportional to the difference from ambient temperature. The equation is dT/dt = k(T − Tₐ), and its solutions approach ambient asymptotically without ever arriving.

Where proportionality fails

Populations run out of resources, so unbounded exponential growth is always a short-term approximation. Recognising the limits of the model is part of using it responsibly.

Step 2: Try It Yourself

Tap and try it out.

Make the exponent negative and growth becomes decay. The same equation covers both.
-8-8-6-6-4-4-2-222446688
y = 1 · 0.5^x + 0
  • Point(1, 0.50)

Step 3: Watch an Example

One step at a time.

Watch Amara Use a Half-Life

Amara has 80 grams of a substance with a half-life of 5 years and wants the amount after 15 years.

  1. Step 1

    She divides 15 by 5, finding that three half-lives have passed.

Step 4: Your Turn

Practice makes it stick.

The Sample

Problem 1 of 2

200 grams with a half-life of 4 days. How many grams remain after 12 days?

The Coffee

Problem 2 of 2

Coffee cooling in a room at 21 degrees. What temperature does it approach in the long run?

Model It

1 of 8

100 grams with a half-life of 3 hours. How much remains after 6 hours?

2 of 8

64 grams with a half-life of 2 years. How much remains after 8 years?

3 of 8

dy/dt = ky with k = −0.3. Is this growth or decay? Enter 1 for growth, 2 for decay.

4 of 8

dT/dt = −k(T − 18). What is the ambient temperature?

5 of 8

A population doubles every 10 years. How many times larger is it after 30 years?

6 of 8

An object at exactly the ambient temperature. What is dT/dt?

7 of 8

Match each situation to the sign of its rate constant.

Tap a card on the left to start.

8 of 8

32 grams with a half-life of 1 year. How much remains after 5 years?

Step 5: Quick Check

Show what you know.

Question 1 of 2

160 grams with a half-life of 5 days. How much remains after 15 days?

Question 2 of 2

Why is Newton's law of cooling the same equation as decay?

What You Learned

  • dy/dt = ky solves to y = y₀e^(kt): positive k grows, negative k decays.
  • A half-life is fixed regardless of the starting amount.
  • Newton's law of cooling is exponential decay of the gap to the ambient temperature.