Newton's law says y′ = −k·(y − A): the object changes at a rate proportional to how far it is from the surrounding temperature A. Substituting u = y − A turns it into the pure decay u′ = −k·u, so the gap shrinks geometrically in Euler's method and exponentially in the solution, and the temperature settles at A. The rate is large while the gap is large and small near the end, which is not a constant rate of cooling.
Core ideas
Three ideas: what the rate is proportional to, the substitution that reduces the equation to one already solved, and where the temperature ends up.
The object changes fastest when it is far from the surroundings and slowly when it is near them. The equation y′ = −k·(y − A) has a negative rate while y > A and a positive one while y < A, so it covers cooling and warming alike.
Put u = y − A. Since A is a constant, u′ = y′, and the equation becomes u′ = −k·u: the decay of the previous lesson, for the gap. So u = u₀·e^(−kt) and y = A + (y₀ − A)·e^(−kt).
The gap u₀·e^(−kt) is never zero. The value A is an equilibrium: start there and y′ = 0. Start anywhere else and the temperature moves towards A forever and arrives at no time.
The material
The law, the substitution, the coffee's first four steps in exact fractions beside the closed form, and a cold cup warming by the same equation.
A cup of coffee at 100 degrees stands in a room at 20 degrees. Newton's observation was that it loses heat at a rate proportional to how much hotter it is than the room. Write y for its temperature and take the constant of proportionality to be k = 1/4 per minute.
A body at temperature y in surroundings at the constant temperature A obeys y′ = −k·(y − A) for a constant k > 0.
It assumes the surroundings stay at A and the body has one temperature throughout. The lab computes what follows from those assumptions; it does not check them for a real cup.
The check is by differentiating: the rate of 20 + 80·e^(−t/4) is −20·e^(−t/4), and −(1/4)·(y − 20) = −(1/4)·80·e^(−t/4), the same. The temperature at time t is a rounded number, since e^(−t/4) is irrational for t ≠ 0, which is why a check by differentiating is the right check and an evaluation is not.
Run Euler's method with a step of one minute. The first step uses the starting rate −(1/4)·(100 − 20) = −20, so y₁ = 80. The gap is now 60 and the rate −15, so y₂ = 65. Then the gap is 45 and the rate is −45/4, so y₃ = 215/4. The rate changes at every step because the gap does.
The gap is multiplied by 1 − kh = 1 − 1/4 = 3/4 at every step (the lab's factor tile, which takes the signed rate −1/4, writes the same number as 1 + kh), so yₙ = 20 + 80·(3/4)ⁿ. At n = 4 that is 20 + 80·81/256 = 725/16, which is 45.3125, and the lab prints it exactly. The closed form at t = 4 is 20 + 80·e^(−1), which the lab prints rounded: ≈ 49.4304. The two differ because a step of one minute assumes the rate stays put for the whole minute, while the true rate keeps falling. The lab's error tile prints the gap as ≈ 4.11786, rounded, and it is the method's error at this step size and no other.
A cup at 5 degrees in a room at 20 has a gap of 5 − 20 = −15, a negative gap, and the rate −k·(−15) is positive. Nothing else changes. With k = 1/2 and a step of 1 the factor is 1/2, so the gap is cut in half at every step: −15, −15/2, −15/4, …. The temperatures 5, 25/2, 65/4, 145/8, 305/16, … climb towards 20 and none of them passes it; the lab's closed form at the fourth step is ≈ 17.97, rounded. The third preset is a drink at 25 degrees in a fridge at 4: the gap is 21, the factor is 9/10, and the steady-state tile reads 4.
The gap follows u₀·e^(−kt), so it halves in ln 2/k, the half-life of the previous lesson. For k = 1/4 that is 4·ln 2 ≈ 2.77259 minutes, and then the coffee is at 20 + 40 = 60 degrees. With the target set to 60, the lab prints the first step of the exact sequence at or below 60: step 3 (t = 3), where the sequence is at 215/4 = 53.75, having been at 65 one step earlier.
Interactive
A geometric sequence beside an exponential
Method
Five moves, and the substitution in the second is the one that reduces a new equation to an old one.
Read A and k off the equation y′ = −k·(y − A). The lab takes the signed coefficient, −k.
u = y − A gives u′ = −k·u with u₀ = y₀ − A. The sign of u₀ says whether the object cools (positive) or warms (negative).
u = u₀·e^(−kt), so y = A + (y₀ − A)·e^(−kt). Check it by differentiating.
The factor 1 − kh multiplies the gap each step, so yₙ = A + (y₀ − A)·(1 − kh)ⁿ, an exact fraction.
The Euler value is exact, the closed form is rounded and marked ≈, and the step size goes beside the pair.
The equation is y′ = −(1/4)·(y − 20) with y(0) = 100 and h = 1. Everything is exact until the closed-form line.
Euler is below the curve here, by the amount the lab's error tile prints. The reason is the one in the table above: each step took the rate at the start of the minute, and the real rate falls during it, so the real coffee had cooled less than the sequence says.
The steady state is 20. In the closed form the term 80·e^(−t/4) goes to zero without reaching it, and in the exact sequence 80·(3/4)ⁿ is positive for every n.
You should be able to read the surroundings and the rate off the equation, substitute u = y − A, write yₙ as A plus a power of the factor, name the steady state, and explain why the rate is not constant.
Something to change about this lesson? Write it down, tick it off when it is handled, and export the list when you want to act on it. Kept in this browser only: nothing is sent anywhere, and nobody is notified.
No recommendations yet for this lesson.