Newton's Law of Cooling

Calculus 2 · Axiom Academy

EXAMPLE Newton's Law of Cooling Solve a first-order differential equation by separation of variables to model how temperature changes over time. A cup of coffee starts at in a room held at , with cooling constant k = 0.1 per minute. Newton's Law of Cooling gives the differential equation . Solve it for T(t) and use the initial condition to find the particular solution. Nice work — you separated variables, integrated, and pinned down the constant with an initial condition. Here is what carried the solution: Separation of variables: Gather every T term on one side and every t term on the other before integrating: . The log pattern: — exponentiating that log is what frees T . Initial condition fixes A: T(0) = 90 with forces A = 90 - 20 = 70 . Exponential decay toward the room: , so the temperature gap shrinks exponentially and . Our coffee follows T(t) = 20 + 70e^ -0.1t : about after 10 minutes, leveling off at room temperature as . The same separation-of-variables move solves growth, decay, and mixing problems throughout Calculus 2.

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