Newton's Law of Cooling states that an object's temperature changes at a rate proportional to the difference between its temperature and the surrounding (ambient) temperature, written as the differential equation dT/dt = k(T − A). It's the classic 'verbal statement to differential equation' setup in AP Calc Topic 7.1.
Newton's Law of Cooling is a physical principle that AP Calc turns into a differential equation. The idea is intuitive. A cup of 95°C coffee in a 20°C room cools fast at first because the temperature gap is huge, then cools slower and slower as it approaches room temperature. The rate of cooling depends on that gap.
In math terms, if T(t) is the object's temperature at time t and A is the ambient (room) temperature, then dT/dt = k(T − A), where k is a constant of proportionality. Notice what this equation actually says. It relates a function (temperature) to its own derivative (rate of temperature change), which is exactly what a differential equation is. The phrase "rate of cooling is proportional to the temperature difference" translates word-for-word into the equation: "rate" becomes dT/dt, "proportional to" becomes k times, and "temperature difference" becomes (T − A). That translation skill is the whole point of Topic 7.1.
Newton's Law of Cooling lives in Unit 7: Differential Equations, specifically Topic 7.1: Modeling Situations with Differential Equations. It directly supports learning objective 7.1.A, which asks you to interpret verbal statements as differential equations involving a derivative expression. Cooling problems are the single most common real-world scenario the exam uses to test this skill, right alongside population growth.
It also matters because it's not a one-topic concept. Once you set up dT/dt = k(T − A) in 7.1, you'll solve it later in Unit 7 using separation of variables, and the solution is an exponential function that levels off at room temperature. So one cooling problem can quietly test modeling, solving, and interpreting all at once.
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Visual cheatsheet
view galleryExponential Growth and Decay Models, dy/dt = ky (Unit 7)
Newton's Law of Cooling is the exponential decay model with the thermostat moved. Instead of decaying toward zero like dy/dt = ky, temperature decays toward the ambient temperature A. Substitute y = T − A and the cooling equation literally becomes dy/dt = ky.
First Derivative (Unit 2)
The dT/dt in the cooling equation is just a first derivative, the instantaneous rate of change of temperature. Unit 7 doesn't introduce new derivative machinery; it asks you to recognize when a sentence like 'the rate at which heat transfers' is describing one.
Independent Variable (Unit 7)
In cooling problems, time t is the independent variable and temperature T is the dependent function T(t). Getting this straight tells you the derivative must be dT/dt, not dt/dT, which is a classic setup trap in multiple choice.
Slope Fields (Unit 7)
Before you ever solve dT/dt = k(T − A), a slope field shows you the story visually. Slopes are steep when T is far from A and flatten to zero at T = A, which is the horizontal line every solution curve approaches.
Newton's Law of Cooling shows up in three predictable ways. First, translation questions hand you the verbal statement ('the rate of cooling is proportional to the difference between the object's temperature and the ambient temperature') and ask which differential equation matches. The answer is dT/dt = k(T − A), and the wrong choices swap the proportionality, drop the difference, or use T alone. Second, solve-for-k questions give you data, like a coffee cup going from 95°C to 70°C in 5 minutes in a 20°C room, and ask for k in dT/dt = k(T − 20). Third, prediction questions ask for the temperature at a later time after you've solved the equation, such as where a 90°C object that hit 60°C in 10 minutes will be later. No released FRQ has used the phrase verbatim, but contextual differential equation FRQs in Unit 7 regularly use this exact structure, so practice writing the equation, separating variables, and using initial conditions to pin down constants.
Both produce exponential behavior, but they decay toward different targets. Plain exponential decay dy/dt = ky heads toward zero. Newton's Law of Cooling, dT/dt = k(T − A), heads toward the ambient temperature A, not zero, because a coffee cup stops cooling at room temperature, not at absolute zero. If an answer choice is missing the (T − A) difference, it's modeling the wrong situation.
Newton's Law of Cooling translates to the differential equation dT/dt = k(T − A), where T is the object's temperature and A is the ambient temperature.
The phrase 'proportional to the difference' means the equation must contain k times (T − A), not k times T alone.
Cooling is fastest when the temperature gap is largest, and the rate approaches zero as T approaches room temperature, so solution curves level off at T = A.
This is a Topic 7.1 skill (LO 7.1.A): turning a verbal rate statement into a differential equation, before you ever solve anything.
To find k, solve the differential equation by separation of variables, then plug in a given data point like 'the coffee is 70°C after 5 minutes.'
Time t is the independent variable and temperature T(t) is the dependent function, which is why the derivative is dT/dt.
It's the principle that an object cools at a rate proportional to the difference between its temperature and the surrounding temperature, modeled by the differential equation dT/dt = k(T − A). In AP Calc it's the go-to scenario for Topic 7.1, translating verbal statements into differential equations.
It depends on how the equation is written. In dT/dt = k(T − A), a cooling object has T > A and a negative dT/dt, so k must be negative. If the book writes dT/dt = −k(T − A) instead, then k is positive. Always check the sign convention in the problem rather than memorizing one answer.
Regular exponential decay dy/dt = ky decays toward zero, while Newton's cooling decays toward the ambient temperature A. The cooling model is exponential decay applied to the temperature difference T − A, which is why solutions look like T = A + Ce^(kt).
No. Topic 7.1 only asks you to set up the equation from the verbal description (LO 7.1.A). Actually solving dT/dt = k(T − A) with separation of variables comes later in Unit 7, often paired with an initial condition like 'a 95°C coffee in a 20°C room.'
Because the object stops cooling at room temperature, not at 0 degrees. The rate depends on how far the object is from its surroundings, so the gap T − A drives the cooling. An equation with k times T alone would wrongly predict the coffee keeps cooling past room temperature toward zero.
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