Temperature and Its Rate of Change
A temperature model describes how the temperature of an object or environment varies with time. Its derivative describes something different: the instantaneous rate at which the modeled temperature is changing at a particular time.
That rate can tell us whether an object is warming or cooling, and how quickly. It also helps distinguish a rapid change from a slow one. To interpret it correctly, keep the temperature and its rate separate: temperature is measured in degrees, while temperature change per unit time is measured in degrees per unit time.
Reading the Sign and Magnitude
The sign of \(T'(t)\) describes the direction of change as time moves forward. A positive derivative means the model’s temperature is increasing at that time, so the object is warming. A negative derivative means the temperature is decreasing, so the object is cooling. A zero derivative means the instantaneous rate is zero at that particular time.
The magnitude, \(\lvert T'(t)\rvert\), describes the pace of change. For instance, a rate of \(-4\) degrees Celsius per minute indicates faster cooling than a rate of \(-1\) degree Celsius per minute. The minus sign gives the direction; the magnitude gives the size of the rate.
| Derivative value | Interpretation as time advances |
|---|---|
| \(T'(t)>0\) | The temperature is increasing: warming. |
| \(T'(t)<0\) | The temperature is decreasing: cooling. |
| \(T'(t)=0\) | The instantaneous rate is zero at that time. |
A derivative describes a local rate, not necessarily the behaviour of the model over an entire interval. A temperature can be warming at one time and cooling later. To describe an extended period, check the derivative throughout that period or use the temperature values at its endpoints to find the average rate.
Why the Derivative Gives a Local Rate
For two times \(t\) and \(t+h\), the average temperature-change rate over the interval between them is the temperature difference divided by the elapsed time. When \(h\ne0\), this average rate is \(\frac{T(t+h)-T(t)}{h}\). The derivative is the limit of these average rates as the time interval shrinks toward zero.
This definition explains why the derivative has the interpretation of an instantaneous rate: it is the limiting value of the average rate over shorter and shorter intervals around the time of interest. In particular, when \(T'(t)\) is close to a fixed number \(r\), the temperature change over a small time interval \(\Delta t\) is approximately \(r\Delta t\).
The approximation is local. It does not promise that the same rate continues for a long period. If the rate changes with time, the estimate becomes less reliable as the time interval grows.
Worked Example: Constant Warming
Worked Example: Constant Warming
A temperature model, measured in degrees Celsius, is \(T(t)=20+3t\), where \(t\) is measured in minutes and \(0\le t\le8\). The model’s derivative is \(T'(t)=3\).
At \(t=4\) minutes, the temperature is
The instantaneous rate at that time is
Because the rate is positive, the model describes warming. Its magnitude is \(3\text{ °C/min}\), so the modeled temperature is increasing by 3 degrees Celsius per minute. Here the rate is constant: for example, from \(t=4\) to \(t=5\), the temperature changes from \(32\text{ °C}\) to \(35\text{ °C}\), a change of \(3\text{ °C}\) in one minute.
Check: The average rate over those endpoints is \(\frac{35-32}{5-4}=3\text{ °C/min}\), agreeing with the derivative.
Worked Example: Warming, a Level Moment, and Cooling
Worked Example: Warming, a Level Moment, and Cooling
Consider \(T(t)=12+8t-t^2\), measured in degrees Celsius for \(0\le t\le6\) minutes. Its derivative is \(T'(t)=8-2t\), in degrees Celsius per minute.
At \(t=1\),
The model is warming at \(6\text{ °C/min}\) at that time.
At \(t=4\),
The temperature has zero instantaneous rate at \(t=4\). To interpret the surrounding behaviour, note that \(T'(t)=8-2t\) is positive when \(t<4\) and negative when \(t>4\). On the model’s domain, it is warming before \(t=4\) and cooling after \(t=4\). Thus, \(t=4\) is a moment when the model changes from warming to cooling; the temperature there is \(28\text{ °C}\).
At \(t=5\),
The negative rate means cooling at \(2\text{ °C/min}\). As a check on the change across the interval from minute 4 to minute 5, the model gives \(27-28=-1\text{ °C}\). The average rate is \(\frac{27-28}{5-4}=-1\text{ °C/min}\). This differs from the instantaneous rate at minute 5 because the rate varies during that minute.
Zero and Small Rates
A zero derivative has a precise but limited meaning: the instantaneous temperature-change rate is zero at that time. It does not, by itself, mean the temperature model is constant, or that the temperature has not changed at nearby times. In the quadratic example, the rate is zero at minute 4 even though the model is warming just before then and cooling just after.
A small derivative magnitude indicates slow instantaneous change. For example, \(\lvert T'(t)\rvert=0.2\text{ °C/min}\) describes a slower change than \(\lvert T'(t)\rvert=2\text{ °C/min}\). To find times when the rate is zero, solve \(T'(t)=0\); to find times when the temperature is changing slowly, look for times when \(\lvert T'(t)\rvert\) is small. These conditions are not the same: a small nonzero rate still describes ongoing warming or cooling.
Worked Example: Cooling More Slowly Over Time
Worked Example: Cooling More Slowly Over Time
Suppose an object’s temperature is modeled by \(T(t)=22+18e^{-0.2t}\), in degrees Celsius, for \(t\ge0\), where \(t\) is in minutes. The associated derivative is \(T'(t)=-3.6e^{-0.2t}\), in degrees Celsius per minute.
At \(t=5\), using \(e^{-1}\approx0.367879\),
Rounded to three significant figures, the temperature is \(28.6\text{ °C}\). The rate is
The negative sign indicates cooling, and the magnitude indicates a rate of about \(1.32\text{ °C/min}\) at minute 5.
At \(t=10\), using \(e^{-2}\approx0.135335\),
The object is still cooling at minute 10, but the rate’s magnitude is smaller than at minute 5. The model therefore describes slower cooling at minute 10. As a check on the model’s long-term temperature level, the term \(18e^{-0.2t}\) approaches zero as \(t\) increases, so \(T(t)\) approaches \(22\text{ °C}\); the derivative likewise approaches zero from below.
Common Pitfall: Confusing Temperature With Temperature Rate
A positive temperature does not imply warming, and a negative derivative does not mean the temperature itself is below zero. For example, in the cooling model at minute 5, the temperature is about \(28.6\text{ °C}\), while the derivative is about \(-1.32\text{ °C/min}\). The first value is the current modeled temperature; the second describes how it is changing.
A mistaken interpretation would be: “Since \(T'(5)<0\), the temperature is below zero.” That conclusion confuses the rate with the temperature. The correction is to read the two quantities separately: \(T(5)\approx28.6\text{ °C}\), so the modeled temperature is above zero; \(T'(5)\approx-1.32\text{ °C/min}\), so it is cooling.
Using a Temperature Derivative Carefully
When interpreting a model, state the time and units along with the rate. “The object is cooling” gives a direction, but “at minute 5, it is cooling at approximately \(1.32\text{ °C/min}\)” gives both the direction and pace at a specified time. If the derivative varies, avoid describing one instantaneous value as the rate for a whole interval.
Also distinguish an instantaneous rate from an average rate. The derivative concerns behaviour at one time. The average rate between times \(a\) and \(b\), with \(a\ne b\), is \(\frac{T(b)-T(a)}{b-a}\). It summarizes the net temperature change over the interval and can differ from the derivative at either endpoint.
Check Your Understanding
For each question, include units in your interpretation whenever the information allows.
- A temperature model has \(T'(3)=2.5\text{ °C/min}\). Is the object warming or cooling, and at what rate?
- At one time, a model has derivative \(-0.4\text{ °C/min}\); at another, it has derivative \(-1.6\text{ °C/min}\). At which time is the cooling faster?
- What does \(T'(t)=0\) tell you, and what does it not tell you on its own?
- Explain the difference between an instantaneous rate and an average rate over a time interval.
- A model gives \(T(5)=28.6\text{ °C}\) and \(T'(5)=-1.32\text{ °C/min}\). State what each value describes.
