Modified exponential growth

The modified exponential model dN/dt = k(N − P) and its solution N = P + Ae^{kt}, Newton's law of cooling, finding constants from data, limiting values, and solving for times using logarithms.

The model

$$\frac{dN}{dt} = k(N - P) \quad\Longleftrightarrow\quad N = P + Ae^{kt}$$

The rate of change is proportional to the difference between $N$ and a fixed value $P$. If $k \lt 0$, $N$ approaches $P$ as $t \to \infty$. Check by differentiating: $\dfrac{dN}{dt} = kAe^{kt} = k(N - P)$.

Newton's law of cooling is the main example: an object's temperature changes at a rate proportional to the difference between its temperature and the surrounding temperature.

Cooling towards room temperature

Tea starts at 90°. The dashed line is the room temperature A, which the curve approaches but never reaches.

Try this

What does a larger $k$ mean physically?

A cooling problem

Example 1●●●

Tea at $90^\circ$C is left in a room at $20^\circ$C. After $5$ minutes it is $60^\circ$C. Find its temperature after $10$ minutes, and when it reaches $30^\circ$C.

$T = 20 + Ae^{-kt}$, and $t = 0$ gives $A = 70$.

$60 = 20 + 70e^{-5k}$, so $e^{-5k} = \tfrac47$ and $k = \tfrac15\ln\tfrac74 \approx 0.112$.

$T(10) = 20 + 70\big(\tfrac47\big)^2 = 20 + \tfrac{160}{7} \approx 42.9^\circ$C.

$30 = 20 + 70e^{-kt}$ gives $e^{-kt} = \tfrac17$, so $t = \dfrac{\ln 7}{k} = \dfrac{5\ln 7}{\ln\frac74} \approx 17.4$ minutes.

Flashcards

$dfrac{dN}{dt} = k(N - P)$ has solution

$N = P + Ae^{kt}$

Newton's law of cooling

The rate of change of temperature is proportional to the difference from the surrounding temperature.

Limiting value when $k lt 0$

$N \to P$

Practice

In Example 1, find $A$.
In Example 1, find the temperature after $10$ minutes, correct to two decimal places.
In Example 1, when does the tea reach $30^\circ$C? Answer in minutes, correct to two decimal places.
In Example 1, what temperature does the tea approach in the long run?
$T = 20 + 80e^{-kt}$ and $e^{-5k} = \tfrac12$. Find $T$ when $t = 10$.
$\dfrac{dT}{dt} = -0.5(T - 20)$. Find $\dfrac{dT}{dt}$ when $T = 50$.
Which satisfies $\dfrac{dN}{dt} = 3(N - 10)$?
In $T = 20 + 70e^{-kt}$, the number $70$ is: