Source: MasterMathMentor Ch. 41–42, Advanced Calculus I (UCF)
Tags: exponential growth, exponential decay, dP/dt = kP, half-life, doubling time, Newton's law of cooling, bounded growth, proportional to current value, growth constant, decay constant, differential equation applications
Difficulty: Intermediate to Advanced Prerequisites: Separable differential equations (Chapter 40), natural logarithm and exponential properties, comfort with solving for unknowns in exponential equations. You must be able to separate and integrate dy/dx = ky fluently before starting.
Chapters 41 and 42 are where differential equations meet the real world. The central idea is deceptively simple: if the rate of change of something is proportional to its current value, you get exponential growth or decay. That one sentence, translated into the DEQ dP/dt = kP, governs bacteria populations, radioactive decay, compound interest, air pressure at altitude, and much more. Chapter 42 then extends the idea: small changes in wording (proportional to "the distance from a limit," proportional to "the square root") produce different DEQs with different solution shapes, including bounded growth (Newton's cooling) and square-root decay (draining tanks). Recognising which model fits the wording is the skill being tested.
"Rate of change proportional to current value" translates to dP/dt = kP, solved by P = P_0 * e^(kt). Positive k means growth, negative k means decay. Variations on the wording (proportional to P - n, proportional to sqrt(P), inversely proportional to P) give different DEQs and solution forms. Every problem boils down to: identify the DEQ, solve it, then use given data to find the constants.
Exponential growth
A process where a quantity increases at a rate proportional to its current size. The larger it gets, the faster it grows. Modelled by P = P_0 * e^(kt) with k > 0.
Exponential decay
A process where a quantity decreases at a rate proportional to its current size. The smaller it gets, the slower it shrinks. Modelled by P = P_0 * e^(kt) with k < 0.
Growth constant (k)
The proportionality constant in dP/dt = kP. When k > 0 the quantity grows; when k < 0 it decays. Its units are "per time" (e.g. per hour, per year). Think of it as a percentage growth rate expressed in continuous terms.
Initial value (P_0)
The value of the quantity at time t = 0. In the solution P = P_0 * e^(kt), this is the starting point. Sometimes written as C.
Half-life
The time it takes for a decaying quantity to reduce to half its current value. For P = P_0 * e^(kt), set P = P_0/2 and solve: half-life = ln(2)/|k|. This is constant regardless of the starting amount.
Doubling time
The time it takes for a growing quantity to double. Doubling time = ln(2)/k. Like half-life, it does not depend on the starting amount.
Newton's Law of Cooling
The rate of temperature change of an object is proportional to the difference between the object's temperature and the ambient (surrounding) temperature. DEQ: dR/dt = k(R - T_ambient). The solution is R(t) = T_ambient + (R_0 - T_ambient) * e^(kt), where k < 0.
Bounded growth (limited growth)
A model where a quantity approaches a carrying capacity or limit L. DEQ: dP/dt = k(L - P). Solution: P(t) = L - (L - P_0) * e^(kt) with k < 0, which rises from P_0 toward L but never exceeds it.
The derivation you should memorise:
Starting from dP/dt = kP:
Separate: dP/P = k dt.
Integrate: ln|P| = kt + C.
Exponentiate: P = e^(kt + C) = e^C * e^(kt).
Rename e^C as P_0 (the value when t = 0).
Result: P = P_0 * e^(kt).
Solving a problem, step by step:
Translate the word problem into dP/dt = kP.
Identify P_0 from the initial condition.
Use a second data point (a known value at a known time) to solve for k.
Answer whatever the question asks (value at a future time, time to reach a target value, etc.).
Finding k from two data points:
If P(0) = P_0 and P(t_1) = P_1, then P_1 = P_0 e^(k t_1), so k = ln(P_1/P_0) / t_1.
Finding k from a half-life:
P_0/2 = P_0 e^(k t_half), so k = -ln(2)/t_half. Note k is negative for decay.
Finding k from a rate of change at a specific time:
If at t = 0, dP/dt = r and P = P_0, then k = r/P_0. This is used in the tire-pressure problem: dP/dt = -0.36 and P = 45, so k = -0.36/45 = -0.008.
Changing the wording changes the DEQ. Here are the four main variants:
a) dP/dt = k(P - n) (proportional to the distance above n)
Separate: dP/(P - n) = k dt.
Solution: P - n = Ce^(kt), or P = n + Ce^(kt).
Behaviour: if k > 0, P moves away from n exponentially. If k < 0, P approaches n.
b) dP/dt = k(n - P) (proportional to the distance below n)
This is the bounded growth model.
Separate: dP/(n - P) = k dt. Integrate: -ln|n - P| = kt + C.
Solution: P = n - Ce^(-kt) (rewritten with the constant absorbed).
Behaviour: P approaches n from below as t increases (when the quantity starts below n and k > 0).
Used for fish populations approaching carrying capacity, students adopting a new lunch menu, Newton's cooling.
c) dP/dt = k * sqrt(P) (proportional to the square root)
Separate: dP/sqrt(P) = k dt, i.e. P^(-1/2) dP = k dt.
Integrate: 2 * sqrt(P) = kt + C.
Solution: sqrt(P) = (kt + C)/2, so P = ((kt + C)/2)^2.
Behaviour: not exponential. The quantity can reach zero in finite time (the draining-tank scenario).
d) dP/dt = k/P (inversely proportional to P)
Separate: P dP = k dt.
Integrate: P^2/2 = kt + C.
Solution: P = sqrt(2kt + C).
DEQ: dR/dt = k(R - T_a) where T_a is ambient temperature.
Solution: R(t) = T_a + (R_0 - T_a) * e^(kt), with k < 0 since the object is cooling.
To find k, use a known temperature at a known time after removal from the heat source.
The object's temperature approaches T_a asymptotically; it never quite reaches it.
DEQ: dF/dt = k(L - F), where L is the maximum population.
Solution: F(t) = L - (L - F_0) * e^(-kt) (with k > 0 in this form).
As t approaches infinity, F(t) approaches L.
Two data points are needed: one for F_0 and one for k.
Standard exponential model: dP/dt = kP, solution P = P_0 * e^(kt)
Half-life to k: k = -ln(2) / t_half
Doubling time to k: k = ln(2) / t_double
Newton's cooling: R(t) = T_ambient + (R_0 - T_ambient) * e^(kt), k < 0
Bounded growth: P(t) = L - (L - P_0) * e^(-kt), k > 0
Square-root proportionality: 2 * sqrt(P) = kt + C
3 million bacteria at t = 0, doubled to 6 million at t = 3 hours. P_0 = 3,000,000. At t = 3: 6,000,000 = 3,000,000 e^(3k). e^(3k) = 2, so k = ln(2)/3 ≈ 0.2310/hr. Population after 24 hours: P(24) = 3,000,000 e^(24 ln(2)/3) = 3,000,000 2^8 = 768,000,000. Time to grow from 1 billion to 2 billion: always ln(2)/k = 3 hours, regardless of starting amount.
Half-life = 24,360 years. k = -ln(2)/24,360. 8 grams decaying to 1 gram: 1 = 8 e^(kt), so e^(kt) = 1/8, kt = ln(1/8) = -3ln(2). t = -3ln(2)/k = 3 24,360 = 73,080 years.
Roast at 160 degrees removed to a 75-degree room. After 30 min, internal temp is 120 degrees. R(t) = 75 + 85 e^(kt). At t = 30: 120 = 75 + 85 e^(30k). 45 = 85 e^(30k), e^(30k) = 9/17, k = ln(9/17)/30 ≈ -0.02107. Temperature after 45 min: R(45) = 75 + 85 e^(45k).
Height h = 20 cm at t = 0, dh/dt = k sqrt(h), and dh/dt = -5 at t = 0. So -5 = k sqrt(20), giving k = -5/sqrt(20) = -sqrt(5)/2. Separate: h^(-1/2) dh = k dt. Integrate: 2 sqrt(h) = kt + C. At t = 0: 2 sqrt(20) = C, so C = 4 sqrt(5). 2 sqrt(h) = (-sqrt(5)/2) t + 4 sqrt(5). Can empties when h = 0: 0 = (-sqrt(5)/2) t + 4 sqrt(5), so t = 8 seconds.
Exponential growth models compound interest, population biology, and the early stages of disease spread. Exponential decay models radioactive isotopes (carbon dating, nuclear waste), drug metabolism (ibuprofen leaving the bloodstream), and depreciation of assets. Newton's cooling is used by forensic scientists to estimate time of death and by engineers designing cooling systems. Bounded growth models market saturation, technology adoption, and wildlife management (fish stocking in lakes).
Students confuse the decay constant k with the percentage decrease. If a quantity drops 20% per year, it does not mean k = -0.20. The relationship is e^k = 0.80, so k = ln(0.80) ≈ -0.2231. The continuous rate and the discrete percentage are different numbers.
When a problem says "proportional to the difference between P and some limit," students sometimes write dP/dt = kP instead of dP/dt = k(P - n) or dP/dt = k(n - P). The exact wording matters enormously.
Students sometimes assume the half-life depends on the starting amount. It does not. Half-life is a property of the decay constant alone.
In Newton's cooling problems, students forget that the temperature approaches the ambient temperature, not zero. If the room is 75 degrees, the roast will never cool below 75.
⚠️ "Directly proportional to its current value" is the trigger phrase for dP/dt = kP. Memorise this translation.
⚠️ You will be given two conditions (either two data points or P_0 plus a rate) and asked to find k. This is mechanical but must be done carefully.
⚠️ Half-life and doubling-time problems appear frequently. Know the shortcut: t_half = ln(2)/|k|.
⚠️ Newton's cooling and bounded growth problems require you to identify the ambient value or carrying capacity first, then set up the correct DEQ. The wrong DEQ gives the wrong solution shape entirely.
⚠️ AP free-response questions love asking you to "write and solve the differential equation." You must show the separation, integration, and application of the initial condition to earn full marks.
⚠️ Ibuprofen and drug-metabolism problems combine exponential decay with piecewise reasoning (a new dose resets the amount). Work each interval separately.
Fill in the blank: if dP/dt = kP and k > 0, the quantity is ______ exponentially.
True or false: the half-life of a substance depends on how much you start with.
The general solution to dP/dt = kP is P = ______.
True or false: in Newton's Law of Cooling, the object's temperature eventually drops to zero.
Fill in the blank: if a fish population approaches a maximum of 900 and the rate of change is proportional to (900 - F), the DEQ is dF/dt = ______.
Answers: 1) Growing. 2) False. 3) P_0 * e^(kt) (or Ce^(kt)). 4) False (it approaches the ambient temperature). 5) k(900 - F).
Q: A car's trade-in value is $9,500 and drops to $9,000 in 6 months, with the rate of depreciation proportional to its current value. Write and solve the DEQ.
A: dV/dt = kV, so V = 9500 e^(kt). At t = 6 months (t = 0.5 years): 9000 = 9500 e^(0.5k). Solve: k = ln(9000/9500)/0.5 = 2 * ln(18/19) ≈ -0.1082/year (or use months: k = ln(18/19)/6 per month).
Q: Carbon-14 has a half-life of 5,750 years. What is the decay constant k?
A: k = -ln(2)/5750 ≈ -0.00012054 per year.
Q: A pork roast at 160 degrees is placed in a 75-degree room. Write the general form of the cooling equation.
A: R(t) = 75 + (160 - 75) e^(kt) = 75 + 85 e^(kt), where k < 0 and t is measured in minutes.
Q: Fish are introduced to a lake with carrying capacity 900. Initially there are 400 fish, and after 3 years there are 600. Find k in the model F(t) = 900 - 500 * e^(-kt).
A: At t = 3: 600 = 900 - 500 e^(-3k). So 500 e^(-3k) = 300, e^(-3k) = 3/5, -3k = ln(3/5), k = -ln(3/5)/3 = ln(5/3)/3 ≈ 0.1703/year.
Q: A water tank drains at a rate proportional to the square root of the remaining volume. Is this exponential decay?
A: No. The DEQ is dV/dt = k * sqrt(V), which is not dV/dt = kV. The solution involves a squared term, not an exponential. Notably, the tank can empty completely in finite time, unlike exponential decay which only approaches zero asymptotically.
These models build directly on the separable DEQ techniques from Chapter 40. Exponential growth connects to continuously compounded interest in finance and to logistic growth models (covered in more advanced courses), where a carrying capacity constrains the exponential. Newton's cooling connects to the study of heat transfer in physics and engineering. The bounded-growth model is a simplified version of the logistic equation dP/dt = kP(L - P), which you may encounter in a college differential equations course.
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