The differential equation whose rate is proportional to the amount, its solution by separating variables, doubling times and half-lives, Newton's law of cooling applied to the temperature difference, and the limits of the model.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 6 — Applications of Integration
Exponential Growth and Decay
Objectives
Five outcomes. One equation, one solution, and a great deal of care about when it applies.
OpenStax Calculus Volume 1, §6.8 Exponential Growth and Decay §6.8, pp. 636-645 — the section these objectives are drawn from
Warm-up
Section 6.7 established the exponential securely, and Section 5.6 noticed it is the only function that is its own derivative.
Discussion prompt
What function has a rate of change proportional to its own size?
Hint: Try the exponential and see what its derivative gives.
Answer:
Differentiating the exponential of k times t brings down a factor of k and leaves the function unchanged — so its rate of change is k times itself, which is exactly proportional to its size.
\[ y = Ae^{kt} \;\Longrightarrow\; \frac{dy}{dt} = kAe^{kt} = ky \]
That single property makes the exponential the solution of the most common differential equation in the applied sciences. Populations, radioactive decay, compound interest, drug elimination and cooling all reduce to it, and this section is about solving it and knowing when it applies.
Concept
When a quantity's rate of change is proportional to how much of it there is, the quantity is exponential. The constant of proportionality decides whether it grows or decays and how fast.
the exponential model — The differential equation stating that a quantity's derivative is a constant multiple of itself. Its solutions are the exponential functions, with the constant appearing in the exponent.
\[ \frac{dy}{dt}=ky \;\Longrightarrow\; y=y_{0}e^{kt} \]
The model is ubiquitous because the proportionality assumption is so often nearly true: twice as many bacteria divide twice as fast, twice as many atoms decay twice as often, twice the money earns twice the interest.
Figure (svg): The defining property: the rate of change is proportional to the amount
OpenStax Calculus Volume 1, §6.8 Exponential Growth and Decay §6.8, pp. 762-772
Section
Section 1
Concept
The equation is solved by collecting the quantity on one side and the time on the other, integrating both, and exponentiating. The logarithm appears because the reciprocal is integrated.
separation of variables — Rearranging a differential equation so that each variable appears with its own differential, then integrating both sides. It is the standard first technique for such equations.
\[ \frac{dy}{y}=k\,dt \;\Longrightarrow\; \ln|y|=kt+C \]
The constant of integration becomes a multiplicative constant after exponentiating, and an initial condition identifies it as the starting amount.
Figure (svg): Solving the equation by separating and integrating
OpenStax Calculus Volume 1, §6.8 Exponential Growth and Decay §6.8, pp. 762-770 — exponential growth model
Picture it
Separate, integrate, exponentiate.
Figure (svg): Solving the equation by separating and integrating
The logarithm in the third line is not a coincidence: integrating the reciprocal is the one case Section 4.10's power rule excluded, and this is where that exception earns its keep.
Worked example
Example 6.48. Separation of variables.
\[ \text{Solve } \frac{dy}{dt}=ky \text{ with } y(0)=y_{0}. \]
Separate the variables
Why: Divide by y, multiply by dt.
\[ \,dy / y = k \,dt \]
Integrate both sides
Why: The reciprocal and a constant.
\[ \ln | y | = k t + C \]
Exponentiate
Why: Undo the logarithm.
\[ | y | = e ^{C} e ^{k t} \]
Absorb the constant
Why: Call it A.
\[ y = A e ^{k t} \]
Impose the condition
Why: Substitute t = 0.
\[ A = y _{0} \]
Figure (svg): Solving the equation by separating and integrating
\[ y(t)=y_{0}e^{kt} \]
Verify: check the solution satisfies both requirements
Why: Differentiating gives k times the initial amount times the exponential, which is k times y — so the differential equation holds. Substituting t equal to zero gives the initial amount, so the condition holds too. Both must be checked, exactly as in Section 4.10's initial-value problems, and they fail in different ways: an antidifferentiation slip breaks the first and an arithmetic slip breaks the second.
OpenStax Calculus Volume 1, §6.8 Exponential Growth and Decay §6.8, pp. 765-767
Fill the middle
The exponential model, rearranged.
Fill in the blanks
\fracy___=ky \;\Longrightarrow\; \frac______}=k\,dt
Why: Dividing by y puts each variable with its own differential, which is what makes both sides integrable. Integrating the left gives a logarithm, since it is the reciprocal.
Worked example
Checkpoint 6.48. Two measurements fix everything.
\[ \text{A culture of } 100 \text{ grows to } 340 \text{ in } 3 \text{ hours. Find } k. \]
Write the model
Why: With the initial amount known.
\[ y = 100 e ^{k t} \]
Impose the second measurement
Why: At t = 3.
\[ 340 = 100 e ^{3 k} \]
Divide
Why: The initial amount cancels.
\[ 3.4 = e ^{3 k} \]
Take logarithms
Why: To bring k down.
\[ \ln 3.4 = 3 k \]
Solve
Why: Divide.
\[ k\text{ about } 0.408\text{ per hour} \]
Figure (svg): The solution to Worked example finding the constant from data shown as a ladder of expressions, one row per legal move
\[ k = \frac{\ln 3.4}{3} \approx 0.408 \]
Verify: check the fitted model reproduces the data
Why: Substituting back gives 100 times the exponential of 0.408 times 3, which is 100 times 3.4, exactly 340 — as it must, since k was chosen to make it so. The genuine test is a third measurement: if the model predicts the count at 6 hours as 1156 and the culture actually reaches that, the exponential assumption is supported; if the culture falls short, the model is already breaking down.
OpenStax Calculus Volume 1, §6.8 Exponential Growth and Decay §6.8, pp. 767-769
Trap
\[ y = e^{kt}+C \]
Add the constant after exponentiating
Why: The student places it as in an ordinary antiderivative.
The constant was added before exponentiating, and exponentiating a sum produces a product. It becomes a multiplicative factor.
\[ y = e^{kt+C} = e^{C}e^{kt} = Ae^{kt} \]
Exponentiate the whole right side, splitting the product
Why: The constant of integration multiplies, it does not add.
Checking the proposed solution against the equation catches it: differentiating the wrong form gives k times the exponential, which is k times y minus C rather than k times y.
Ranking
Solving the exponential model.
Put in order
Why: Step d is where the constant changes character, from additive to multiplicative, because it was added inside an exponent. Step e then identifies it as the starting amount, which is the interpretation that makes the solution usable.
Two truths and a lie
All three are about the solution.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Differentiating it gives k times the exponential, which is k times the quantity minus the constant — not k times the quantity. Only the multiplicative form satisfies the equation, and checking is what reveals it.
Prediction
Commit before reasoning.
Predict first
Why does solving this equation produce a logarithm?
Correct: Because the reciprocal is integrated.
\[ \int\frac{dy}{y} = \ln|y| \quad \text{(the power rule's exception)} \]
Why: Dividing by y leaves dy over y on the left, which is exactly the integrand Section 4.10 excluded from the power rule and Section 6.7 defined the logarithm by. So the logarithm's appearance is traceable to a specific earlier result rather than to a vague affinity between the two functions — and it is why the exponential emerges when the logarithm is undone.
Section
Section 2
Concept
The time to multiply by a given factor depends only on the rate constant, because the initial amount cancels when a ratio is imposed. That is why half-lives are tabulated.
doubling time and half-life — The time for a quantity to double or halve. Both are determined by the rate constant alone and are independent of how much is present.
\[ T_{2} = \frac{\ln 2}{k}, \qquad T_{1/2} = \frac{\ln 2}{|k|} \]
The independence is the model's most distinctive prediction and its easiest test: measuring the doubling time at two different population sizes and getting the same answer is evidence the model applies.
Figure (svg): Doubling time and half-life: fixed intervals, independent of the amount
OpenStax Calculus Volume 1, §6.8 Exponential Growth and Decay §6.8, pp. 766-776 — doubling time and half-life
Picture it
A quantity doubling repeatedly.
Figure (svg): Doubling time and half-life: fixed intervals, independent of the amount
The gaps between the dashed lines are equal in time and the heights multiply by the same factor across each. That is a constant ratio rather than a constant difference, which is what distinguishes exponential from linear.
Worked example
Example 6.50. The initial amount cancels.
\[ \text{A substance decays with } k=-0.139 \text{ per hour. Find its half-life.} \]
Write the model
Why: With a general initial amount.
\[ y = y _{0} e ^{-0.139 t} \]
Impose the halving
Why: Half remains.
\[ y _{0} / 2 = y _{0} e ^{-0.139 t} \]
Cancel the initial amount
Why: It divides out.
\[ \frac{1}{2} = e ^{-0.139 t} \]
Take logarithms
Why: To bring t down.
\[ -\ln 2 = -0.139 t \]
Solve
Why: Divide.
\[ \text{about } 5\text{ hours} \]
Figure (svg): Doubling time and half-life: fixed intervals, independent of the amount
\[ T_{1/2} = \frac{\ln 2}{0.139} \approx 5 \]
Verify: check by tracking a sample
Why: Starting with 100 units, after 5 hours the model gives 100 times the exponential of negative 0.695, about 49.9 — half, as required. Starting with 8000 units instead gives about 3990 after the same 5 hours, again half. The independence from the starting amount is not an approximation but exact, and it is what makes a single tabulated half-life useful across every sample of a substance.
OpenStax Calculus Volume 1, §6.8 Exponential Growth and Decay §6.8, pp. 769-771
Fill the middle
A decay constant given.
Fill in the blanks
T_2 = \frac___}}___
Why: The logarithm of the factor divided by the rate constant gives the time. For tripling the 2 becomes a 3, and the structure is otherwise identical.
Worked example
Checkpoint 6.50. The general version.
\[ \text{How long for a quantity growing at } k \text{ to triple?} \]
Write the ratio condition
Why: Three times the start.
\[ 3 y _{0} = y _{0} e ^{k t} \]
Cancel
Why: The initial amount divides out.
\[ 3 = e ^{k t} \]
Take logarithms
Why: Bring t down.
\[ \ln 3 = k t \]
Solve
Why: Divide.
\[ t = \ln(3) / k \]
Compare with doubling
Why: The ratio of logarithms.
\[ \text{about } 1.58 \times\text{ as long} \]
Figure (svg): The solution to Worked example any factor, not just two shown as a ladder of expressions, one row per legal move
\[ t = \frac{\ln 3}{k} \]
Verify: check the tripling time against successive doublings
Why: Two doublings quadruple, so tripling should take between one and two doubling times — and 1.58 sits inside. More precisely, the time to multiply by any factor is proportional to that factor's logarithm, so multiplying by 4 takes exactly twice as long as doubling. That logarithmic relationship between factor and time is the model's characteristic signature.
OpenStax Calculus Volume 1, §6.8 Exponential Growth and Decay §6.8, pp. 771-773
Error analysis
A student reasons about two half-lives.
Annotate
On: \( \text{half-life } 5\text{ h} \;\Longrightarrow\; \text{gone after } 10\text{ h} \)
Exponential decay has a constant ratio, not a constant difference. Two half-lives leave a quarter, three an eighth, and no finite number of them empties the sample.
Sorting
Each half-life halves what is left.
Sort into buckets
Sort each elapsed time by the fraction remaining, for a 5-hour half-life.
After 100 hours, twenty half-lives have passed and about one millionth remains — small but not zero. The model never reaches zero, which is both its mathematical character and one of its physical limitations.
Two truths and a lie
All three are about characteristic times.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Each half-life halves what remains, so two leave a quarter and no finite number empties the sample. Constant ratio rather than constant difference is the defining feature of exponential change.
Prediction
Commit before reasoning.
Predict first
Why can one half-life be quoted for a substance regardless of the sample size?
Correct: Because the initial amount cancels.
\[ \frac{y_{0}}{2}=y_{0}e^{kt} \;\Longrightarrow\; \tfrac12=e^{kt} \]
Why: Setting half the initial amount equal to the initial amount times the exponential lets the starting quantity divide out of both sides, leaving an equation in the rate constant and the time alone. If the question asked instead how long until a fixed number of grams remained, the initial amount would not cancel and the answer would depend on the sample. The independence belongs to ratio questions specifically.
Section
Section 3
Concept
An object cools at a rate proportional to how far its temperature exceeds its surroundings. It is the difference that satisfies the exponential model, not the temperature itself.
Newton's law of cooling — The rate of temperature change is proportional to the difference between the object's temperature and the ambient temperature. The difference decays exponentially toward zero.
\[ \frac{dT}{dt}=k(T-T_{a}) \;\Longrightarrow\; T=T_{a}+(T_{0}-T_{a})e^{kt} \]
Applying the model to the temperature directly predicts cooling toward absolute zero, which is the standard error here and is immediately visible as absurd.
Figure (svg): Newton's law of cooling: the DIFFERENCE from ambient decays exponentially
OpenStax Calculus Volume 1, §6.8 Exponential Growth and Decay §6.8, pp. 772-780 — Newton's law of cooling
Picture it
A cooling object.
Figure (svg): Newton's law of cooling: the DIFFERENCE from ambient decays exponentially
The curve flattens toward the ambient line without meeting it. Subtracting the ambient temperature is what turns this into the standard exponential model.
Worked example
Example 6.52. Work with the difference.
\[ \text{Coffee at } 90 \text{ degrees in a } 20 \text{ degree room cools to } 60 \text{ in } 10 \text{ minutes. Find when it reaches } 40. \]
Work with the difference
Why: Subtract ambient.
\[ \text{starts at } 70 ^\circ\text{ above} \]
Write the model for the difference
Why: It decays exponentially.
\[ D = 70 e ^{k t} \]
Use the ten-minute measurement
Why: The difference is then 40.
\[ 40 = 70 e ^{10 k} \]
Solve for k
Why: Take logarithms.
\[ k\text{ about } -0.0560 \]
Solve for the difference of 20
Why: The target minus ambient.
\[ \text{about } 22.4\text{ minutes} \]
Figure (svg): Newton's law of cooling: the DIFFERENCE from ambient decays exponentially
\[ t = \frac{\ln(20/70)}{k} \approx 22.4 \]
Verify: check the answer is consistent with the cooling slowing
Why: The first 30 degrees of cooling took 10 minutes and the next 20 took a further 12.4 — slower per degree, which is right because the temperature difference driving the cooling has shrunk. Note also that reaching room temperature exactly would take infinitely long, so any question about cooling to 20 degrees has no finite answer under this model. Both observations are consequences of working with the difference.
OpenStax Calculus Volume 1, §6.8 Exponential Growth and Decay §6.8, pp. 775-777
Fill the middle
Newton's law of cooling, solved.
Fill in the blanks
T = T_T_a+(T____-___)e^___
Why: The quantity that decays exponentially is the difference from ambient, so the ambient temperature is subtracted at the start and added back at the end.
Worked example
Checkpoint 6.52. The error made visible.
\[ \text{What happens if the exponential model is applied to } T \text{ directly?} \]
Write the wrong model
Why: The temperature itself decaying.
\[ T = 90 e ^{k t} \]
Fit to the ten-minute data
Why: 60 degrees.
\[ k\text{ about } -0.0405 \]
Extrapolate to one hour
Why: Substitute.
\[ \text{about } 7.9 ^\circ \]
Extrapolate further
Why: Two hours.
\[ \text{about } 0.7 ^\circ \]
Note the absurdity
Why: It approaches zero.
Figure (svg): The solution to Worked example what the wrong model predicts shown as a ladder of expressions, one row per legal move
\[ T \to 0 \text{ rather than } T \to T_{a} \]
Verify: identify why the error is easy to miss early
Why: In the first ten minutes the wrong model fits the data exactly, because it was fitted to them — and even at twenty minutes it is only a few degrees off. The absurdity appears only on extrapolation, which is exactly when models are most trusted and least checked. Note also that the prediction depends on the temperature scale: in Fahrenheit the wrong model would approach zero degrees Fahrenheit instead, which is a different absurdity and a clear sign the model has no physical basis.
OpenStax Calculus Volume 1, §6.8 Exponential Growth and Decay §6.8, pp. 777-779
Trap
\[ T = T_{0}e^{kt} \]
Apply the exponential model to the temperature
Why: The student uses the standard form directly.
This predicts cooling toward zero on whatever scale is being used, which has no physical meaning.
\[ T = T_{a}+(T_{0}-T_{a})e^{kt} \]
Apply the model to the DIFFERENCE from ambient
Why: It is the difference that decays.
A scale test settles it: the correct model gives the same physical predictions in Celsius and Fahrenheit, and the wrong one does not — which shows the wrong one depends on an arbitrary choice of zero.
Sorting
Identify what satisfies the model.
Sort into buckets
Sort each quantity.
The pattern is that the model applies to a quantity measured from the value it is heading toward. For decay to zero that is the quantity itself; for cooling it is the difference, and getting the reference level right is the whole subtlety.
Two truths and a lie
All three are about cooling.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. The temperature approaches the ambient value, not zero, so it is the difference that decays. Applying the model to the temperature predicts cooling below room temperature and eventually toward absolute zero, which no object does.
Prediction
Commit before reasoning.
Predict first
Why does the exponential model apply to the temperature difference rather than the temperature?
Correct: Because the difference is what heads to zero.
\[ T-T_{a} \to 0, \quad \text{not } T \to 0 \]
Why: Heat flows because the object and its surroundings are at different temperatures, and the flow stops when they match — so the quantity approaching zero is the difference. The exponential model always describes a quantity decaying toward zero, so it must be applied to whatever quantity actually does. A hot object in a hot room does not cool at all, which the difference correctly predicts and the temperature alone does not.
Section
Section 4
Concept
The model gives the amount from the time directly; getting the time from the amount requires the logarithm, which is what inverts the exponential.
inverting the model — Solving the exponential equation for the time by dividing out the initial amount and taking logarithms. The initial amount cancels whenever a ratio is specified.
\[ t = \frac{1}{k}\ln\frac{y}{y_{0}} \]
The cancellation explains why doubling times and half-lives are amount-independent while questions about reaching a specific absolute amount are not.
Figure (svg): Solving for a time: the logarithm is what inverts the model
OpenStax Calculus Volume 1, §6.8 Exponential Growth and Decay §6.8, pp. 768-780 — solving exponential equations
Picture it
The model inverted.
Figure (svg): Solving for a time: the logarithm is what inverts the model
The division in the second line is what allows the initial amount to cancel for a ratio question. When the target is an absolute amount instead, it does not.
Worked example
Example 6.49. The initial amount cancels.
\[ \text{How long for a population growing at } 0.408 \text{ per hour to reach five times its size?} \]
Write the ratio condition
Why: Five times the start.
\[ 5 y _{0} = y _{0} e ^{0.408 t} \]
Cancel
Why: The initial amount divides out.
\[ 5 = e ^{0.408 t} \]
Take logarithms
Why: Bring t down.
\[ \ln 5 = 0.408 t \]
Solve
Why: Divide.
\[ \text{about } 3.94\text{ hours} \]
Note the independence
Why: No starting count was used.
Figure (svg): Solving for a time: the logarithm is what inverts the model
\[ t = \frac{\ln 5}{0.408} \approx 3.94 \]
Verify: cross-check against the doubling time
Why: The doubling time is the logarithm of 2 over 0.408, about 1.70 hours, and reaching five times should take between two and three doublings — between 3.40 and 5.10 hours. The answer of 3.94 sits inside, nearer the low end because five is nearer four than eight. Expressing one characteristic time in terms of another is a quick and reliable check.
OpenStax Calculus Volume 1, §6.8 Exponential Growth and Decay §6.8, pp. 767-769
Fill the middle
Solving the exponential equation for the time.
Fill in the blanks
t = \frac\ln___\,___\frac______}
Why: The logarithm inverts the exponential, so it is what brings the time down out of the exponent. Every question about how long something takes reduces to this.
Worked example
Checkpoint 6.49. The starting amount matters here.
\[ \text{How long for a culture of } 100 \text{ growing at } 0.408 \text{ to reach } 1000? \]
Write the condition
Why: A specific target.
\[ 1000 = 100 e ^{0.408 t} \]
Divide by the initial amount
Why: It does not cancel away.
\[ 10 = e ^{0.408 t} \]
Take logarithms
Why: Bring t down.
\[ \ln 10 = 0.408 t \]
Solve
Why: Divide.
\[ \text{about } 5.64\text{ hours} \]
Note the dependence
Why: A different start gives a different answer.
Figure (svg): The solution to Worked example an absolute question shown as a ladder of expressions, one row per legal move
\[ t = \frac{\ln 10}{0.408} \approx 5.64 \]
Verify: see how the answer shifts with the starting amount
Why: Starting from 200 instead, reaching 1000 is a factor of 5 rather than 10, taking about 3.94 hours — nearly two hours less. So absolute targets give amount-dependent answers while ratio targets do not, and the difference is exactly whether the initial amount cancels. Reading which kind of question is being asked is the first step, exactly as it was for net against total in Chapter 5.
OpenStax Calculus Volume 1, §6.8 Exponential Growth and Decay §6.8, pp. 769-771
Error analysis
A student solves for a time.
Annotate
On: \( 1000 = 100e^{0.408t} \;\Longrightarrow\; \ln 1000 - \ln 100 = \ln(1000-100) \)
The property proved in Section 6.7 was about products, and no corresponding property exists for sums or differences. Mistaking one for the other is the commonest logarithm error in any applied setting.
Sorting
Ratio, or absolute target?
Sort into buckets
Sort each question.
The distinction is visible in the wording: a factor or a fraction gives independence, a number of units does not. Spotting which kind of question has been asked determines whether the starting amount is needed data or not.
Two truths and a lie
All three are about solving for time.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Section 6.7 proved the property for products, and no analogue exists for sums or differences — the logarithm of 1000 minus 100 is not the logarithm of 1000 minus the logarithm of 100. Confusing them is the commonest error when solving these equations.
Prediction
Commit before reasoning.
Predict first
Why does solving for a time always involve a logarithm?
Correct: Because the time sits in an exponent.
\[ y_{0}e^{kt} = y \;\Longrightarrow\; kt = \ln\frac{y}{y_{0}} \]
Why: The model puts the time inside an exponential, so extracting it requires the exponential's inverse — which Section 6.7 established is the logarithm. Every question of the form how long until something reaches a given level therefore reduces to a logarithm, whatever the application. That structural fact is why logarithmic scales appear wherever exponential processes are studied.
Section
Section 5
Concept
Unbounded growth is never physically realisable. Real populations meet limits and the model must be replaced, most often by one that levels off at a carrying capacity.
the limits of the model — The exponential model assumes the rate stays proportional to the amount, which fails once resources, space or susceptible individuals become scarce. The logistic model is the standard replacement.
\[ \text{exponential: unbounded}; \quad \text{logistic: levels off} \]
The two models agree closely at small amounts, which is why fitting an exponential to early data and extrapolating is both easy and unreliable.
Figure (svg): Where the model breaks down
OpenStax Calculus Volume 1, §6.8 Exponential Growth and Decay §6.8, pp. 776-782 — limitations of the exponential model
Picture it
Two models on one plot.
Figure (svg): Where the model breaks down
The curves are nearly identical for the first stretch and then separate completely. Early data cannot distinguish them, which is exactly why extrapolation from it is so hazardous.
Worked example
Example 6.53. Following the model too far.
\[ \text{A bacterial culture doubles hourly. What does the model predict after } 3 \text{ days?} \]
Count the doublings
Why: Seventy-two hours.
\[ 72\text{ doublings} \]
Compute the factor
Why: Two to the seventy-second.
\[ \text{about } 4.7 \times 10 ^{21} \]
Estimate the mass
Why: At a picogram per cell.
\[ \text{about } 4.7\text{ million tonnes} \]
Compare with reality
Why: From one cell.
Identify the failure
Why: Resources run out long before.
Figure (svg): Where the model breaks down
\[ 2^{72} \approx 4.7\times 10^{21} \text{ cells} \]
Verify: identify when the model actually fails
Why: Real cultures follow the exponential closely for perhaps twenty to thirty doublings and then level off as nutrients are exhausted — well before any absurdity appears in the arithmetic. So the model's failure is gradual and begins long before the prediction becomes visibly ridiculous, which means the ridiculous case is a poor guide to when to stop trusting it. Only measurement establishes the range of validity.
OpenStax Calculus Volume 1, §6.8 Exponential Growth and Decay §6.8, pp. 779-781
Sorting
Is the rate still proportional to the amount?
Sort into buckets
Sort each situation.
Radioactive decay is the case where the model holds essentially exactly, because atoms decay independently and never run out of anything. Wherever units interact, the proportionality is only an approximation valid at small amounts.
Worked example
Checkpoint 6.53. The logistic model.
\[ \text{How is the model modified for a limited environment?} \]
Identify the failed assumption
Why: Rate proportional to amount.
Introduce a carrying capacity
Why: The environment's limit.
Modify the rate
Why: Multiply by the free fraction.
\[ k y(1 - \frac{y}{M}) \]
Check the small case
Why: When y is small.
Check the large case
Why: When y approaches M.
Figure (svg): The solution to Worked example what replaces it shown as a ladder of expressions, one row per legal move
\[ \frac{dy}{dt}=ky\left(1-\frac{y}{M}\right) \]
Verify: confirm the modification does what is wanted at both ends
Why: For a small population the bracketed factor is close to one and the model reduces to the exponential, which is right because crowding is irrelevant then. As the population approaches capacity the factor approaches zero and growth stops, which is the behaviour the exponential lacked. A good modification must agree with the old model where the old model worked, and this one does — which is the test any replacement should be held to.
OpenStax Calculus Volume 1, §6.8 Exponential Growth and Decay §6.8, pp. 781-782
Trap
\[ \text{fitted } k \text{ from 3 hours} \;\Longrightarrow\; \text{predict 3 days} \]
Extend the fit far past the measured range
Why: The student trusts the model outside its evidence.
The fit says nothing about behaviour twenty-four times beyond the data, and the exponential's unbounded growth guarantees eventual absurdity.
\[ \text{predict within the measured range; test before extending} \]
Treat the model's range of validity as an empirical question
Why: It cannot be settled from the fit alone.
Two models that agree on the data can diverge completely beyond it, as the logistic and exponential do. A good fit constrains behaviour only where there are measurements.
Fill the middle
The exponential model modified for a capacity M.
Fill in the blanks
\fracM___=ky\left(1-\frac______}\right)
Why: The bracketed factor is the fraction of capacity still free. It is near one for a small population, recovering the exponential, and near zero as capacity is reached.
Two truths and a lie
All three are about the model's limits.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Two models agreeing on the measured range can diverge completely beyond it, so a good fit constrains behaviour only where there are measurements. Establishing the range of validity is an empirical question that the fit itself cannot answer.
Prediction
Commit before reasoning.
Predict first
Why can early measurements not distinguish exponential from logistic growth?
Correct: Because the logistic reduces to the exponential when the amount is small.
\[ y \ll M \;\Longrightarrow\; 1-\frac{y}{M} \approx 1 \]
Why: The logistic's extra factor is the fraction of capacity still free, which is close to one while the population is small — so the two models are genuinely almost identical there, not merely hard to tell apart with noisy data. Distinguishing them requires measurements taken as the capacity is approached, which is exactly when they are hardest to obtain and most urgently wanted.
Comparison
Fill the blanks. One model, two signs.
Comparison matrix
| Growth | Decay | |
|---|---|---|
| The constant | positive | negative |
| Characteristic time | doubling time | half-life |
| Formula | ln(2)/k | ln(2)/|k| |
| Long-term behaviour | unbounded: physically impossible | approaches zero without reaching it |
Both long-term behaviours are suspect. Unbounded growth is never realised and a decaying quantity made of discrete units does eventually reach zero, which the continuous model cannot represent.
Pattern
Given a quantity changing at a rate proportional to its amount.
Steps one and five bracket the mathematics with modelling judgement. The middle three are mechanical; the outer two decide whether the answer means anything.
Stewart, Calculus: Early Transcendentals 8e, §3.8 Exponential Growth and Decay §3.8, pp. 237-244
Check
The solution.
Check your understanding
What solves dy/dt = ky with y(0) = y0?
Answer: A
Why: The constant of integration becomes a multiplicative factor, identified as the initial amount.
Check
Half-lives.
Check your understanding
After two half-lives, how much of a sample remains?
Answer: A
Why: Each half-life halves what remains, so two leave a quarter.
Check
Cooling.
Check your understanding
In Newton's law of cooling, which quantity decays exponentially?
Answer: A
Why: Heat flows because of the difference, and the flow stops when it vanishes.
Real world
Public health officials watching the first weeks of an outbreak fit an exponential curve to reported cases and must decide what to project and what to say publicly.
Discussion prompt
Explain why the exponential fits early, what the fitted constant means, and why long projections are unreliable.
Hint: Each infected person infects others at a roughly constant rate while almost everyone is susceptible.
Answer:
Early on, almost everyone is susceptible, so each infected person passes the infection on at a roughly constant rate — making new cases proportional to current cases, which is exactly the model. The fitted constant gives a doubling time, and that is the number officials actually use, because it is independent of the case count and therefore comparable between regions with different reporting.
\[ \frac{dI}{dt}=kI, \qquad T_{2}=\frac{\ln 2}{k} \]
Long projections are unreliable for a reason built into the model: it assumes the susceptible pool never depletes. Once a substantial fraction has been infected or vaccinated, the rate falls and the curve bends — the logistic behaviour of the last idea. An exponential extrapolated a few doubling times too far predicts more cases than there are people.
The public communication problem is the one the figure showed. Exponential and logistic curves are nearly identical in the early data, so the fit cannot distinguish them, and the difference between them a month out is enormous. Officials therefore quote doubling times, which the data does support, rather than case counts at a distant date, which it does not.
Note that behaviour change makes this harder still: the rate constant is not a constant of nature but a summary of contact patterns, which respond to the announcements themselves. A projection that changes behaviour invalidates its own assumptions, which is a difficulty the model has no way of representing.
Commit first
Answer, then rate your confidence honestly.
Predict first
Why is a half-life the same for every sample of a substance?
Correct: Because the initial amount cancels.
\[ \tfrac{y_{0}}{2}=y_{0}e^{kt} \;\Longrightarrow\; y_{0} \text{ cancels} \]
Why: Setting half the initial amount equal to the initial amount times the exponential divides the starting quantity out of both sides, leaving an equation in the rate constant and time alone. Had the question asked how long until a fixed mass remained, the initial amount would not cancel and the answer would depend on the sample. The independence belongs to ratio questions, and it is what makes a single tabulated half-life useful.
Explain it
They applied the exponential model to a cooling object's temperature and got it cooling below room temperature.
Discussion prompt
In four sentences or fewer, show them the fix.
Hint: Ask what the object is heading toward.
Answer:
Ask them where the object's temperature is heading: toward the room's temperature, not toward zero. The exponential model always describes something decaying toward zero, so it has to be applied to whatever quantity actually goes to zero — here the DIFFERENCE between the object and the room.
So subtract the room temperature at the start, apply the model to that difference, and add the room temperature back at the end. A quick test: their version gives different physical predictions in Celsius and Fahrenheit, which shows it cannot be right.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For solving, separate and remember the constant becomes multiplicative. For characteristic times, impose a ratio and watch the initial amount cancel. For cooling, apply the model to the difference from ambient. For judging, ask whether the units interact and whether the prediction is inside the data's range. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, write the differential equation and solve it in five lines, marking where the constant changes from additive to multiplicative and why. Below, sketch a growing and a decaying exponential side by side with the sign of the constant labelled, and mark equal time intervals showing the constant ratio on each. In the middle of the page, derive the half-life formula in four lines, circling the step where the initial amount cancels, and write one sentence on why absolute-amount questions differ. To the right, sketch a cooling curve approaching an ambient line and write the solution with the ambient temperature in both places it appears. In the lower half, sketch an exponential and a logistic curve on the same axes, marking where they agree and where they diverge. At the bottom, write what the fitted constant means and one sentence on the range within which it should be trusted.
If your cooling curve reaches the ambient line, redraw it approaching without touching — the difference decays exponentially and never reaches zero, which is why questions about reaching room temperature have no finite answer.
Recap
Five things, and the last is the one that decides whether the others mean anything.
| If you see | Then |
|---|---|
| A rate proportional to an amount | The solution is exponential |
| A constant of integration in an exponent | It becomes a multiplicative factor |
| A ratio target | The initial amount cancels |
| An absolute target | It does not |
| A cooling object | Model the difference from ambient |
| A 'how long until' question | It reduces to a logarithm |
| A long extrapolation | Check it lies within the data's support |
Section 6.9 closes the course with the hyperbolic functions, built from the exponential this chapter secured — and named for the curve their identity describes, just as the circular functions are named for theirs.
OpenStax Calculus Volume 1, §6.8 Exponential Growth and Decay §6.8, pp. 636-645 — everything on these slides traces back here
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