Exponential Models with Differential Equations: Study Guide (Calculus AB 7.8)

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.

A note on notation

This page has no equation renderer, so powers of e are written in a compact form:

e^(kt) means “e raised to the power kt”. y₀ (say “y nought”) is the value of y when t = 0. ln is the natural logarithm, so ln(e^a) = a and e^(ln a) = a for a > 0.

On paper, write e^(kt) with kt as a raised exponent.

From a sentence to a differential equation

Many quantities change at a rate that depends on how much of them there is. A large population has more parents, so it gains more new members per day. A large dose of a drug is cleared faster than a small one. The common sentence is:

A quantity changes at a rate proportional to its current size.

“Proportional to” means “equals a constant times”. So if y is the quantity and t is time, the sentence becomes

dy/dt = ky

Read each symbol in context:

• y is the amount (people, milligrams, cells, metres per second).
• t is time (hours, days, years).
• dy/dt is the rate of change of the amount, in units of y per unit of t.
• k is the constant of proportionality. Because ky must have the units of dy/dt, k has units of 1 per unit of time (for example, per hour).

The sign of k tells you the direction. If the amount is positive and k > 0, then dy/dt > 0, so y grows. If k < 0, y decays. Some questions write decay as dy/dt = −ky with k > 0. Both forms are correct; just be clear which one you are using.

Not every sentence about rates gives this equation. Translate carefully:

Sentence	Differential equation	Exponential model?
Rate of change of P is proportional to P	dP/dt = kP	Yes
Q decreases at a rate proportional to the amount present	dQ/dt = −kQ, k > 0	Yes (decay)
Rate of change of y is proportional to time	dy/dt = kt	No: y = (k/2)t² + C
Rate of change of y is proportional to the square of y	dy/dt = ky²	No
Rate of change of T is proportional to the difference between T and 20	dT/dt = k(T − 20)	Shifted version (see below)

Solving dy/dt = ky by separation of variables

You met this method in Topics 7.6 and 7.7. Suppose y > 0.

1. Separate: (1/y) dy = k dt.
2. Integrate both sides: ln|y| = kt + C.
3. Exponentiate: |y| = e^(kt + C) = e^C · e^(kt).
4. Tidy the constant: e^C is just a positive constant. Allowing for negative y as well, write y = A·e^(kt), where A is any constant. (A = 0 gives the constant solution y = 0, which also satisfies the equation.)
5. Use the initial condition y = y₀ when t = 0: y₀ = A·e⁰ = A.

So the particular solution is

y = y₀e^(kt)

Notice step 3. The constant C ends up multiplying e^(kt), not added to it. Writing y = e^(kt) + C is a very common error. It does not satisfy dy/dt = ky.

A domain note. Dividing by y in step 1 assumes y ≠ 0. That is safe here: if y₀ > 0, then y₀e^(kt) is positive for every t, so y never reaches 0. The solution is valid for all real t, although a model is only trusted over the times it was built for.

Confirm the solution before you use it

You can always check a proposed solution, without repeating the algebra.

Does it satisfy the differential equation? Differentiate y = y₀e^(kt) using the chain rule: dy/dt = y₀ · k e^(kt) = k(y₀e^(kt)) = ky. ✓

Does it satisfy the initial condition? At t = 0, y = y₀e⁰ = y₀. ✓

Then check that your numbers are reasonable:

• If the situation is growth, k must come out positive and later values must be larger. For decay, k is negative and the amount falls but stays positive.
• The units must match: k in “per hour” if t is in hours.
• Exponential decay never reaches exactly 0. If your model predicts a negative amount, something has gone wrong.

What y₀ and k mean

y₀ is the starting amount.

k is the relative rate of change. Divide both sides of dy/dt = ky by y:

(dy/dt)/y = k

So k is the rate of change as a fraction of the current amount. If k = 0.04 per year, the amount is changing at 4% of its current size per year, at every instant.

Two consequences matter in questions:

1. The rate is not constant. dy/dt = ky changes as y changes. When y doubles, the rate doubles. A model with dy/dt = ky is not “y increases by the same amount each hour”.

2. Equal time steps multiply y by the same factor. From y = y₀e^(kt), moving forward Δt units of time multiplies y by e^(kΔt), whatever the current value. This gives the two famous times:

• Doubling time (k > 0): solve e^(kT) = 2, so T = ln 2 / k.
• Half-life (k < 0): solve e^(kT) = 1/2, so T = ln 2 / |k| (equivalently k = −ln 2 / T).

Neither depends on y₀. A quantity with a half-life of 2 hours halves every 2 hours, whether you start with 64 mg or 6 mg.

Worked example 1: finding k from two measurements

Question. A laboratory yeast culture grows at a rate proportional to the number of cells present. At t = 0 hours there are 1200 cells; at t = 3 hours there are 2700 cells. (The data are invented.)

(a) Write and solve a differential equation for the number of cells N(t). (b) Estimate N(5). (c) Find dN/dt at t = 3 and interpret it. (d) When will there be 10 000 cells?

1. Translate. “Rate proportional to number present” gives dN/dt = kN, with N(0) = 1200. So N(t) = 1200e^(kt).
2. Use the second data point to find k. 2700 = 1200e^(3k), so e^(3k) = 2700/1200 = 9/4. Take ln of both sides: 3k = ln(9/4), so k = (1/3) ln(9/4) ≈ 0.270 per hour.
3. Check reasonableness. The culture grew, and k > 0. ✓
4. (b) Predict. N(5) = 1200e^(5k) ≈ 4636.093, so about 4636 cells.
5. (c) Rate. Use the differential equation directly: dN/dt = kN = 0.27031 × 2700 ≈ 729.837 cells per hour. At t = 3 hours the culture is gaining cells at about 730 cells per hour. This is an instantaneous rate, not a count for the next hour.
6. (d) Time. Solve 1200e^(kt) = 10 000: e^(kt) = 10 000/1200, so t = ln(10 000/1200)/k ≈ 7.844 hours.

Check. Substituting t = 7.844 back gives N ≈ 10 000. The doubling time is ln 2 / k ≈ 2.564 hours, so in about 7.8 hours the culture doubles about three times: 1200 → 2400 → 4800 → 9600. That is close to 10 000, so the answer is sensible.

Tip. Keep k stored to full accuracy in your calculator. Rounding k to 0.27 too early shifts later answers.

Worked example 2: a half-life without a calculator

Question. A tracer substance (invented for this example) leaves the body at a rate proportional to the amount present. Its half-life is 6 hours. A patient receives 40 mg at t = 0.

(a) Find the amount A(t) in milligrams. (b) Find A(15) exactly. (c) After how many hours is there 5 mg left?

1. Model. dA/dt = kA with A(0) = 40, so A(t) = 40e^(kt).
2. Find k from the half-life. A(6) = 20, so e^(6k) = 1/2. Then 6k = ln(1/2) = −ln 2, so k = −ln 2 / 6 ≈ −0.116 per hour. Negative, as decay requires. ✓
3. Rewrite. e^(kt) = e^(−(ln 2)t/6) = (e^(ln 2))^(−t/6) = 2^(−t/6). So A(t) = 40 · 2^(−t/6). This form shows the halving directly.
4. (b) A(15) = 40 · 2^(−15/6) = 40 · 2^(−5/2) = 40/(4√2) = 10/√2 = 5√2 ≈ 7.071 mg.
5. (c) 40 → 20 → 10 → 5 takes three half-lives, so t = 18 hours. Algebraically: 2^(−t/6) = 5/40 = 1/8 = 2^(−3), so t/6 = 3.

Check. 15 hours is between two half-lives (12 hours, 10 mg) and three (18 hours, 5 mg), and 7.071 mg is between 10 and 5. ✓

Worked example 3: motion along a line

Finding particular solutions also answers questions about motion. Here the velocity, not an amount, decays.

Question. A test cart on a straight track coasts after its motor switches off at t = 0. Its velocity v (m/s) satisfies dv/dt = −0.5v, with v(0) = 12. Its position is x(0) = 0 metres. (Invented data.)

(a) Find v(t). (b) Find the position x(t). (c) When is the velocity 3 m/s, and where is the cart then? (d) How far can the cart travel in total?

1. (a) This is dv/dt = kv with k = −0.5 per second. So v(t) = 12e^(−0.5t). Confirm: dv/dt = 12 × (−0.5)e^(−0.5t) = −0.5v ✓, and v(0) = 12 ✓.
2. Interpret the equation. dv/dt is acceleration. At t = 0 the acceleration is −0.5 × 12 = −6 m/s². The faster the cart moves, the harder it slows.
3. (b) Position is the accumulated velocity: x(t) = 0 + ∫ (0 to t) 12e^(−0.5s) ds = [−24e^(−0.5s)] from 0 to t = 24 − 24e^(−0.5t) metres.
4. (c) 12e^(−0.5t) = 3 gives e^(−0.5t) = 1/4, so t = 2 ln 4 = 4 ln 2 ≈ 2.773 seconds. Then x = 24 − 24 × (1/4) = 18 m.
5. (d) As t → ∞, e^(−0.5t) → 0, so x(t) → 24. The cart moves forward for all time but never passes 24 m.

Check. v stays positive, so x should always increase; it does. The acceleration when v = 3 is −0.5 × 3 = −1.5 m/s², a quarter of the starting value, just as the velocity is a quarter of its start. ✓

A related model: rate proportional to a difference

Sometimes the rate is proportional to the gap between y and a fixed value M: dy/dt = k(y − M). Let u = y − M. Then du/dt = dy/dt = ku, which is the exponential model again. So

y − M = (y₀ − M)e^(kt)

This is solved with the same separation of variables from Topic 7.6. Example (invented): a metal part at 90 °C cools in a 20 °C room with dT/dt = −0.1(T − 20), t in minutes. Then T = 20 + 70e^(−0.1t). The gap above room temperature halves from 70 to 35 when T = 55 °C, at t = 10 ln 2 ≈ 6.931 minutes. Note that T itself does not halve: only the gap behaves exponentially.

Common misconceptions

• “y = e^(kt) + C.” The constant from integrating multiplies e^(kt). Check: e^(kt) + C does not satisfy dy/dt = ky unless C = 0.
• “k is the amount gained each hour.” k is a relative rate (a fraction of the current amount per unit time). The amount gained per hour changes as y changes.
• “The rate is constant, so I can use rate × time.” In Worked example 1, the rate at t = 0 is about 324 cells per hour and at t = 3 about 730. Only the relative rate is constant.
• “k = 0.05 means 5% growth per year exactly.” Over one year, y is multiplied by e^0.05 ≈ 1.0513, a rise of about 5.13%. The 5% is an instantaneous rate.
• “Half-life depends on how much you start with.” It does not: T = ln 2 / |k|.
• Sign errors with decay. If you write dy/dt = −ky, then k > 0 and y = y₀e^(−kt). Do not put a second minus sign in.
• “Proportional to t” is the same as “proportional to y”. dy/dt = kt gives a parabola, not an exponential. Read the sentence carefully.
• Rounding k early. Store k exactly (for example as ln(9/4)/3) and round only the final answer.

Where this leads

Exponential models are the simplest models built from differential equations. BC students next meet the logistic model in Topic 7.9, where growth slows as the amount approaches a limit. Unit 8 then uses definite integrals for averages and accumulations, beginning with the average value of a function. If separating variables still feels slow, go back to Topic 7.7. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.