AP® Calculus AB review sheet from Aim for Five (aimforfive.com/calc-ab/units/7/7-8)
Unit 7 · Topic 7.8
7.8 Exponential Models with Differential Equations
When a quantity changes at a rate proportional to its own size, dy/dt = ky, and the solution is y = y₀eᵏᵗ. This one model covers population growth, radioactive decay, continuous interest and more.
Key terms
- exponential growth
- exponential decay
- dy/dt = ky
- y = y₀eᵏᵗ
- growth constant k
Deriving the model
“The rate of change of y is proportional to y” translates to dy/dt = ky. Separate and integrate: (1/y) dy = k dt, so ln|y| = kt + C, and y = Aeᵏᵗ. At t = 0, y = A, so A is the starting amount y₀:
y = y₀eᵏᵗ.
You can use this result directly on the exam. You don't have to derive it each time unless the question asks you to solve the differential equation.
What k tells you
- k > 0: exponential growth. The bigger y gets, the faster it grows.
- k < 0: exponential decay. The amount shrinks toward 0, and shrinks more slowly as it gets smaller.
- The relative growth rate (dy/dt)/y equals k, which is constant. That's the defining feature of this model.
- Doubling time: y doubles when eᵏᵗ = 2, so t = ln 2/k. Half-life: y halves when eᵏᵗ = ½, so t = ln(½)/k = −ln 2/k (positive, since k < 0).
Finding k from data
Usually you're told the starting amount and the amount at a later time. Substitute both into y = y₀eᵏᵗ, isolate eᵏᵗ, and take the natural log. Keep k exact (like k = ln(2.8)/3) until the final step, so rounding doesn't build up.
Reading the equation directly also works. If dy/dt = −0.05y, you know k = −0.05 without solving anything: at every instant, the amount is decreasing at a rate equal to 5% of its current value per unit of time.
Where the model shows up
- Populations with plenty of resources: dP/dt = kP with k > 0.
- Radioactive decay and drug elimination from the body: dA/dt = kA with k < 0.
- Money with continuously compounded interest at annual rate r: dB/dt = rB, so B = B₀eʳᵗ.
- Any statement like “grows at a rate of 4% of its current size per year (continuously)”: dy/dt = 0.04y.
A close cousin: shifted exponential models
Some equations look like dy/dt = k(y − M), for example a hot drink cooling toward room temperature M. This is still separable, and the solution is y = M + Ceᵏᵗ: the difference y − M grows or decays exponentially. Don't write y = y₀eᵏᵗ for this equation. Solve it with separation of variables, or recognize the shift.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Growth from two data points
A bacteria culture grows at a rate proportional to its size. It has 500 bacteria at t = 0 and 1400 at t = 3 hours. Find k, and find when the culture reaches 5000.
Show the solutionHide the solution
- Step 1: Model: P = 500eᵏᵗ.
- Step 2: At t = 3: 1400 = 500e³ᵏ, so e³ᵏ = 2.8 and k = ln(2.8)/3 ≈ 0.343 per hour.
- Step 3: Set P = 5000: 5000 = 500eᵏᵗ, so eᵏᵗ = 10 and t = ln 10/k = 3 ln 10/ln 2.8.
- Step 4: t ≈ 6.709 hours.
Answer: k = ln(2.8)/3 ≈ 0.343; the culture reaches 5000 at t ≈ 6.709 hours.
- Example 2Calculator allowed
Half-life
A substance has a half-life of 8 days. Starting with 120 mg, how much is left after 20 days?
Show the solutionHide the solution
- Step 1: Half-life gives e^(8k) = ½, so k = −ln 2/8.
- Step 2: y = 120e^(−(ln 2/8)·20) = 120 · 2^(−20/8) = 120 · 2^(−2.5).
- Step 3: 2^(−2.5) ≈ 0.17678, so y ≈ 21.213 mg.
- Step 4: Sanity check: 16 days is two half-lives (30 mg), and 24 days is three (15 mg). 21.2 mg at 20 days fits between them.
Answer: About 21.213 mg
- Example 3
Trap: the model shifted by room temperature
Coffee cools according to dT/dt = k(T − 70), with T in °F and t in minutes. T(0) = 190 and T(5) = 150. Find T(10).
Show the solutionHide the solution
- Step 1: This is not dT/dt = kT, so T = 190eᵏᵗ is wrong.
- Step 2: Separate: dT/(T − 70) = k dt, so ln|T − 70| = kt + C and T = 70 + Ceᵏᵗ.
- Step 3: T(0) = 190 gives C = 120. T(5) = 150 gives 120e⁵ᵏ = 80, so e⁵ᵏ = 2/3.
- Step 4: T(10) = 70 + 120e¹⁰ᵏ = 70 + 120(e⁵ᵏ)² = 70 + 120(4/9) = 70 + 53.33… ≈ 123.333.
Answer: T(10) = 370/3 ≈ 123.333°F
Common mistakes
- Using y = y₀eᵏᵗ for an equation like dy/dt = k(y − M).
- Rounding k early. A k of 0.34 instead of 0.343206… can change the answer in the second decimal place.
- Writing a positive k for a decay problem, or getting a negative time from a sign slip.
- Confusing “proportional to y” (exponential) with “increases by a constant amount” (linear).
On the exam
- Multiple-choice questions often describe a situation in words and ask for the function or the time to reach a value. Recognize dy/dt = ky and go straight to y = y₀eᵏᵗ.
- On free response, if asked to solve the differential equation, show the separation of variables, even though you know the answer form.
Connected topics
Videos
Check yourself
4 questions on 7.8 Exponential Models with Differential Equations. Pick an answer to see if you got it, and why.
A radioactive substance decays at a rate proportional to the amount present, so its amount is y = y₀eᵏᵗ, where t is in years. The substance has a half-life of 6 years. What is the value of k?
| t (hours) | P(t) (bacteria) |
|---|---|
| 0 | 500 |
| 3 | 1200 |
| 6 | 2880 |
Invented data for practice
The number of bacteria P in a culture grows at a rate proportional to the number present, so dP/dt = kP, where t is in hours. Selected values of P are shown in the table. What is the value of k?
The population in the table follows the model dP/dt = kP. According to the model, at what time t, in hours, does the population reach 3,000 bacteria?
The population in the table follows the model dP/dt = kP. According to the model, at what rate is the population growing at time t = 5 hours?
0 of 4 answered