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AP® Calculus AB · Unit 7 · Topic 7.8

Exponential Models with Differential Equations: Revision Notes (Calculus AB 7.8)

One-page recap of dy/dt = ky: translating the sentence, the solution y = y₀e^(kt), finding k from data or a half-life, checking answers and the common slips.

Short on time? This page is the recap. For the derivation, the graph and worked examples, use the full study guide. This topic is shared by Calculus AB and Calculus BC.

Notation: e^(kt) means e to the power kt; y₀ is the value of y at t = 0.

Recap

  • “The rate of change of y is proportional to y” translates to dy/dt = ky.
  • k > 0: growth. k < 0: decay. Units of k: per unit of time.
  • Separating variables gives ln|y| = kt + C, so y = y₀e^(kt). The constant multiplies; it is not added.
  • k is the relative rate: (dy/dt)/y = k. The rate dy/dt itself changes as y changes.
  • Over any time step Δt, y is multiplied by e^(kΔt), whatever its current value.
  • Motion: dv/dt = kv (k < 0) gives v = v₀e^(kt); integrate v to get position.

Key relationships

Situation Result
dy/dt = ky, y(0) = y₀ y = y₀e^(kt)
Two values y(0) = y₀, y(T) = y₁ k = ln(y₁/y₀)/T
Doubling time D k = ln 2 / D
Half-life H k = −ln 2 / H; y = y₀ · 2^(−t/H)
Rate at a given moment dy/dt = ky (use the equation, no need to differentiate)
Time to reach a value Y t = ln(Y/y₀)/k
dy/dt = k(y − M) y − M = (y₀ − M)e^(kt)

How to confirm a solution

  1. Differentiate your y and check that dy/dt equals k times your y.
  2. Substitute t = 0 and check that you get y₀.
  3. Check reasonableness: sign of k, units, growth or decay in the right direction, and no negative amounts.

Mistakes to avoid

  1. Writing y = e^(kt) + C instead of y = Ce^(kt).
  2. Treating k as a fixed amount per unit time. It is a fraction of the current amount.
  3. Using rate × time for a non-constant rate.
  4. Double minus signs: if you write dy/dt = −ky with k > 0, the solution is y₀e^(−kt).
  5. Confusing “proportional to t” with “proportional to y”.
  6. Rounding k early. Store it at full accuracy.

Quick self-check

  1. y = 3e^(kt) with k = (ln 2)/2. Find y(4) without a calculator. (12: the doubling time is 2, so 3 → 6 → 12)
  2. dQ/dt = −0.1Q, t in days. Find the half-life. (ln 2 / 0.1 ≈ 6.931 days)
  3. A quantity triples every 5 years. Find k. (ln 3 / 5 ≈ 0.220 per year)

Next: practice questions.

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