AP Calculus: Exponential Growth and Decay Models (dy/dt = ky)

Exponential Growth and Decay Models (dy/dt = ky)

Concept Summary

• Differential equation (dy/dt = ky): A first-order linear ODE where the rate of change of y is proportional to y itself, modeling unbounded growth or decay.
• Proportionality constant (k): Determines growth (k > 0) or decay (k < 0); units are inverse time (e.g., 1/year).
• General solution (y = Ce): The unique family of solutions to dy/dt = ky, where C is the initial value y(0).
• Half-life (t?/?): Time for a decaying quantity to reduce to half its value; t?/? = ln(2)/|k| (independent of initial amount).
• Doubling time (t?): Time for a growing quantity to double; t? = ln(2)/k (only for k > 0).

Core Questions

WHAT (definitional)

Q: What is the differential equation for exponential growth/decay? A: dy/dt = ky, where k is the constant of proportionality linking rate of change to the current quantity y. Trap/Clarification: The equation assumes k is constant; if k varies with time, the model no longer applies.

Q: What does the solution y = Ce represent? A: The general solution to dy/dt = ky, where C is the initial value y(0) and e captures the exponential behavior. Trap/Clarification: C is not the "rate"; it’s the starting quantity, while k controls the rate.

WHY (causal/explanatory)

Q: Why does the solution to dy/dt = ky involve e? A: The function e is the only function whose derivative is proportional to itself (dy/dt = ke), satisfying the ODE. Trap/Clarification: Students often guess y = k? or y = Ck?, but these fail the derivative test.

Q: Why is the half-life independent of the initial amount? A: The decay process is proportional to the current quantity, so the time to halve is constant (e.g., 50%-25% takes the same time as 100%-50%). Trap/Clarification: This only holds for k < 0; growth (k > 0) has a constant doubling time instead.

HOW (process/application)

Q: How do you solve an initial-value problem for dy/dt = ky? A: 1) Write the general solution y = Ce, 2) plug in t = 0 to find C = y(0), 3) substitute C back into the solution. Trap/Clarification: Forgetting to solve for C is a common error; the general solution alone is not the answer.

Q: How is the half-life calculated from k? A: Use t?/? = ln(2)/|k|; for decay, k is negative, but the formula uses its absolute value. Trap/Clarification: Mixing up ln(2) with log?(2) (which equals 1) leads to incorrect answers.

CAN (conditions/possibilities)

Q: Can k be zero in dy/dt = ky? A: Yes, but the solution degenerates to y = C (constant function), as the rate of change is zero. Trap/Clarification: This is a trivial case; exams focus on k-0 for meaningful growth/decay.

Q: Under what conditions does the model dy/dt = ky fail? A: When the rate of change depends on factors other than y (e.g., limited resources, external forces) or if k is not constant. Trap/Clarification: Real-world scenarios often require logistic models (dy/dt = ky(1 – y/L)), not pure exponential.

Quick Facts & Traps

• Fact: The solution y = Ce is unique for dy/dt = ky with a given initial condition (by the Existence-Uniqueness Theorem).
• Trap: Confusing k with C-Reality: k is the rate constant; C is the initial value y(0).
• Fact: For decay, k is negative, but half-life formulas use |k| (e.g., t?/? = ln(2)/|k|).
• Trap: Assuming y = Ce works for dy/dt = ky + 5-Reality: This is a nonhomogeneous ODE; the solution is y = Ce – 5/k.
• Fact: Doubling time and half-life are inversely proportional to |k| (e.g., larger |k| = faster change).
• Trap: Using log instead of ln in formulas-Reality: AP Calculus requires natural log (ln) for e-based models.

Rapid-Fire True/False

• Statement: If k = 0.05, the quantity doubles every ln(2)/0.05 years. Answer: TRUE Why the common mistake happens: Students forget the formula or misapply it to decay (k < 0).

• Statement: The solution to dy/dt = 3y with y(0) = 4 is y = 4e³. Answer: FALSE (y = 4e³?) Why the common mistake happens: Omitting the t in the exponent is a frequent algebra error.

• Statement: A substance with k = –0.1 decays faster than one with k = –0.05. Answer: TRUE Why the common mistake happens: Students compare magnitudes incorrectly, thinking –0.05 is "more negative."