Drill 1 · Math · Exponential Growth and Decay Models
AP Calculus AB: Exponential Growth and Decay Models (Drill 1) is a Math practice drill covering Exponential Growth and Decay Models. It contains 5 original questions created by Brian Stewart, a Barron's test prep author with over 20 years of tutoring experience.
Practice applying the differential equation \( \dfrac{dy}{dt} = ky \) and its solution \( y = Ce^{kt} \) to model exponential growth and decay, including half-life, doubling time, and interpreting the constant k in context. These skills appear on the AP Calculus AB exam in both multiple-choice and free-response questions.
Question 1. A quantity \( y \) satisfies \( \dfrac{dy}{dt} = -0.3y \). If \( y(0) = 50 \), which of the following gives \( y(t) \)?
Explanation: Choice C is correct. The general solution to \( \dfrac{dy}{dt} = ky \) is \( y = Ce^{kt} \). Here \( k = -0.3 \) and \( C = y(0) = 50 \), giving \( y = 50e^{-0.3t} \). Choice A is incorrect because it treats the equation as \( \dfrac{dy}{dt} = -0.3 \) (a constant rate), producing a linear model instead of an exponential one, this confuses proportional decay with constant decay. Choice B is incorrect because the student used the correct exponential form but applied a positive exponent, ignoring the negative sign on \( k \) and producing exponential growth instead of decay. Choice D is incorrect because the student confused the roles of \( C \) and \( k \), swapping the initial value and the rate constant in the formula.
Question 2. A population grows according to \( \dfrac{dP}{dt} = 0.04P \). Approximately how long does it take for the population to double?
Explanation: Choice B is correct. Setting \( 2P_0 = P_0 e^{0.04t} \) gives \( \ln 2 = 0.04t \), so \( t = \dfrac{\ln 2}{0.04} \approx \dfrac{0.693}{0.04} \approx 17.3 \) years. Choice A is incorrect because the student read \( k = 0.04 \) and interpreted it directly as a time value, confusing the growth rate constant with the doubling time. Choice C is incorrect because the student computed \( \dfrac{1}{0.04} = 25 \), which gives the e-folding time (the time to multiply by \( e \approx 2.718 \)), not the doubling time. Choice D is incorrect because the student computed \( \dfrac{1}{0.02} = 50 \), apparently halving \( k \) under the mistaken belief that "doubling" requires dividing the rate by 2.
Question 3. A radioactive substance decays according to \( A(t) = A_0 e^{-0.0231t} \), where \( t \) is in years. What is the half-life of the substance, to the nearest year?
Explanation: Choice B is correct. The half-life satisfies \( \dfrac{1}{2} = e^{-0.0231\, t_{1/2}} \). Taking \( \ln \) of both sides: \( -\ln 2 = -0.0231\, t_{1/2} \), so \( t_{1/2} = \dfrac{\ln 2}{0.0231} \approx \dfrac{0.693}{0.0231} \approx 30 \) years. Choice A is incorrect because the student confused the half-life formula with the e-folding formula, computing \( \dfrac{1}{0.0231} \approx 43 \), then halved that result to get approximately 20, conflating the two formulas rather than using \( \dfrac{\ln 2}{k} \) directly. Choice C is incorrect because the student computed \( \dfrac{1}{0.0231} \approx 43 \), which gives the e-folding time (the time for the quantity to fall to \( \dfrac{1}{e} \) of its original value) rather than the half-life. Choice D is incorrect because the student likely misplaced a decimal in the exponent, treating the decay constant as \( 0.00231 \) instead of \( 0.0231 \), and computed \( \dfrac{\ln 2}{0.00231} \approx 300 \), then rounded or truncated to 100.
Question 4. A cup of coffee cools according to Newton's Law of Cooling: \( T(t) = 22 + 68e^{-0.05t} \), where \( T \) is temperature in °C and \( t \) is time in minutes. Which of the following best interprets the value \( k = -0.05 \)?
Explanation: Choice A is correct. In the model \( T(t) - 22 = 68e^{-0.05t} \), the quantity \( T - 22 \) is the temperature excess above room temperature. The value \( k = -0.05 \) means this excess decays exponentially at a rate of approximately 5% per minute, the rate of cooling is proportional to the current temperature difference, which is exactly Newton's Law of Cooling. Choice B is incorrect because \( k \) governs a proportional rate of change, not a constant one, this choice confuses exponential decay with linear (constant) cooling. Choice C is incorrect because at \( t = 20 \) minutes, \( T(20) = 22 + 68e^{-1} \approx 47 \)°C, which is far above room temperature; setting \( t = 1/|k| \) does not give the time to reach room temperature. Choice D is incorrect because \( k \) determines the rate of exponential decay, not a ratio between temperature values; the initial temperature is \( T(0) = 22 + 68 = 90 \)°C, which has no multiplicative relationship to room temperature through \( k \).
Question 5. At time \( t = 0 \), a bacterial culture contains \( 100 \) bacteria. At time \( t = 2 \) hours, it contains \( 900 \) bacteria. Assuming exponential growth \( y = Ce^{kt} \), what is the approximate value of \( k \)?
Explanation: Choice B is correct. With \( y(0) = 100 \), we have \( C = 100 \). Substituting \( y(2) = 900 \): \( 900 = 100e^{2k} \), so \( e^{2k} = 9 \), giving \( 2k = \ln 9 \) and \( k = \dfrac{\ln 9}{2} = \ln 3 \approx 1.10 \). Choice A is incorrect because the student divided \( \ln 9 \) by 4 instead of 2, perhaps doubling the time value in the denominator by mistake, giving \( k \approx 0.55 \). Choice C is incorrect because the student forgot to divide by \( t = 2 \) as the final step, stopping at \( k = \ln 9 \approx 2.20 \), a very common error of omitting the last division. Choice D is incorrect because the student computed \( \dfrac{\ln 900}{2} \approx 3.35 \), failing to first divide the final population by the initial population before taking the logarithm, this reflects a misunderstanding of how the exponential model is solved for \( k \).