Differential Equation Models: Exponential, Logistic, Cooling & Data Interpretation

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| Questions: 15 | Updated: Dec 16, 2025

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1) A population of bacteria grows at a rate proportional to its current size. Which differential equation correctly models this situation?

Dy/dt = k(y - 1000)

Dy/dt = ky

Dy/dt = k/y

Dy/dt = k(1000 - y)

When a quantity grows (or decays) at a rate proportional to its own size, the phrase "proportional to the size of the quantity" directly translates to the term ky on the right-hand side, where k is the constant of proportionality. No carrying capacity or external factor is mentioned, so the standard exponential growth model dy/dt = ky is the correct equation.

Explanation

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About This Quiz

Differential Equation Models: Exponential, Logistic, Cooling & Data Interpretation - Quiz

Can you spot the difference between unlimited growth and growth with limits? In this quiz, you’ll learn how logistic models describe populations that grow quickly at first but slow down as they approach a maximum sustainable size. You’ll interpret equations like dy/dt = ky(a − y), identify the carrying capacity,... see moreand understand why growth levels off. By the end, you’ll be able to explain how logistic growth behaves and what the key parameters mean in real situations.
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2) A radioactive substance decays such that its rate of decay is proportional to the amount present. If Q(t) represents the amount at time t, what is the correct differential equation?

DQ/dt = kQ²

DQ/dt = -kQ

DQ/dt = k(Q - 100)

DQ/dt = k/Q

Decay means the quantity is decreasing, so the rate of change is negative. The rate being proportional to the current amount gives -kQ, where k > 0. The negative sign ensures that the solution will decrease over time, matching radioactive decay.

Explanation

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3) The rate of cooling of an object is proportional to the difference between its temperature T and the surrounding temperature 30°C. Which equation models Newton's law of cooling?

A. dT/dt = kT

B. dT/dt = k(T - 30)

C. dT/dt = -k(T - 30)

D. dT/dt = k(30 - T)

Newton's law of cooling states that the object cools (rate is negative) when T > 30, and the rate is proportional to the temperature difference T - 30. Using k > 0, we write dT/dt = -k(T - 30) so that when T > 30 the derivative is negative, and when T

Explanation

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4) A population follows the logistic growth model dP/dt = 0.02 P (1 - P/5000), where the carrying capacity is 5000. What is the carrying capacity of the environment?

A. 0.02

B. 5000

C. 0.02 × 5000

D. 1/0.02

In the logistic equation dP/dt = kP(1 - P/L), the constant L in the denominator of the second term is the carrying capacity. Here the equation is written as 0.02 P (1 - P/5000), so L = 5000, meaning the population approaches 5000 as t → ∞.

Explanation

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5) A quantity y satisfies dy/dt = 0.05 y and y(0) = 200. What is the general solution?

A. y = 200 e^(-0.05t)

B. y = 200 e^(0.05t)

C. y = 200 e^(5t)

D. y = 200 / (1 + 0.05t)

Separate variables: dy/y = 0.05 dt. Integrate both sides: ln|y| = 0.05 t + C. Exponentiate: y = e^C e^(0.05t). Let A = e^C, so y = A e^(0.05t). Apply initial condition y(0) = 200: 200 = A e⁰→ A = 200. Thus y = 200 e^(0.05t).

Explanation

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6) A sample of 100 mg of a radioactive isotope decays according to dQ/dt = -0.012 Q. How much remains after 50 years?

A. 54.9 mg

B. 60.2 mg

C. 45.1 mg

D. 18.3 mg

The solution is Q(t) = Q0 e^(-kt) = 100 e^(-0.012 × 50). First compute the exponent: -0.012 × 50 = -0.6. Then e^(-0.6) ≈ 0.5488. So Q(50) = 100 × 0.5488 ≈ 54.9 mg.

Explanation

The solution is Q(t) = Q0 e^(-kt) = 100 e^(-0.012 × 50). First compute the exponent: -0.012 × 50 = -0.6. Then e^(-0.6) ≈ 0.5488. So Q(50) = 100 × 0.5488 ≈ 54.9 mg.

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7) The half-life of a substance is 8 days. What is the value of the decay constant k in dQ/dt = -kQ?

A. k = ln(2)/8

B. k = 8/ln(2)

C. k = 0.693/8

D. k = 8/0.693

At half-life t = 8, Q(8) = Q0/2. Using Q(t) = Q0 e^(-kt), we have Q0/2 = Q0 e^(-8k). Divide by Q0: 1/2 = e^(-8k). Take natural log: ln(½) = -8k → -ln(2) = -8k → k = ln(2)/8.

Explanation

At half-life t = 8, Q(8) = Q0/2. Using Q(t) = Q0 e^(-kt), we have Q0/2 = Q0 e^(-8k). Divide by Q0: 1/2 = e^(-8k). Take natural log: ln(½) = -8k → -ln(2) = -8k → k = ln(2)/8.

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8) A population obeys logistic growth dP/dt = 0.03 P (1 - P/10000) with P(0) = 2000. What is the limiting population as t → ∞?

A. 2000

B. 0.03

C. 10000

D. 3000

In any logistic equation dP/dt = rP(1 - P/L), the population approaches the carrying capacity L as time goes to infinity, regardless of the initial population (as long as P0 > 0). Here L = 10000.

Explanation

In any logistic equation dP/dt = rP(1 - P/L), the population approaches the carrying capacity L as time goes to infinity, regardless of the initial population (as long as P0 > 0). Here L = 10000.

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9) For the logistic equation dP/dt = kP(1 - P/C), at what population is the growth rate maximum?

A. P = 0

B. P = C

C. P = C/2

D. P = 2C

The growth rate is the right-hand side f(P) = kP(1 - P/C). This is a quadratic that opens downward. The vertex occurs at P = C/2. Alternatively, take derivative df/dP = k(1 - 2P/C) = 0 → P = C/2.

Explanation

The growth rate is the right-hand side f(P) = kP(1 - P/C). This is a quadratic that opens downward. The vertex occurs at P = C/2. Alternatively, take derivative df/dP = k(1 - 2P/C) = 0 → P = C/2.

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10) A town’s population follows dP/dt = 0.0004 P (12000 - P) and currently P = 3000. Is the population currently increasing or decreasing?

A. Decreasing because P > 6000

B. Increasing because P < 12000

C. Decreasing because P < 6000

D. Increasing because P > 0

The carrying capacity is 12000. When P 0, so the right-hand side is positive and the population is still increasing. It will continue increasing until it reaches 12000.

Explanation

The carrying capacity is 12000. When P 0, so the right-hand side is positive and the population is still increasing. It will continue increasing until it reaches 12000.

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11) A chemical dissolves in water at a rate proportional to the amount still undissolved. If 20 g are initially present and 5 g dissolve in the first 10 minutes, when will only 1 g remain undissolved?

A. 26.6 minutes

B. 33.2 minutes

C. 40.0 minutes

D. 104.1 minutes

Let A(t) be the amount undissolved. Then dA/dt = -kA, so A(t) = 20 e^(-kt). After 10 min, A(10) = 15 g → 15 = 20 e^(-10k) → e^(-10k) = 15/20 = 0.75 → -10k = ln(0.75) → k = -ln(0.75)/10 ≈ 0.02878. When A(t) = 1: 1 = 20 e^(-0.02878 t) → e^(-0.02878 t) = 0.05 → -0.02878 t = ln(0.05) → t = ln(0.05)/(-0.02878) ≈ 104.1 minutes.

Explanation

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12) A logistic model is dP/dt = 0.08 P (1 - P/8000) with P(0) = 1000. Approximately what time will the population first reach 4000 (half the carrying capacity)?

A. t ≈ 24.3 years

B. t ≈ 8.7 years

C. t ≈ 17.3 years

D. t ≈ 20.1 years

The solution to the logistic differential equation dP/dt = rP(1 - P/K) with initial condition P(0) = P0 is given by P(t) = (K P0) / (P0 + (K - P0) e^{-rt}). Here, K = 8000, P0 = 1000, r = 0.08. We set P(t) = 4000 and solve for t: 4000 = (8000 * 1000) / (1000 + 7000 e^{-0.08t}). This simplifies to 1000 + 7000 e^{-0.08t} = 2000, so 7000 e^{-0.08t} = 1000, e^{-0.08t} = 1/7. Taking natural logarithms gives -0.08t = ln(1/7) = -ln(7), thus t = ln(7)/0.08. Evaluating, ln(7) ≈ 1.94591, so t ≈ 1.94591/0.08 ≈ 24.3239 years.

Explanation

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13) A bacteria culture follows logistic growth with carrying capacity 1,000,000 and growth rate constant 0.1 per hour. At what population size is the culture growing fastest?

A. 0

B. 500,000

C. 1,000,000

D. 2,000,000

The growth rate function is r(P) = 0.1 P (1 - P/1000000). This quadratic reaches its maximum at the vertex P = 500,000, which is half the carrying capacity.

Explanation

The growth rate function is r(P) = 0.1 P (1 - P/1000000). This quadratic reaches its maximum at the vertex P = 500,000, which is half the carrying capacity.

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14) Carbon-14 has a half-life of 5730 years. An organism that died 10,000 years ago had 12% of its original C-14 remaining. Is this consistent with the known half-life?

A. Yes, expected ≈ 12.1%

B. No, expected ≈ 30%

C. No, expected ≈ 5%

D. Yes, expected exactly 12%

Fraction remaining = e^(-kt) with k = ln(2)/5730. After 10,000 years: e^(-(ln2 × 10000)/5730) = 2^(-10000/5730) ≈ 2^(-1.745) ≈ 0.298 or 29.8%.

Explanation

Fraction remaining = e^(-kt) with k = ln(2)/5730. After 10,000 years: e^(-(ln2 × 10000)/5730) = 2^(-10000/5730) ≈ 2^(-1.745) ≈ 0.298 or 29.8%.

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15) A drug is eliminated from the bloodstream according to dA/dt = -0.2 A (first-order kinetics). After how many hours will 90% of the drug be eliminated?

A. 6.93 hours

B. 11.51 hours

C. 15.33 hours

D. 23.05 hours

A(t) = A0 e^(-0.2t). We want A(t)/A0 = 0.1 → e^(-0.2t) = 0.1 → -0.2t = ln(0.1) → t = -ln(0.1)/0.2 = 4.605/0.2 ≈ 11.51 hours.

Explanation

A(t) = A0 e^(-0.2t). We want A(t)/A0 = 0.1 → e^(-0.2t) = 0.1 → -0.2t = ln(0.1) → t = -ln(0.1)/0.2 = 4.605/0.2 ≈ 11.51 hours.

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