Real-World DE Modelling: Interest, Half-Life & Logistic Population Dynamics

1) The rate of spread of a rumor is proportional to the product of the number who know it and the number who don’t. If the town has 10,000 people, which differential equation models the number P(t) who have heard the rumor?

DP/dt = kP

DP/dt = kP(10000 - P)

DP/dt = -kP(10000 - P)

DP/dt = k(10000 - P)

The rate is jointly proportional to the number who know (P) and the number who don’t (10000 - P). This is exactly the logistic form dP/dt = kP(10000 - P), where the constant k absorbs the proportionality factor.

Explanation

The rate is jointly proportional to the number who know (P) and the number who don’t (10000 - P). This is exactly the logistic form dP/dt = kP(10000 - P), where the constant k absorbs the proportionality factor.

2) A lake is stocked with fish. The rate of growth of the fish population is proportional to the current population. Which equation is appropriate?

DP/dt = k(P - 1000)

DP/dt = -kP

DP/dt = kP

DP/dt = k/P

Pure growth with no limiting factors mentioned means the rate is proportional only to the current population, giving the exponential model dP/dt = kP with k > 0.

Explanation

Pure growth with no limiting factors mentioned means the rate is proportional only to the current population, giving the exponential model dP/dt = kP with k > 0.

3) The temperature T of a cup of coffee satisfies dT/dt = -0.09 (T - 20), where room temperature is 20°C. This is an example of:

Exponential growth

Newton's law of cooling

Logistic growth

Radioactive decay

The rate is proportional to the difference between the object's temperature and the ambient temperature, and the negative sign ensures cooling toward 20°C. This is the standard form of Newton's law of cooling.

Explanation

The rate is proportional to the difference between the object's temperature and the ambient temperature, and the negative sign ensures cooling toward 20°C. This is the standard form of Newton's law of cooling.

4) A population P satisfies dP/dt = 0.05 P (1 - P/2000). What is the carrying capacity?

0.05

2000

1/0.05

0.05 × 2000

The logistic equation is written in the form rP(1 - P/L). Here the term (1 - P/2000) shows that L = 2000 is the carrying capacity.

Explanation

The logistic equation is written in the form rP(1 - P/L). Here the term (1 - P/2000) shows that L = 2000 is the carrying capacity.

5) Solve dy/dt = -0.03 y with y(0) = 500.

Y = 500 e^(-0.03t)

Y = 500 e^(0.03t)

Y = 500 / (1 + 0.03t)

Y = 500 e^(-3t)

Separate variables: dy/y = -0.03 dt. Integrate: ln|y| = -0.03 t + C. Exponentiate: y = e^C e^(-0.03t). Let A = e^C. Then y = A e^(-0.03t). Using y(0) = 500 gives A = 500, so y = 500 e^(-0.03t).

Explanation

Separate variables: dy/y = -0.03 dt. Integrate: ln|y| = -0.03 t + C. Exponentiate: y = e^C e^(-0.03t). Let A = e^C. Then y = A e^(-0.03t). Using y(0) = 500 gives A = 500, so y = 500 e^(-0.03t).

6) An investment grows continuously at 6% per year. If $10,000 is invested, how long until it doubles?

Approximately 11.55 years

Approximately 16.67 years

Approximately 20.00 years

Approximately 23.10 years

Model is dA/dt = 0.06 A, so A(t) = 10000 e^(0.06t). Set 20000 = 10000 e^(0.06t) → 2 = e^(0.06t) → ln2 = 0.06 t → t = ln2 / 0.06 ≈ 0.693147 / 0.06 ≈ 11.55 years.

Explanation

Model is dA/dt = 0.06 A, so A(t) = 10000 e^(0.06t). Set 20000 = 10000 e^(0.06t) → 2 = e^(0.06t) → ln2 = 0.06 t → t = ln2 / 0.06 ≈ 0.693147 / 0.06 ≈ 11.55 years.

7) The decay constant for a substance is k = 0.02 per minute. What is its half-life?

0.02 minutes

34.7 minutes

50 minutes

0.02 / ln2 minutes

Half-life t½ = ln2 / k = ln2 / 0.02 ≈ 0.693 / 0.02 ≈ 34.65 ≈ 34.7 minutes.

Explanation

Half-life t½ = ln2 / k = ln2 / 0.02 ≈ 0.693 / 0.02 ≈ 34.65 ≈ 34.7 minutes.

8) In the logistic model dP/dt = rP(1 - P/L), the population grows fastest when:

P = 0

P = L/2

P = L

P = 2L

The per-capita growth rate decreases linearly, but the absolute growth rate rP(1 - P/L) is a quadratic with maximum at P = L/2.

Explanation

The per-capita growth rate decreases linearly, but the absolute growth rate rP(1 - P/L) is a quadratic with maximum at P = L/2.

9) A bacteria population obeys dP/dt = 0.12 P (1 - P/1000000) with P(0) = 200000. What is the long-term behavior?

Dies out to 0

Grows without bound

Approaches 1000000

Stabilizes at 200000

The carrying capacity is 1000000. Since the initial population is positive and less than the carrying capacity, the population will increase and approach 1000000 as t → ∞.

Explanation

The carrying capacity is 1000000. Since the initial population is positive and less than the carrying capacity, the population will increase and approach 1000000 as t → ∞.

10) For any logistic equation dP/dt = kP(a - P), the limiting value as t → ∞ is:

0

A

K

P(0)

When P is close to a, the term (a - P) is close to zero, so the growth rate approaches zero. The equilibrium points are P = 0 and P = a, and P = a is stable.

Explanation

When P is close to a, the term (a - P) is close to zero, so the growth rate approaches zero. The equilibrium points are P = 0 and P = a, and P = a is stable.

11) A 50 g sample of radioactive material has a half-life of 12 hours. How much remains after 48 hours?

3.125 g

6.25 g

12.5 g

25 g

48 hours is exactly 4 half-lives (48/12 = 4). After 4 half-lives the amount is 50 × (½)⁴ = 50 × 1/16 = 3.125 g.

Explanation

48 hours is exactly 4 half-lives (48/12 = 4). After 4 half-lives the amount is 50 × (½)⁴ = 50 × 1/16 = 3.125 g.

12) A drug is administered continuously at a rate of 30 mg/h and is eliminated according to dA/dt = -0.15 A. What is the steady-state amount in the body?

100 mg

150 mg

200 mg

300 mg

At steady state dA/dt = 0, so 30 - 0.15 A = 0 → 0.15 A = 30 → A = 30 / 0.15 = 200 mg.

Explanation

At steady state dA/dt = 0, so 30 - 0.15 A = 0 → 0.15 A = 30 → A = 30 / 0.15 = 200 mg.

13) A population follows dP/dt = 0.01 P (8000 - P) with P(0) = 1000. At what population will the growth rate be maximum?

1000

4000

8000

7000

Maximum growth rate occurs at half the carrying capacity, which is 8000/2 = 4000.

Explanation

Maximum growth rate occurs at half the carrying capacity, which is 8000/2 = 4000.

14) Caffeine has a half-life of 6 hours. If you drink a cup containing 200 mg at 8:00 AM, approximately how much caffeine remains at 8:00 PM (12 hours later)?

25 mg

50 mg

70 mg

100 mg

12 hours is exactly 2 half-lives. After 2 half-lives: 200 × (½)² = 200 × 1/4 = 50 mg.

Explanation

12 hours is exactly 2 half-lives. After 2 half-lives: 200 × (½)² = 200 × 1/4 = 50 mg.