Module 3 Quiz

10 questions covering exponential growth/decay, logistic models, mixing problems, and Newton's cooling law.

What is the general solution to dP/dt = kP with P(0) = P₀?

Answer: P(t) = P₀e^kt.

A substance has k = -0.04 per day. Find its half-life.

Answer: t_1/2 = ln(2)/0.04 ≈ 17.33 days.

In the logistic equation dP/dt = rP(1 - P/K), what happens when P = K?

Answer: When P = K, the factor (1 - P/K) = 0, so dP/dt = 0. The population stays at the carrying capacity -- it is a stable equilibrium.

A population of 300 follows dP/dt = 0.4P(1 - P/2000). Write the explicit solution.

Answer: A = (2000 - 300)/300 = 17/3. P(t) = 2000/(1 + (17/3)e^-0.4t).

For a mixing problem with a 200-L tank, inflow 4 L/min at 6 g/L, outflow 4 L/min, and Q(0) = 0, write the ODE and solve.

Answer: dQ/dt = 24 - Q/50. Solving: Q(t) = 1200(1 - e^-t/50).

In the mixing problem above, what is the steady-state concentration?

Answer: As t → ∞, Q → 1200 g. Concentration = 1200/200 = 6 g/L (equals the inflow concentration).

Write the general solution to Newton's cooling law dT/dt = k(T - T_env).

Answer: T(t) = T_env + (T₀ - T_env)e^kt, where T₀ = T(0) and k < 0 for cooling.

A pie at 175°C is placed in a 22°C room. After 20 min it is 100°C. Find k.

Answer: 100 = 22 + 153e^20k ⇒ 78 = 153e^20k ⇒ e^20k = 78/153 ⇒ k = ln(78/153)/20 ≈ -0.03376 per min.

$12,000 is invested at 3.5% compounded continuously. What is the balance after 8 years?

Answer: A(8) = 12000e^0.035(8) = 12000e^0.28 ≈ 12000(1.3231) ≈ $15,877.06.

For the logistic model with r = 0.5 and K = 10000, what is the maximum sustainable harvest rate?

Answer: h_max = rK/4 = 0.5(10000)/4 = 1250 per unit time.