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

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.

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.

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

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 → ∞.

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).

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.

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.

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.

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.

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.

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.

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.

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.

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%.

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.