Exponential Growth, Decay & Logistic Basics (k, half-life, carrying capacity)

When a quantity grows (or decays) at a rate proportional to itself, the word "proportional to the size of the quantity" directly translates to multiplying the quantity y by a constant rate k. This gives the standard exponential growth/decay model dy/dt = ky where k > 0 means growth and k < 0 means decay.

Separate variables: dy/y = k dt. Integrate both sides: ∫(1/y) dy = ∫k dt which gives ln|y| = kt + C. Exponentiate both sides: y = e^(kt + C) = e^C e^kt. The constant e^C is positive and absorbed into the initial condition, so with y(0) = y0 we get y0 = e^C and therefore y = y0 e^kt.

Doubling every 3 hours means the growth rate is such that y = y0 * 2^(t/3). After 9 hours we have three doubling periods (9/3 = 3), so 500 * 2³ = 500 * 8 = 4000 bacteria. Alternatively, using continuous model: doubling time gives k = ln(2)/3, so y = 500 e^(ln(2)/3 * 9) = 500 e^(3 ln 2) = 500 * 8 = 4000.

Half-life means y(t) = y0 / 2 when t = 10. Using y = y0 e^kt we get 1/2 = e^(10k). Take natural log: ln(½) = 10k → -ln 2 = 10k → k = -ln 2 / 10. The negative sign is required because the substance is decaying.

Model is y = y0 e^kt with y0 = 200. At t = 4, y = 150, so 150 = 200 e^(4k) → 0.75 = e^(4k) → ln(0.75) = 4k → k = ln(0.75)/4. Then at t = 8, y = 200 e^(8k) = 200 [e^(4k)]² = 200 * (0.75)² = 200 * 0.5625 = 112.5 mg.

The sign of k determines growth or decay. When k > 0, the derivative dy/dt is positive whenever y > 0, meaning the quantity is increasing, and the solution y = y0 e^kt grows without bound as t increases, which is exponential growth.

Continuous growth at 2% means k = 0.02, so y = 50000 e^(0.02 * 10) = 50000 e^0.2. Using e^0.2 ≈ 1.2214, we get 50000 * 1.2214 ≈ 61070.

From 80 to 20 is a reduction to 1/4, so in 12 days it multiplies by 1/4 = e^(12k) → 12k = ln(1/4) = -2 ln 2 → k = -ln 2 / 6. To reach 10 grams (1/8 of 80), we need e^(tk) = 1/8 = 2^(-3), so t(-ln 2 / 6) = -3 ln 2. This gives us t = 18 / (ln 2) * ln 2 = 18 days from 80 to 10 grams.

Decay requires the rate of change to be negative when y is positive. Only choice B has a negative constant of proportionality with the quantity itself, giving dy/dt

We can rewrite the differential equation dy/dt = -0.1y as dy/y = -0.1 dt. Integrating both sides yields ln|y| = -0.1t + C. Exponentiating gives us y = eC e-0.1t = Ae-0.1t, where A is a constant. Using the initial condition y(0) = 50, we find 50 = Ae0, so A = 50. Thus, the solution is y(t) = 50e-0.1t. After 5 days, y(5) = 50e-0.5.

In the logistic equation dy/dt = ky(a - y), when y approaches a, the term (a - y) approaches zero, making dy/dt approach zero. This means the population stops changing and stabilizes at y = a. This equilibrium value represents the maximum sustainable population, which is called the carrying capacity.

The growth rate dy/dt = ky(a - y) is a quadratic function in y that opens downward. The maximum occurs at the vertex y = a/2. At that point the product y(a - y) is largest, so the population increases most rapidly when it is half the carrying capacity.

in the logistic equation dy/dt = ky(1000 - y), when y > 1000 the term (1000 - y) becomes negative, making dy/dt

The form dy/dt = k y (C - y) is the standard logistic equation. The growth rate is jointly proportional to the current population y and how far it is from the carrying capacity C = 200.

The growth rate dy/dt = k y (5000 - y) is maximized when the expression y(5000 - y) is maximized. This quadratic has its vertex at y = 5000/2 = 2500. At that point the population is growing fastest.