Ejercicios resueltos de crecimiento y decrecimiento exponencial

$\frac{dN}{dt} = -kN$ donde $N(t)$ es la cantidad de sustancia en el tiempo $t$.

$N(t) = N_0 e^{-kt} = 200e^{-kt}$

$N(8) = 100 = 200e^{-8k}$

$\frac{1}{2} = e^{-8k}$

$\ln\left(\frac{1}{2}\right) = -8k$

$k = \frac{\ln 2}{8} \approx 0.0866$ h$^{-1}$

$N(24) = 200e^{-0.0866 \times 24} = 200e^{-2.078} = 200 \times 0.125 = 25$ gramos

$P(t) = P_0 e^{kt} = 5000e^{0.4t}$

$P(3) = 5000e^{0.4 \times 3} = 5000e^{1.2} = 5000 \times 3.32 = 16600$ bacterias

$P(t) = 15000 = 5000e^{0.4t}$

$3 = e^{0.4t}$

$\ln 3 = 0.4t$

$t = \frac{\ln 3}{0.4} = \frac{1.099}{0.4} = 2.75$ horas

$P(2.75) = 5000e^{0.4 \times 2.75} = 5000e^{1.1} = 5000 \times 3.004 \approx 15020$ bacterias $\checkmark$

$T(t) = T_a + (T_0 - T_a)e^{-kt} = 25 + 75e^{-kt}$

$T(10) = 75 = 25 + 75e^{-10k}$

$50 = 75e^{-10k}$

$\frac{2}{3} = e^{-10k}$

$\ln\left(\frac{2}{3}\right) = -10k$

$k = \frac{\ln(3/2)}{10} = \frac{0.405}{10} = 0.0405$ min$^{-1}$

$30 = 25 + 75e^{-0.0405t}$

$5 = 75e^{-0.0405t}$

$\frac{1}{15} = e^{-0.0405t}$

$\ln\left(\frac{1}{15}\right) = -0.0405t$

$t = \frac{\ln 15}{0.0405} = \frac{2.708}{0.0405} = 66.9$ minutos

$T(66.9) = 25 + 75e^{-0.0405 \times 66.9} = 25 + 75e^{-2.708} = 25 + 75 \times 0.067 = 30^\circ C$ $\checkmark$

La solución de la ecuación logística es:

$P(t) = \frac{K}{1 + \left(\frac{K-P_0}{P_0}\right)e^{-rt}} = \frac{10000}{1 + 19e^{-0.1t}}$

$5000 = \frac{10000}{1 + 19e^{-0.1t}}$

$1 + 19e^{-0.1t} = 2$

$19e^{-0.1t} = 1$

$e^{-0.1t} = \frac{1}{19}$

$-0.1t = \ln\left(\frac{1}{19}\right)$

$t = \frac{\ln 19}{0.1} = 29.4$ días

$\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right)$

Para encontrar el máximo, derivamos: $\frac{d^2P}{dt^2} = r\frac{dP}{dt}\left(1 - \frac{2P}{K}\right) = 0$

Esto ocurre cuando $P = \frac{K}{2} = 5000$.

$\left(\frac{dP}{dt}\right)_{max} = r \cdot \frac{K}{2} \cdot \left(1 - \frac{1}{2}\right) = 0.1 \times 5000 \times 0.5 = 250$ individuos/día