미분방정식 모델링

(a) Let the skydiver's downward velocity at time $t$ be given by $v(t)$. Since gravity causes her downward velocity to increase at the rate of $g\approx 9.8$, there is a factor of $+9.8$ meters per second squared in the differential equation for $\frac{dv}{dt}$, while the factor of wind resistence causing the velocity to decrease contributes the negative factor $-kv$. Thus, the differential equation is $$\frac{dv}{dt} = 9.8 - kv$$ (b) To find the equilibrium solutions, we set $\frac{dv}{dt} = 0$ and solve for $v$. Doing so, $0 = 9.8 - kv$, so $v = \frac{9.8}{k}$. Moreover, we note that for $v < \frac{9.8}{k}$, $\frac{dv}{dt} > 0$, while for $\frac{dv}{dt} > \frac{9.8}{k}$, $\frac{dv}{dt} < 0$. Thus, for velocities just above the equilibrium, the velocity will decrease and for velocities just below the equilibrium, the velocity will decrease, making $v = \frac{9.8}{k}$ a stable equilibrium. (c) We assume that $v(0) = 0$ and solve the IVP with the differential equation we developed. Taking the differential equation $$\frac{dv}{dt} = 9.8 - kv$$ and separating the variables, we have $$\frac{1}{9.8 - kv} \frac{dv}{dt} = 1\,.$$ Integrating with respect to $t$, $$\int \frac{1}{9.8 - kv} \, dv = \int 1 \, dt\,,$$ and using a standard substitution, we have $$-\frac{1}{k} \ln | 9.8 - kv | = t + C\,.$$ Solving for $v$, we have $\ln | 9.8 - kv | = -kt + C_1$ (where $C_1 = -Ck$, or $$|9.8 - kv| = e^{-kt + C_1} = C_2 e^{-kt}\,.$$ Incorporating the absolute value into the constant $C_2$ in the usual way, we have for some constant $A$ that $$9.8 - kv = Ae^{-kt}$$ and thus $$v = \frac{9.8 - Ae^{-kt}}{k}\,.$$ Setting $v(0) = 0$, we can also determine $A$, since then $$0 = \frac{9.8 - A}{k}$$ so $A = 9.8$. The solution to the IVP is therefore $$v(t) = \frac{9.8 - 9.8e^{-kt}}{k}$$ (d) If the skydiver's terminal velocity is 54 meters per second, this means that $$\lim_{t \to \infty} v(t) = 54$$ Using the solution we found in (c), we observe that $$\lim_{t \to \infty} \frac{9.8 - 9.8e^{-kt}}{k} = \frac{9.8}{k}$$ since $e^{-kt} \to 0$ as $t \to \infty$. Thus, $54 = \frac{9.8}{k}$, so $k = 9.8/54 \approx 0.181481$. (e) To find how long does it take to reach 50% of the terminal velocity, we set $v(t) = \frac{9.8 - 9.8e^{-0.181481t}}{0.181481} = 0.5(54)$ and solve for $t$. We have $$9.8 - 9.8e^{-0.181481t} = 0.5(54)(0.181481)\,,$$ so $$9.8-9.8e^{-0.181481t} = 4.9\,,$$ and thus $9.8e^{-0.181481t} = 4.9$, which tells us $$e^{-0.181481t} = 0.5\,.$$ Solving for $t$, $t = \frac{\ln(0.5)}{-0.181481} \approx 3.1894$ seconds.