미분방정식 모델링

$y = 10 + Ce^{-kx}$

13.66 days

4.43 days

13,304.65 years old

$x = \begin{cases} \displaystyle\frac{ab(1 - e^{(a-b)kt})}{b-ae^{(a-b)kt}} & \text{ if } a \neq b\\ \displaystyle \frac{a^2kt}{1+akt} & \text{ if } a = b \end{cases}$

24.57 minutes

$\displaystyle y = 60 - 3.69858e^{-\frac{1}{4}t} + 43.69858e^{-0.0390169 t}$

0.06767 g/gal

$y = 8(1-e^{-\frac{1}{2}t})$ g/cm$^2$

$y = \displaystyle 20 - \frac{10}{17}\left (4\cos(2t)- \sin(2t)\right) - \frac{300}{17}e^{-\frac{1}{2}t}$ g

11.00075 g

pond 1: 50.4853 grams per million gallons

pond 2: 32.8649 grams per million gallons

(a) Let $A(t)$ be the amount of money (in millions) in the account at time $t$. We know that the rate of change of $A$ is affected by the addition of $1$ million dollars a year, plus the interest earned, which is $0.05A$. Thus, $$\frac{dA}{dt} = 1 + 0.05A$$ (b) If there is no amount of money in the account when you open it, that means $A(0) = 0$. Solving the initial value problem, we can find how much money will be in the account after 25 years. Note that $1 + 0.05A = 0.05(20 + A)$, so we have $$\frac{1}{20+A} \frac{dA}{dt} = 0.05$$ and thus $$\int \frac{1}{20+A} \, dA = \int 0.05 \, dt\,.$$ Integrating (and noting that $A(t) > 0$ for all $t$, $$\ln (20 + A) = 0.05t + C\,,$$ so solving for $A$ in the standard way, $$20 + A = Ke^{0.05t}$$ and thus $A(t) = Ke^{0.05t} - 20$. Since $A(0) = 0$, $K = 20$, and we've shown $$A(t) = 20(e^{0.05t} - 1)\,.$$ It follows that $A(25) = 49.80686$ million dollars. (c) For this option, we know that $\frac{dA}{dt} = 0.05A$, $A(0) = 10$, and thus by the standard solution to this IVP, $$A(t) = 10e^{0.05t}$$ from which we see that $A(25) = 34.90343$ million dollars. (d) The first option is obviously better because it leads to more money in the long term. (e) The amount of money in the account under the first option overtakes the amount of money in the account under the second option when $20(e^{0.05t} - 1) = 10e^{0.05t}$. Solving for $t$, we find that $10e^{0.05t} = 20$, so $e^{0.05t} = 2$, and thus $$t = 20 \ln(2) \approx 13.86 \ \text{years}\,.$$

(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.

(a) Since the baby's weight, $w(t)$ (in pounds at month $t$), grows at a rate proportional to its reciprocal, we have $$\frac{dw}{dt} = \frac{k}{w}\,.$$ (b) Given that $W(0) = 8$ and $W(1) = 9$, we can both solve the IVP and determine $k$. To start, we separate variables and solve the differential equation by writing $$w \frac{dw}{dt} = k$$ and integrating to get $$\int w \frac{dw}{dt} \, dt = \int k \, dt\,.$$ It follows that $$\frac{1}{2}w^2 = kt + C$$ and therefore $w(t) = \sqrt{2kt + C_1}$ (where $C_1 = 2C$ and we take the positive square root because we know $w(t) > 0$). Applying the initial condition $w(0) = 8$, we find that $8 = \sqrt{C_1}$, so $C_1 = 64$ and $$w(t) = \sqrt{2kt + 64}\,.$$ To find $k$, we use the additional information that $w(1) = 9$, so $$9 = \sqrt{2k + 64}$$ which implies that $81 = 2k + 64$ and thus $k = 17/2$. Thus, $$w(t) = \sqrt{17t+64}\,.$$ At one year, according to the model the baby weighs $w(12) = \sqrt{268} \approx 16.37$ pounds. (c) While the function $w(t) = \sqrt{7t+64}$ grows without bound as $t$ increases, the model is unrealistic more because the function initially doesn't grow fast enough. For example, the model predicts that at age $10$, $w(120) \approx 30.06$ pounds, and at age $20$, $w(120) = 41.76$ pounds. A realistic model would need to grow much more quickly early in life and then level off around age $20$.

(a) The volume of solution in the tank remains constant because the inflow and outflow are at the same rate. (b) Salt enters the tank with the inflow, which occurs at a rate of $3$ gallons per minute at a concentration of $20$ grams per gallon. Observing that $$20 \frac{\text{grams}}{\text{gallon}} \cdot 3 \frac{\text{gallons}}{\text{minute}} = 60 \frac{\text{grams}}{\text{minute}}\,,$$ we see that salt is entering the tank at a rate of $60$ grams per minute. (c) Letting $S(t)$ denotes the number of grams of salt in the tank at time $t$, since there are $100$ gallons of solution in the tank at any time $t$, the concentration of salt in the tank in grams per gallon is $$\frac{S(t)}{100} \frac{\text{grams}}{\text{gallon}}$$ at any time $t$. (d) With water leaving the tank at $3$ gallons per minute, it follows that salt is leaving the tank at a rate of $$\frac{S(t)}{100} \frac{\text{grams}}{\text{gallon}} \cdot 3 \frac{\text{gallons}}{\text{minute}} = \frac{3S(t)}{100} \frac{\text{grams}}{\text{minute}}\,.$$ (e) Taking into account the rates at which salt enters and leaves the tank as established in (b) and (d) and noting that the units on $\frac{dS}{dt}$ are grams per minute, we have shown that $$\frac{dS}{dt} = 60 - \frac{3}{100} S\,.$$ (f) Setting $60 - \frac{3}{100} S = 0$, we find that $S = 2000$ is the only equilibrium solution. This makes sense because the inflow is carrying $20$ grams of salt per gallon and thus when the tank reaches this concentration, there will be $20 \cdot 100$ total grams of salt. This equilibrium is stable (both intuitively and because $\frac{dS}{dt}$ is positive for $S < 2000$ and negative for $S > 2000$. (g) Assuming that $S(0) = 0$, we can solve the IVP with $\frac{dS}{dt} = 60 - \frac{3}{100} S$. For somewhat simpler computations, we write $60 - \frac{3}{100}S = -\frac{3}{100}(S-2000)$. Separating variables, $$\frac{1}{S-2000} \frac{dS}{dt} = -\frac{3}{100}$$ and integrating with respect to $t$, $$\int \frac{1}{S-2000} \, dS = -\int \frac{3}{100} \, dt$$ so that $$\ln|S-2000| = -\frac{3}{100}t + C\,.$$ Solving for $S$ in the usual way, $$S = 2000 + Ke^{-\frac{3}{100}t}$$ and using the initial condition that $S(0) = 0$, we find $K = -2000$ so that $$S(t) = 2000 - 2000e^{-\frac{3}{100}t}\,.$$ (h) As $t \to \infty$, we know $e^{-\frac{3}{100}t} \to 0$, and thus $S(t) \to 2000$. This makes sense because the inflow carries $20$ grams of salt per liter continuously. As time goes on, the concentration of solution in the tank should tend to that same concentration, and with $100$ gallons of solution, the amount of salt will tend to $20 \cdot 100 = 2000$ grams. This value also matches the stable equilibrium that we found in (f).