Modeling with First-Order Differential Equations
Physical and biological formulation of first-order differential equations: radioactive decay, Newton's law of cooling with time-dependent ambient variations, cascading mixing tank balances, logistic population growth with harvesting bifurcation, falling bodies under linear and quadratic aerodynamic drag, and Cartesian and polar orthogonal trajectories.
ยง4.1 Exponential Growth, Decay & Newton's Law of Cooling
1. Law of Exponential Growth and Decay
When the rate of change of a physical quantity $N(t)$ is directly proportional to the amount currently present, it obeys the foundational differential equation:
If $k > 0$, the process represents exponential growth (e.g., unconstrained bacterial colonies, continuous compound interest). If $k < 0$, writing $k = -\lambda$ ($\lambda > 0$) models exponential decay (e.g., radioactive isotopes). The half-life $t_{1/2}$ is the time required for half the original atoms to decay:
2. Newton's Law of Cooling/Warming
Newton's empirical law states that the time rate of heat loss of a body is proportional to the difference in temperatures between the body and its surroundings:
where $T(t)$ is the temperature of the object and $T_m(t)$ is the ambient surrounding temperature. If $T_m$ is constant, this is a separable equation yielding:
If $T_m(t)$ varies dynamically with time (e.g., daily solar oscillations $T_m(t) = T_0 + A \sin(\omega t)$), the equation is solved as a first-order linear ODE via integrating factor $\mu(t) = e^{kt}$.
ยง4.2 Mixing Tank Balances & Logistic Harvesting Bifurcation
1. Principle of Mass Balance in Well-Stirred Mixing Tanks
Consider a tank containing volume $V(t)$ of liquid into which a solute (e.g., salt) is pumped. Let $Q(t)$ denote the mass of solute in the tank at time $t$. The conservation of mass requires:
Let fluid enter at volumetric rate $r_{\text{in}}$ with concentration $c_{\text{in}}$, and leave at volumetric rate $r_{\text{out}}$ with uniform concentration $c_{\text{out}}(t) = \frac{Q(t)}{V(t)}$. Then:
This yields the first-order linear differential equation:
2. Logistic Population Dynamics Under Harvesting
The Verhulst logistic growth model accounts for resource limitation with carrying capacity $K$ and intrinsic reproductive rate $r$. When subject to a constant harvesting rate $H$ (e.g., commercial fishing), the governing equation is:
The equilibrium populations are given by the roots of the quadratic RHS:
Bifurcation Analysis:
- If $H < \frac{rK}{4}$: There are two distinct equilibria $P_1^* < P_2^*$. $P_2^*$ is stable (carrying state) while $P_1^*$ is unstable (threshold extinction boundary).
- If $H = \frac{rK}{4}$: A saddle-node bifurcation occurs at the critical harvesting limit $P^* = K/2$.
- If $H > \frac{rK}{4}$: $\frac{dP}{dt} < 0$ for all $P$, causing catastrophic population collapse to zero in finite time.
ยง4.3 Falling Bodies Under Air Drag & Orthogonal Trajectories
1. Motion of a Falling Body Under Drag
By Newton's second law, $m \frac{dv}{dt} = \sum F = m g - F_{\text{drag}}$.
- Linear Drag (Stokes Regime, Low Reynolds Number): $F_{\text{drag}} = k v$. $$\frac{dv}{dt} = g - \frac{k}{m} v \implies v(t) = \frac{m g}{k}\left(1 - e^{-(k/m)t}\right) + v_0 e^{-(k/m)t}$$ The terminal velocity is $v_T = \lim_{t\to\infty} v(t) = \frac{mg}{k}$.
- Quadratic Drag (Newtonian Regime, High Reynolds Number): $F_{\text{drag}} = k v^2$. $$\frac{dv}{dt} = g - \frac{k}{m} v^2 = g\left(1 - \frac{v^2}{v_T^2}\right), \quad \text{where } v_T = \sqrt{\frac{mg}{k}}$$ Separating variables: $\int \frac{dv}{v_T^2 - v^2} = \frac{g}{v_T^2} dt \implies v(t) = v_T \tanh\left(\frac{g t}{v_T}\right)$ (for $v(0) = 0$).
2. Theory of Orthogonal Trajectories
An orthogonal trajectory is a curve that intersects every member of a given family of curves at right angles ($90^\circ$).
- Cartesian Coordinates: Given a family $F(x, y, C) = 0$, differentiate to find the differential equation $\frac{dy}{dx} = f(x, y)$. Since the product of perpendicular slopes is $-1$, the differential equation of the orthogonal family is: $$\left(\frac{dy}{dx}\right)_{\text{orth}} = -\frac{1}{f(x, y)}$$
- Polar Coordinates: Given a family $F(r, \theta, C) = 0$, differentiate to find $r \frac{d\theta}{dr} = f(r, \theta)$. The orthogonal family satisfies: $$\left(r \frac{d\theta}{dr}\right)_{\text{orth}} = -\frac{1}{f(r, \theta)} \implies \left(\frac{dr}{d\theta}\right)_{\text{orth}} = -r^2 \left(\frac{d\theta}{dr}\right)_{\text{orig}}$$
Step-by-Step Solved Examination Problems
Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.
A coroner arrives at a crime scene at 10:00 AM and measures a body's temperature to be $30^\circ\text{C}$. At 11:00 AM, the temperature has dropped to $28^\circ\text{C}$. The ambient room temperature is held strictly constant at $20^\circ\text{C}$. Assuming normal body temperature at death is $37^\circ\text{C}$, determine the exact time of death.
Step 1: Formulate the cooling model
$\frac{dT}{dt} = -k(T - T_m)$ with $T_m = 20^\circ\text{C}$.
The general solution is $T(t) = 20 + (T(0) - 20) e^{-kt}$.
Let $t = 0$ correspond to 10:00 AM, so $T(0) = 30^\circ\text{C}$.
Step 2: Determine cooling constant $k$
At 11:00 AM ($t = 1\text{ hr}$), $T(1) = 28^\circ\text{C}$:
Step 3: Solve for time of death $t_d < 0$
At time of death $t_d$, $T(t_d) = 37^\circ\text{C}$:
Step 4: Convert to clock time
$2.378\text{ hours} \approx 2\text{ hours and } 23\text{ minutes}$ prior to 10:00 AM.
Subtracting from 10:00 AM yields approximately 7:37 AM.
Estimated time of death: approximately 7:37 AM ($t_d \approx -2.38\text{ hr}$).
A tank initially holds $100\text{ gallons}$ of pure water. Brine containing $2\text{ lb/gal}$ of salt flows into the tank at $5\text{ gal/min}$, and the well-stirred mixture flows out at $3\text{ gal/min}$. Find the amount of salt in the tank at any time $t$, and calculate the exact salt concentration when the tank reaches $200\text{ gallons}$.
Step 1: Formulate the balance equation
Initial conditions: $V_0 = 100\text{ gal}$, $Q(0) = 0\text{ lb}$.
Flow rates: $r_{\text{in}} = 5\text{ gal/min}$, $c_{\text{in}} = 2\text{ lb/gal}$, $r_{\text{out}} = 3\text{ gal/min}$.
Volume at time $t$: $V(t) = 100 + (5 - 3)t = 100 + 2t\text{ gallons}$.
The differential equation for salt mass $Q(t)$ is:
Step 2: Integrating factor
Step 3: Solve for $Q(t)$
Integrating both sides:
Dividing by $(100 + 2t)^{3/2}$:
Step 4: Initial condition & concentration calculation
At $t = 0, Q(0) = 0$: $0 = 2(100) + C(100)^{-3/2} \implies C = -200 \times 1000 = -200000$.
The tank reaches $200\text{ gallons}$ when $100 + 2t = 200 \implies t = 50\text{ min}$.
The concentration is $c(50) = \frac{329.29\text{ lb}}{200\text{ gal}} \approx 1.646\text{ lb/gal}$.
$Q(t) = 2(100 + 2t) - 200000(100 + 2t)^{-3/2}\text{ lbs}$. Concentration at $V = 200\text{ gal}$: $\approx 1.65\text{ lb/gal}$.
Prove that the family of confocal parabolas having focus at the origin and axis along the $x$-axis:
is self-orthogonal (that is, its orthogonal trajectories are identical to the original family with an altered parameter).
Step 1: Obtain the differential equation of the given family
Given $y^2 = 4ax + 4a^2$. Differentiate with respect to $x$:
Step 2: Eliminate the parameter $a$
Substitute $a = \frac{1}{2}yp$ into $y^2 = 4ax + 4a^2$:
Divide by $y \neq 0$:
Step 3: Replace $\frac{dy}{dx}$ with $-\frac{dx}{dy}$ for the orthogonal family
For the orthogonal trajectories, replace $p = \frac{dy}{dx}$ with $-\frac{1}{p} = -\frac{dx}{dy}$:
Multiply the entire equation by $-p^2$:
Conclusion:
The differential equation for the orthogonal trajectories is algebraically identical to the original differential equation! Therefore, the family is self-orthogonal.
Proven. The differential equation $y(p^2 - 1) + 2xp = 0$ is invariant under the transformation $p \mapsto -1/p$, proving self-orthogonality.