Modeling with Differential Equations - Calculus 2

A spherical raindrop evaporates at a rate proportional to its surface area. Let $V(t)$ be the volume and $r(t)$ be the radius of the raindrop. Which differential equation describes the rate of change of the radius, $\frac{dr}{dt}$ ?
Answer: $\frac{dr}{dt} = -k$ The rate of evaporation is proportional to the surface area $A = 4\pi r^2$ . So, $\frac{dV}{dt} = -c A = -c(4\pi r^2)$ for some positive constant $c$ . The volume of a sphere is $V = \frac{4}{3}\pi r^3$ . Differentiating with respect to $t$ gives $\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) = 4\pi r^2 \frac{dr}{dt}$ . Equating the two expressions for $\frac{dV}{dt}$ gives $4\pi r^2 \frac{dr}{dt} = -c(4\pi r^2)$ . Dividing both sides by $4\pi r^2$ results in $\frac{dr}{dt} = -c$ . Letting $k=c$ , we find $\frac{dr}{dt} = -k$ , a constant.
A pendulum's angular displacement $\theta(t)$ from vertical satisfies $\frac{d^2\theta}{dt^2} + \frac{g}{L}\sin\theta = 0$ . For small oscillations, this is approximated as $\frac{d^2\theta}{dt^2} + \omega^2\theta = 0$ where $\omega^2 = \frac{g}{L}$ . If a pendulum has period $T = 4$ seconds under this approximation, and we want to double the period to 8 seconds, by what factor must we change the length $L$ ?
Answer: Increase $L$ by a factor of 4 When you encounter pendulum problems involving period and length relationships, you're dealing with the fundamental connection between these quantities in simple harmonic motion. The key insight is recognizing that for small oscillations, the period of a pendulum is $T = 2\pi\sqrt{\frac{L}{g}}$ . This comes from solving the differential equation $\frac{d^2\theta}{dt^2} + \omega^2\theta = 0$ where $\omega = \sqrt{\frac{g}{L}}$ , giving us $T = \frac{2\pi}{\omega}$ . To find how length affects period, let's set up a ratio. If the original period is $T_1 = 2\pi\sqrt{\frac{L_1}{g}}$ and the new period is $T_2 = 2\pi\sqrt{\frac{L_2}{g}}$ , then: $\frac{T_2}{T_1} = \sqrt{\frac{L_2}{L_1}}$ Since we want $T_2 = 8$ seconds and $T_1 = 4$ seconds: $\frac{8}{4} = 2 = \sqrt{\frac{L_2}{L_1}}$ Squaring both sides: $4 = \frac{L_2}{L_1}$ , so $L_2 = 4L_1$ . Choice A suggests doubling the length, but $T \propto \sqrt{L}$ , so doubling $L$ only increases $T$ by $\sqrt{2}$ . Choice B would give a period increase of $\sqrt{\sqrt{2}} \approx 1.19$ , far too small. Choice D represents $\sqrt{8}$ , which would give a period ratio of $\sqrt{\sqrt{8}} = 2^{3/4} \approx 1.68$ . Remember: period scales with the square root of length, so to change period by factor $n$ , you must change length by factor $n^2$ .
A tank initially contains 100 L of pure water. A brine solution with a salt concentration of 50 g/L is pumped into the tank at a rate of 2 L/min. The well-mixed solution is pumped out at a rate of 1 L/min. Let $A(t)$ be the amount of salt in the tank in grams at time $t$ in minutes. Which differential equation models this scenario?
Answer: $\frac{dA}{dt} = 100 - \frac{A}{100+t}$ The rate of change of salt, $\frac{dA}{dt}$ , is the rate of salt in minus the rate of salt out. Rate in = (concentration in) × (flow rate in) = (50 g/L) × (2 L/min) = 100 g/min. The volume of the solution in the tank at time $t$ is $V(t) = 100 + (2-1)t = 100+t$ L. The concentration of salt in the tank at time $t$ is $C(t) = \frac{A(t)}{V(t)} = \frac{A}{100+t}$ g/L. Rate out = (concentration out) × (flow rate out) = $\left(\frac{A}{100+t}\right) \times (1\text{ L/min}) = \frac{A}{100+t}$ g/min. Therefore, the differential equation is $\frac{dA}{dt} = 100 - \frac{A}{100+t}$ .
An object of mass $m$ is dropped from a great height. The force of gravity is $mg$ . The force due to air resistance is proportional to the square of the object's velocity, $v(t)$ . Let $c > 0$ be the constant of proportionality for the air resistance. Using Newton's Second Law ( $F_{net} = ma$ ) and taking the downward direction as positive, which equation models the velocity of the object?
Answer: $m\frac{dv}{dt} = mg - cv^2$ The net force $F_{net}$ is the sum of the forces acting on the object. The force of gravity is $mg$ acting downwards (positive direction). The force of air resistance is $cv^2$ acting upwards (negative direction), opposing the motion. Therefore, $F_{net} = mg - cv^2$ . According to Newton's Second Law, $F_{net} = ma = m\frac{dv}{dt}$ . Equating these gives $m\frac{dv}{dt} = mg - cv^2$ .
In a simple predator-prey model, let $R(t)$ be the population of rabbits and $W(t)$ be the population of wolves. In the absence of wolves, the rabbit population grows at a rate proportional to its current size. The presence of wolves decreases the rabbit population at a rate jointly proportional to the number of rabbits and the number of wolves. Using $a$ and $b$ as positive constants, which equation models the rabbit population?
Answer: $\frac{dR}{dt} = aR - bRW$ The natural growth of the rabbit population is proportional to its size, giving the term $aR$ , where $a > 0$ . The predation by wolves decreases the rabbit population. This interaction rate is 'jointly proportional' to both populations, meaning it is proportional to their product, $RW$ . Since this decreases the rabbit population, the term is $-bRW$ , where $b > 0$ . The complete model is $\frac{dR}{dt} = aR - bRW$ .
The price $P(t)$ of a certain commodity changes at a rate proportional to the excess demand (Demand - Supply). The demand for the commodity is given by $D(P) = 1000 - 2P$ , and the supply is given by $S(P) = 100 + 3P$ . Let $k > 0$ be the proportionality constant. Which differential equation models the price of the commodity?
Answer: $\frac{dP}{dt} = k(900 - 5P)$ The rate of change of price is proportional to the excess demand, so $\frac{dP}{dt} = k(D - S)$ . First, calculate the excess demand: $D - S = (1000 - 2P) - (100 + 3P) = 1000 - 2P - 100 - 3P = 900 - 5P$ . Substituting this into the proportionality relationship gives $\frac{dP}{dt} = k(900 - 5P)$ .
In an SIR model of an epidemic, $S(t)$ is the number of susceptible people, $I(t)$ is the number of infected people, and $R(t)$ is the number of recovered people. The rate of new infections is proportional to the product of $S$ and $I$ . Infected people recover at a rate proportional to the number of infected people. Let $\beta$ be the infection rate constant and $\gamma$ be the recovery rate constant. Which equation models the rate of change of the infected population, $\frac{dI}{dt}$ ?
Answer: $\frac{dI}{dt} = \beta SI - \gamma I$ The infected population $I$ increases as susceptible people get infected. This happens at a rate of $\beta SI$ . The infected population decreases as people recover. This happens at a rate proportional to the number of infected people, $\gamma I$ . The net rate of change for the infected population is the rate of new infections minus the rate of recovery, which gives $\frac{dI}{dt} = \beta SI - \gamma I$ .
A patient receives a drug intravenously at a constant rate of $R$ mg/hour. The body eliminates the drug at a rate proportional to the amount, $A(t)$ , present in the bloodstream. Let $k > 0$ be the elimination rate constant. Which differential equation models the amount of the drug in the bloodstream?
Answer: $\frac{dA}{dt} = R - kA$ The rate of change, $\frac{dA}{dt}$ , is the rate of administration minus the rate of elimination. The drug is administered at a constant rate $R$ . The body eliminates the drug at a rate proportional to the amount present, which means the elimination rate is $kA$ . Since this process removes the drug, it is a negative term. Therefore, the overall model is $\frac{dA}{dt} = R - kA$ .
Water is draining from a hole in the bottom of a large cylindrical tank with a constant base radius $r$ . According to Torricelli's Law, the rate of change of the volume of water, $\frac{dV}{dt}$ , is proportional to the square root of the water's height, $h(t)$ . Let $c > 0$ be the constant of proportionality. Which differential equation models the height of the water in the tank?
Answer: $\frac{dh}{dt} = -\frac{c\sqrt{h}}{\pi r^2}$ We are given that $\frac{dV}{dt} = -c\sqrt{h}$ , where the negative sign indicates the volume is decreasing. The volume of water in a cylinder is $V = \pi r^2 h$ . Differentiating with respect to time $t$ , we get $\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$ since $r$ is constant. Equating the two expressions for $\frac{dV}{dt}$ gives $\pi r^2 \frac{dh}{dt} = -c\sqrt{h}$ . Solving for $\frac{dh}{dt}$ yields $\frac{dh}{dt} = -\frac{c\sqrt{h}}{\pi r^2}$ .
A bucket is filled with water at a constant rate of 30 cm³/s. Simultaneously, water leaks from a small hole in the bottom at a rate proportional to the current volume of water, $V(t)$ , in the bucket. Let $k > 0$ be the proportionality constant for the leak. Which differential equation models the volume of water in the bucket?
Answer: $\frac{dV}{dt} = 30 - kV$ The rate of change of volume, $\frac{dV}{dt}$ , is the rate of inflow minus the rate of outflow. The inflow rate is constant at 30 cm³/s. The outflow (leak) rate is proportional to the volume, so it can be written as $kV$ . Since this is an outflow, it decreases the volume. Therefore, the model is $\frac{dV}{dt} = 30 - kV$ .
According to Kirchhoff's Voltage Law, the sum of voltage drops around a closed circuit equals the source voltage. In a series circuit containing a resistor with resistance $R$ , an inductor with inductance $L$ , and a constant voltage source $E$ , the voltage drop across the resistor is $RI(t)$ and across the inductor is $L\frac{dI}{dt}$ , where $I(t)$ is the current. Which differential equation models the current in this RL circuit?
Answer: $L\frac{dI}{dt} + RI = E$ Kirchhoff's Voltage Law states that the sum of the voltage drops across the components equals the voltage of the source. The voltage drop across the resistor is $V_R = RI$ . The voltage drop across the inductor is $V_L = L\frac{dI}{dt}$ . Therefore, $V_R + V_L = E$ , which gives the equation $RI + L\frac{dI}{dt} = E$ , commonly written as $L\frac{dI}{dt} + RI = E$ .
In a radioactive decay chain, substance A decays into substance B at a rate proportional to the amount of A, with constant $k_1 > 0$ . Substance B then decays into a stable substance C at a rate proportional to the amount of B, with constant $k_2 > 0$ . Let $A(t)$ and $B(t)$ be the amounts of substances A and B at time $t$ . Which equation models the amount of substance B?
Answer: $\frac{dB}{dt} = k_1 A - k_2 B$ The amount of substance B changes due to two processes. It is created from the decay of A, and it is consumed by its own decay into C. The rate of creation of B is the rate at which A decays, which is $k_1 A$ . This is a positive term for $\frac{dB}{dt}$ . The rate of decay of B is proportional to its own amount, which is $k_2 B$ . Since this process consumes B, it is a negative term for $\frac{dB}{dt}$ . Combining these gives $\frac{dB}{dt} = k_1 A - k_2 B$ .
Two chemicals, X and Y, react to form a substance Z. For every 1 gram of Z produced, 0.4 g of X and 0.6 g of Y are consumed. The rate of formation of Z is proportional to the product of the amounts of X and Y remaining. Initially, there are 100 g of X and 150 g of Y. Let $z(t)$ be the amount of Z in grams at time $t$ . Which equation models this reaction?
Answer: $\frac{dz}{dt} = k(100 - 0.4z)(150 - 0.6z)$ Let $z(t)$ be the amount of Z. The amount of X consumed is $0.4z$ , and the amount of Y consumed is $0.6z$ . The amounts remaining are $X(t) = 100 - 0.4z$ and $Y(t) = 150 - 0.6z$ . The rate of formation of Z, $\frac{dz}{dt}$ , is proportional to the product of the remaining amounts: $\frac{dz}{dt} = k(X(t))(Y(t)) = k(100 - 0.4z)(150 - 0.6z)$ .
A predator-prey system is modeled by $\frac{dx}{dt} = ax - bxy$ and $\frac{dy}{dt} = -cy + dxy$ , where $x$ is prey population, $y$ is predator population, and $a, b, c, d > 0$ . At an equilibrium point where both populations are positive and constant, which relationship must hold between the parameters?
Answer: The equilibrium occurs at $x = \frac{c}{d}$ and $y = \frac{a}{b}$ regardless of initial conditions When you encounter a predator-prey system, you're looking at a classic application of differential equations where equilibrium occurs when both derivative equations simultaneously equal zero. To find equilibrium points, set both $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} = 0$ . From the first equation: $ax - bxy = 0$ , which factors as $x(a - by) = 0$ . This gives us either $x = 0$ or $y = \frac{a}{b}$ . From the second equation: $-cy + dxy = 0$ , which factors as $y(-c + dx) = 0$ . This gives us either $y = 0$ or $x = \frac{c}{d}$ . Since we need both populations to be positive (not zero), we must have $x = \frac{c}{d}$ and $y = \frac{a}{b}$ . Notice that these values depend only on the system parameters, not on any relationship between them or initial conditions. Option A incorrectly suggests the equilibrium populations are equal to $\sqrt{ac}$ and requires $\frac{a}{b} = \frac{c}{d}$ , which isn't necessary. Option B provides wrong formulas for both the equilibrium populations and incorrectly states that $ad = bc$ is required. Option D has the equilibrium populations backwards ( $x$ and $y$ are swapped) and unnecessarily requires $ab = cd$ . The key insight is that predator-prey equilibrium depends only on the individual parameter ratios, not on relationships between parameters. Remember: set derivatives to zero and solve systematically, keeping track of which variable corresponds to which population.
A metal object's temperature $T(t)$ changes over time. According to Newton's Law of Cooling, its rate of change is proportional to the difference between its temperature and the ambient temperature. The object is placed in a special chamber where the ambient temperature, $A(t)$ , is not constant but increases linearly with time according to $A(t) = 20 + 0.5t$ , where $t$ is in minutes. If the object is cooling and $k$ is a positive constant of proportionality, which equation models the object's temperature?
Answer: $\frac{dT}{dt} = -k(T - (20 + 0.5t))$ Newton's Law of Cooling states $\frac{dT}{dt}$ is proportional to $T - A$ . So, $\frac{dT}{dt} = C(T - A(t))$ . Since the object is cooling, $T > A(t)$ and $\frac{dT}{dt} < 0$ . This means the constant of proportionality $C$ must be negative. Letting $C = -k$ where $k$ is a positive constant gives $\frac{dT}{dt} = -k(T - A(t))$ . Substituting $A(t) = 20 + 0.5t$ gives $\frac{dT}{dt} = -k(T - (20 + 0.5t))$ .
In a fixed population of $N$ individuals, the rate at which a rumor spreads is jointly proportional to the number of people who have heard the rumor, $P(t)$ , and the number of people who have not yet heard it. Which differential equation describes the spread of the rumor?
Answer: $\frac{dP}{dt} = kP(N-P)$ The number of people who have heard the rumor is $P$ . The number who have not heard it is $N-P$ . The phrase 'jointly proportional' means the rate, $\frac{dP}{dt}$ , is proportional to the product of these two quantities. Therefore, $\frac{dP}{dt} = kP(N-P)$ for some proportionality constant $k$ . This is the standard logistic growth model, often used for such scenarios.
A certain population's growth is modeled by the Gompertz equation, where the relative growth rate decreases over time. Specifically, the rate of change of the population $P(t)$ is proportional to the current population, but the proportionality factor is an exponentially decaying function of time, $k(t) = r e^{-\alpha t}$ , where $r$ and $\alpha$ are positive constants. Which equation describes this model?
Answer: $\frac{dP}{dt} = r P e^{-\alpha t}$ The problem states that the rate of change, $\frac{dP}{dt}$ , is proportional to the population $P$ , so $\frac{dP}{dt} = k(t)P$ . It then gives the specific form of the proportionality factor: $k(t) = r e^{-\alpha t}$ . Substituting this into the equation gives $\frac{dP}{dt} = (r e^{-\alpha t})P$ , or $\frac{dP}{dt} = rPe^{-\alpha t}$ .
A student is memorizing a list of $M$ new terms. The rate of memorization is proportional to the number of terms not yet memorized. Concurrently, the student forgets terms at a rate proportional to the number of terms they have already memorized. Let $N(t)$ be the number of terms memorized at time $t$ , and let $k_1$ and $k_2$ be positive constants for memorization and forgetting, respectively. Which equation models $N(t)$ ?
Answer: $\frac{dN}{dt} = k_1(M-N) - k_2N$ The rate of change $\frac{dN}{dt}$ has two parts. The rate of memorization adds to $N$ . It is proportional to the number of terms not yet memorized, which is $M-N$ . So, this term is $k_1(M-N)$ . The rate of forgetting subtracts from $N$ . It is proportional to the number of terms already memorized, $N$ . So, this term is $-k_2N$ . Combining these two effects gives the overall rate: $\frac{dN}{dt} = k_1(M-N) - k_2N$ .
A skydiver of mass $m$ falls from rest. The force of air resistance is proportional to the product of the air density, $\rho$ , and the square of the velocity, $v$ . As the skydiver descends, the air density increases with time, modeled by $\rho(t) = ho_0 + at$ for positive constants $\rho_0$ and $a$ . Taking downward as positive, which equation models the skydiver's velocity?
Answer: $m\frac{dv}{dt} = mg - k(\rho_0 + at)v^2$ Using Newton's Second Law, $m\frac{dv}{dt} = F_{net}$ . The net force is gravity minus air resistance. Gravity is $mg$ . Air resistance is $F_{drag} = k \rho v^2$ for some constant $k$ . Substituting the time-dependent density gives $F_{drag} = k(\rho_0 + at)v^2$ . The force of air resistance opposes motion, so it is a negative term. Therefore, the complete model is $m\frac{dv}{dt} = mg - k(\rho_0 + at)v^2$ .
A spherical raindrop falls through air and grows by condensation. Its radius increases at a rate proportional to its surface area, and it falls with velocity proportional to its radius. If $r(t)$ is the radius at time $t$ , which system of differential equations best models this situation?
Answer: $\frac{dr}{dt} = kr^2$ , $v = cr$ The radius increases at a rate proportional to surface area: $\frac{dr}{dt} = k \cdot 4\pi r^2 = kr^2$ (absorbing $4\pi$ into $k$ ). The velocity is proportional to radius: $v = cr$ . This is not a rate equation but a direct relationship. Choice A incorrectly models velocity with acceleration. Choice C uses wrong proportionality for radius growth. Choice D assumes constant radius growth rate.

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