Setting Up DEs from Word Problems - Differential Equations

In a population of $N$ individuals, the number of people who have a certain piece of information is $I(t)$ . The information spreads at a rate proportional to the product of those who have the information and those who do not. Additionally, those who have the information can forget it and become uninformed again, at a rate proportional to the square root of the number of informed people. Let $k$ be the transmission rate constant and $\gamma$ be the forgetting rate constant. Which differential equation models $I(t)$ ?
Answer: $\frac{dI}{dt} = kI(N-I) - \gamma \sqrt{I}$ The rate of change, $dI/dt$ , is the rate of spreading minus the rate of forgetting. The rate of spreading is proportional to the product of the number of informed people, $I$ , and the number of uninformed people, $S = N-I$ . So, the spreading term is $kI(N-I)$ . The rate of forgetting is stated to be proportional to the square root of the number of informed people, which is $\gamma\sqrt{I}$ . Since forgetting decreases the number of informed people, this term is subtracted. Combining these gives the equation: $dI/dt = kI(N-I) - \gamma\sqrt{I}$ .
A pendulum's angular displacement $\theta(t)$ from vertical satisfies the nonlinear equation $\frac{d^2\theta}{dt^2} + \frac{g}{L}\sin\theta = 0$ for small oscillations, where $g = 9.8$ m/s² and $L = 2$ m. However, the pendulum also experiences air resistance proportional to angular velocity with damping coefficient $c = 0.5$ , and a small periodic driving force $F_0\cos(\omega t)$ with $F_0 = 0.1$ and $\omega = 2$ . Which equation correctly models this driven, damped pendulum using the small-angle approximation $\sin\theta \approx \theta$ ?
Answer: $\frac{d^2\theta}{dt^2} + 0.5\frac{d\theta}{dt} + 4.9\theta = 0.1\cos(2t)$ When you encounter pendulum problems with multiple forces, you need to systematically build the differential equation by identifying each physical effect and its mathematical representation. Start with the basic pendulum equation $\frac{d^2\theta}{dt^2} + \frac{g}{L}\sin\theta = 0$ . The restoring force coefficient is $\frac{g}{L} = \frac{9.8}{2} = 4.9$ . Since you're told to use the small-angle approximation $\sin\theta \approx \theta$ , this becomes $\frac{d^2\theta}{dt^2} + 4.9\theta = 0$ . Next, add the damping term. Air resistance proportional to angular velocity contributes $+c\frac{d\theta}{dt} = +0.5\frac{d\theta}{dt}$ to the left side. Finally, the periodic driving force $F_0\cos(\omega t) = 0.1\cos(2t)$ appears on the right side as an external forcing term. Combining everything: $\frac{d^2\theta}{dt^2} + 0.5\frac{d\theta}{dt} + 4.9\theta = 0.1\cos(2t)$ , which is answer C. Let's examine why the other options fail: Answer A incorrectly uses $g = 9.8$ instead of calculating $\frac{g}{L} = 4.9$ . Answer B keeps $\sin\theta$ instead of applying the required small-angle approximation $\sin\theta \approx \theta$ . Answer D scrambles the damping and restoring force terms, placing the velocity term with $\theta$ and the displacement term with $\frac{d\theta}{dt}$ . Study tip: For driven, damped oscillator problems, always follow the pattern: acceleration + damping×velocity + restoring×displacement = driving force. Double-check that coefficients match the physical setup and any approximations are properly applied.
An epidemic spreads through a population of 10,000 people. The rate of new infections is proportional to the product of infected individuals $I(t)$ and susceptible individuals $S(t)$ , with transmission rate $\beta = 0.0001$ per person per day. Infected individuals recover at rate $\gamma = 0.05$ per day, becoming immune. Additionally, susceptible individuals can become immune through vaccination at a constant rate of 50 people per day, regardless of infection status. If $R(t)$ represents recovered/immune individuals, which system models this SIR epidemic with vaccination?
Answer: $\frac{dS}{dt} = -0.0001SI - 50, \quad \frac{dI}{dt} = 0.0001SI - 0.05I, \quad \frac{dR}{dt} = 0.05I + 50$ When modeling epidemic dynamics with differential equations, you need to carefully track how individuals flow between compartments (Susceptible → Infected → Recovered) and account for all processes affecting each population. Let's build the system step by step. The susceptible population $S(t)$ decreases through two processes: infection (rate $\beta SI = 0.0001SI$ ) and vaccination (constant 50 people/day). So $\frac{dS}{dt} = -0.0001SI - 50$ . The infected population $I(t)$ increases from new infections ( $+0.0001SI$ ) and decreases from recovery ( $-0.05I$ ). Vaccination doesn't directly affect the infected group since those 50 people come from the susceptible population. Therefore: $\frac{dI}{dt} = 0.0001SI - 0.05I$ . The recovered/immune population $R(t)$ gains people through natural recovery ( $+0.05I$ ) and vaccination ( $+50$ ). So $\frac{dR}{dt} = 0.05I + 50$ . Choice B correctly captures this flow: susceptible individuals leave through infection and vaccination, infected individuals are created by infection and removed by recovery, and recovered individuals are created by both recovery and vaccination. Choice A incorrectly subtracts vaccination from the infection rate rather than the susceptible population. Choice C mistakenly adds vaccination to the infected population, which makes no biological sense. Choice D omits vaccination from the recovered equation entirely. Study tip: Always verify that your SIR system conserves population (the derivatives should sum to zero when there's no birth/death) and that each process appears with correct signs in the appropriate equations.
An object of mass $m$ falls from rest. The force of gravity is $mg$ . Air resistance is proportional to the object's velocity $v(t)$ , with a proportionality constant $k>0$ . The object also has a small rocket attached which provides a constant upward thrust $T$ . Let the downward direction be positive. Which differential equation describes the velocity $v(t)$ of the object?
Answer: $m\frac{dv}{dt} = mg - kv - T$ When solving physics problems involving forces and motion, you need to carefully apply Newton's second law: $F_{net} = ma$ , where the net force equals mass times acceleration. Since acceleration is $\frac{dv}{dt}$ , you're looking for an equation of the form $m\frac{dv}{dt} = F_{net}$ . To find the net force, identify each force and its direction. Since downward is positive, gravity acts as $+mg$ (positive). Air resistance always opposes motion, so when the object moves downward (positive velocity), air resistance acts upward (negative direction), giving $-kv$ . The rocket thrust $T$ acts upward, so it contributes $-T$ to the net force. Therefore: $m\frac{dv}{dt} = mg - kv - T$ , which is answer C. Here's why the other options are wrong: Option A incorrectly shows thrust as $+T$ , treating it as if it acts downward when it actually provides upward force. Option B makes two errors: it shows air resistance as $+kv$ (as if it helps the object fall faster rather than opposing motion) and thrust as $+T$ (downward instead of upward). Option D gets the signs completely backwards, showing gravity as negative when downward should be positive according to the given coordinate system. Study tip: Always establish your coordinate system first, then systematically go through each force, asking "Which direction does this force point?" and "What sign does that direction get in my coordinate system?" This methodical approach prevents sign errors, which are the most common mistakes in force problems.
The concentration of a pollutant in a lake is being reduced by a natural cleaning process. The rate of reduction is proportional to the current concentration. However, a nearby factory discharges more pollutant into the lake at a rate $D(t) = D_0e^{-\alpha t}$ . Let $k > 0$ be the proportionality constant relating the removal rate to the amount of pollutant. Which differential equation models the amount of pollutant $A(t)$ in the lake?
Answer: $\frac{dA}{dt} = -k A + D_0 e^{-\alpha t}$ When modeling population dynamics or concentration changes, you need to carefully account for all inflows and outflows affecting the system. The key is translating each physical process into its mathematical representation. The rate of change $\frac{dA}{dt}$ equals the difference between what's being added and what's being removed. Here, two processes affect the pollutant amount: natural removal (proportional to current concentration) and factory discharge. The natural cleaning removes pollutant at a rate proportional to the current amount, so this contributes $-kA$ to the rate equation (negative because it decreases the total). The factory adds pollutant at rate $D_0e^{-\alpha t}$ , contributing positively. Therefore: $\frac{dA}{dt} = -kA + D_0e^{-\alpha t}$ , which is answer C. Answer A incorrectly makes the factory discharge negative, suggesting it removes rather than adds pollutant. This contradicts the problem statement that the factory "discharges more pollutant into the lake." Answer B introduces an unnecessary volume term $V$ in the denominator. While concentration equals amount divided by volume, the problem states that removal is proportional to concentration, which means it's proportional to $A/V$ . But since we're modeling the amount $A$ , not concentration, and $k$ already incorporates any volume effects, this extra division is incorrect. Answer D has the wrong sign on the removal term, making it $+kA$ instead of $-kA$ . This would mean more pollutant causes faster accumulation rather than faster removal. Study tip: In rate problems, carefully identify each process as adding (+) or removing (-) from your quantity of interest, then sum all contributions.
A predator-prey system involves foxes (F) and rabbits (R). Rabbits grow exponentially at rate $a = 0.8$ per month in the absence of foxes, but are consumed by foxes at a rate proportional to the product $FR$ with constant $b = 0.03$ . Foxes die exponentially at rate $c = 0.6$ per month without rabbits, but their population increases due to consuming rabbits at a rate proportional to $FR$ with efficiency constant $d = 0.02$ . However, environmental carrying capacity limits rabbit growth when $R > 100$ , reducing growth by a factor of $(1 - R/200)$ . Which system correctly models this situation?
Answer: $\frac{dR}{dt} = 0.8R(1 - R/200) - 0.03FR, \quad \frac{dF}{dt} = -0.6F + 0.02FR$ When you encounter predator-prey problems, you need to carefully translate each biological process into mathematical terms, paying close attention to what affects each population's growth rate. Let's build the correct model step by step. For rabbits: they grow exponentially at rate 0.8 (giving us $0.8R$ ), lose individuals through predation at rate $0.03FR$ , and face carrying capacity limits that reduce growth by factor $(1 - R/200)$ . The carrying capacity affects the natural growth term, not predation, so we get: $\frac{dR}{dt} = 0.8R(1 - R/200) - 0.03FR$ . For foxes: they die at rate 0.6 without food (giving us $-0.6F$ ) and gain population through successful predation at rate $0.02FR$ . No carrying capacity affects foxes, so: $\frac{dF}{dt} = -0.6F + 0.02FR$ . Now let's examine why the other options fail: Option A incorrectly applies the carrying capacity factor to the fox equation rather than the rabbit growth term, and doesn't properly model logistic growth for rabbits. Option B adds an incorrect $-R^2/200$ term to rabbits (this isn't how carrying capacity was described) and omits the carrying capacity factor from the natural growth term. Option D uses the wrong carrying capacity parameter ( $R/100$ instead of $R/200$ ), which would make the effective carrying capacity 100 instead of 200. Study tip: In predator-prey problems, always identify which population each constraint affects. Carrying capacity typically modifies natural growth rates through logistic factors like $(1 - N/K)$ , where $K$ is the carrying capacity.
A cylindrical tank of height $H$ and cross-sectional area $A$ is initially full of water. At time $t=0$ , a valve at the bottom is opened. Water flows out at a rate proportional to the current water pressure at the bottom of the tank. The pressure at the bottom is proportional to the depth of the water. Let $h(t)$ be the height of the water at time $t$ . Which differential equation describes $h(t)$ ? Let $k>0$ be the final proportionality constant.
Answer: $\frac{dh}{dt} = -kh$ When modeling fluid flow from tanks, you need to connect physical relationships to create the differential equation. The key insight is translating "rate proportional to pressure" and "pressure proportional to depth" into mathematical form. Start with the physical setup: water flows out at a rate proportional to pressure at the bottom, and this pressure is proportional to the water depth $h(t)$ . So the outflow rate is proportional to $h(t)$ : outflow rate $= kh$ for some constant $k > 0$ . Now convert to the differential equation. The rate of change of water height equals the rate water flows in minus the rate it flows out. Since no water flows in and water flows out at rate $kh$ , you get: $\frac{dh}{dt} = 0 - kh = -kh$ This gives us answer C. Looking at the wrong answers: Answer A gives $\frac{dh}{dt} = -kh^2$ , which would result from outflow rate being proportional to the square of the depth - not what the problem states. Answer B gives $\frac{dh}{dt} = -k\sqrt{h}$ , which actually describes Torricelli's law where outflow depends on $\sqrt{h}$ due to gravitational effects, but that's not the relationship described here. Answer D gives $\frac{dh}{dt} = -k$ , representing constant outflow rate regardless of water level - this ignores the pressure dependence entirely. Study tip: In tank problems, always identify what the outflow rate depends on first, then translate that physical relationship directly into the differential equation. The rate of change of height equals inflow minus outflow.
The population of a species, $P(t)$ , is modeled by the logistic equation with a growth rate $r$ and carrying capacity $K$ . The species is subject to harvesting. The harvesting rate is not constant; it is proportional to the current population, with a proportionality constant $h$ that oscillates seasonally according to $h(t) = h_0(1 + \cos(\omega t))$ . Which differential equation describes the population $P(t)$ ?
Answer: $\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right) - h_0(1+\cos(\omega t))P$ The logistic growth model without harvesting is $dP/dt = rP(1 - P/K)$ . Harvesting reduces the population, so a negative term must be added. The problem states the harvesting rate is proportional to the current population, with the proportionality constant being $h(t) = h_0(1 + \cos(\omega t))$ . Therefore, the harvesting term is $h(t)P = h_0(1 + \cos(\omega t))P$ . Combining these gives the full model: $dP/dt = rP(1 - P/K) - h_0(1 + \cos(\omega t))P$ .
A student is learning a new skill. Let $P(t)$ be their proficiency level at time $t$ , with $P=0$ representing no skill and $P=1$ representing mastery. The rate of learning is proportional to the product of their current proficiency and the amount of proficiency they have yet to gain. Let the constant of proportionality be $k>0$ . Which differential equation models the student's proficiency $P(t)$ ?
Answer: $\frac{dP}{dt} = kP(1-P)$ When you encounter word problems asking you to translate descriptions into differential equations, your key task is converting each phrase into mathematical language. Pay special attention to how rates of change depend on the current state of the system. The problem states that the learning rate is proportional to the product of two quantities: current proficiency $P$ and "the amount of proficiency they have yet to gain." If mastery is $P = 1$ and current proficiency is $P$ , then the remaining proficiency to gain is $(1-P)$ . Since the rate is proportional to their product with constant $k$ , we get $\frac{dP}{dt} = kP(1-P)$ Looking at the wrong answers: Choice A gives $\frac{dP}{dt} = k(1-P)$ , which makes learning rate depend only on remaining proficiency, ignoring current skill level entirely. Choice B has $\frac{dP}{dt} = kP(P-1)$ , where $(P-1)$ is negative since $P < 1$ , making $\frac{dP}{dt} < 0$ and causing proficiency to decrease over time—the opposite of learning. Choice C shows $\frac{dP}{dt} = kP$ , representing exponential growth that ignores the constraint that proficiency can't exceed mastery level. The correct answer D captures the logistic growth model, where learning accelerates as you gain skill but slows as you approach mastery. Notice how $\frac{dP}{dt} = 0$ when $P = 0$ (no learning without some skill) or $P = 1$ (no learning once mastery is achieved). Study tip: When translating rate problems, identify each factor mentioned in the description and ensure your equation includes all of them mathematically.
A drug is administered intravenously at a constant rate of 5 mg/hour into a patient's bloodstream. The drug is eliminated from the body according to first-order kinetics with elimination constant $k = 0.2$ per hour. The drug also binds reversibly to proteins in the blood: free drug becomes bound at rate $k_b[D_f]$ and bound drug becomes free at rate $k_u[D_b]$ , where $[D_f]$ and $[D_b]$ are concentrations of free and bound drug, respectively. If $k_b = 0.8$ per hour and $k_u = 0.3$ per hour, and only free drug is eliminated, which equation correctly models the free drug concentration?
Answer: $\frac{d[D_f]}{dt} = 5 - 0.2[D_f] - 0.8[D_f] + 0.3[D_b]$ When modeling drug concentrations with binding and elimination, you need to carefully track all the processes that add to or remove free drug from the bloodstream. Think systematically about each rate term and its sign. For free drug concentration $[D_f]$ , four processes are occurring simultaneously: - Input: Constant IV infusion adds 5 mg/hour (positive term: +5) - Elimination: Only free drug is eliminated at rate $k[D_f] = 0.2[D_f]$ (negative term: -0.2[ $D_f$ ]) - Binding: Free drug becomes bound at rate $k_b[D_f] = 0.8[D_f]$ (negative term: -0.8[ $D_f$ ]) - Unbinding: Bound drug becomes free at rate $k_u[D_b] = 0.3[D_b]$ (positive term: +0.3[ $D_b$ ]) This gives: $\frac{d[D_f]}{dt} = 5 - 0.2[D_f] - 0.8[D_f] + 0.3[D_b]$ Answer B is correct because it includes exactly these four terms with proper signs. Answer A incorrectly subtracts $0.2[D_b]$ , suggesting bound drug is also eliminated, which contradicts the problem statement that only free drug is eliminated. Answer C factors the free drug loss terms but is mathematically equivalent to B, making B the more standard form. Answer D incorrectly applies elimination to both free and bound drug by writing $0.2([D_f] + [D_b])$ , again violating the condition that only free drug is eliminated. Study tip: In compartmental models, systematically identify every process affecting your variable of interest, assign the correct sign (input = positive, output = negative), and double-check that your terms match the problem's constraints about which processes affect which compartments.
In a reversible chemical reaction, substance A converts to substance B at a rate $k_1[A]$ , and substance B converts back to A at a rate $k_2[B]^2$ , where $[X]$ denotes the concentration of substance X. The total concentration $[A] + [B]$ is a constant $C$ . Let $x(t) = [A]$ . Which differential equation describes the concentration of substance A?
Answer: $\frac{dx}{dt} = -k_1 x + k_2 (C-x)^2$ The rate of change of the concentration of A, $dx/dt$ , is the rate at which A is formed minus the rate at which A is consumed. A is consumed in the forward reaction $A \rightarrow B$ at a rate of $k_1[A] = k_1x$ . A is formed in the reverse reaction $B \rightarrow A$ at a rate of $k_2[B]^2$ . Since $[A] + [B] = C$ , we have $[B] = C - [A] = C-x$ . So, the rate of formation of A is $k_2(C-x)^2$ . Combining these gives the net rate of change for A: $dx/dt = (\text{rate of formation}) - (\text{rate of consumption}) = k_2(C-x)^2 - k_1x$ .
The value of an investment, $V(t)$ , grows due to interest at a rate proportional to its current value, with a constant of proportionality $r$ . In addition, funds are continuously withdrawn at a rate proportional to the square of the investment's value, with a constant $k$ . A constant amount $D$ is also deposited continuously into the account per year. Which differential equation models the value of the investment $V(t)$ ?
Answer: $\frac{dV}{dt} = rV - kV^2 + D$ When modeling real-world scenarios with differential equations, you need to carefully translate each described rate into mathematical terms, paying close attention to whether each effect increases or decreases the quantity. Let's build this equation piece by piece. The investment grows due to interest "at a rate proportional to its current value" with proportionality constant $r$ . This gives us $+rV$ (positive because it's growth). Funds are "continuously withdrawn at a rate proportional to the square of the investment's value" with constant $k$ . Since withdrawal decreases the investment, this contributes $-kV^2$ (negative because it's a loss). Finally, a constant amount $D$ is deposited continuously, contributing $+D$ . Combining these three effects: $\frac{dV}{dt} = rV - kV^2 + D$ Choice A incorrectly uses $-k\sqrt{V}$ instead of $-kV^2$ . The problem specifically states the withdrawal rate is proportional to the square of the value, not the square root. Choice B has $+kV^2$ instead of $-kV^2$ , treating withdrawals as deposits, and $-D$ instead of $+D$ , treating deposits as withdrawals—both signs are backwards. Choice C uses just $r$ instead of $rV$ , missing that interest growth is proportional to the current value, not constant. Study tip: When translating word problems into differential equations, identify each rate of change separately, determine whether it increases (+) or decreases (-) your quantity, then combine all terms. Words like "withdrawn," "lost," or "decays" signal negative terms, while "deposited," "grows," or "increases" signal positive terms.
A tank initially contains 100 gallons of pure water. Brine containing 2 pounds of salt per gallon flows into the tank at a rate of 3 gallons per minute, while the well-mixed solution flows out at a rate of 2 gallons per minute. If $S(t)$ represents the amount of salt in the tank at time $t$ minutes, which differential equation correctly models this situation?
Answer: $\frac{dS}{dt} = 6 - \frac{2S}{100 + t}$ The rate of change of salt equals salt in minus salt out. Salt flows in at (2 lb/gal)(3 gal/min) = 6 lb/min. The tank volume at time t is 100 + (3-2)t = 100 + t gallons. Salt flows out at rate (S/(100+t) lb/gal)(2 gal/min) = 2S/(100+t) lb/min. Thus dS/dt = 6 - 2S/(100+t). Choice B incorrectly uses 100+2t for volume (using outflow rate instead of net rate). Choice C uses wrong outflow rate coefficient. Choice D uses inflow rate instead of salt inflow rate.
A chemical reaction converts substance A to substance B at a rate proportional to the amount of A present. Simultaneously, substance B decomposes back to A at a rate proportional to the amount of B present. If $[A](t)$ and $[B](t)$ represent concentrations at time $t$ , the forward reaction rate constant is $k_f = 0.3$ , the reverse reaction rate constant is $k_r = 0.1$ , and the total concentration $[A] + [B] = C$ (constant), which single differential equation in $[A]$ only captures this system?
Answer: $\frac{d[A]}{dt} = -0.3[A] + 0.1(C - [A])$ When you encounter a reversible chemical reaction problem, you need to carefully track the direction and magnitude of each reaction step, then combine them to find the net rate of change. Let's build the differential equation systematically. Substance A decreases due to the forward reaction (A → B) at rate $k_f[A] = 0.3[A]$ , so this contributes $-0.3[A]$ to $\frac{d[A]}{dt}$ . Simultaneously, A increases due to the reverse reaction (B → A) at rate $k_r[B] = 0.1[B]$ , contributing $+0.1[B]$ to $\frac{d[A]}{dt}$ . Since the total concentration is constant ( $[A] + [B] = C$ ), we can substitute $[B] = C - [A]$ to eliminate B from our equation. This gives us: $\frac{d[A]}{dt} = -0.3[A] + 0.1(C - [A])$ This matches answer choice D exactly. Option A has the wrong signs—it suggests A increases from the forward reaction, which contradicts the problem statement. Option B correctly identifies the individual reaction terms but doesn't eliminate [B] as required for a single equation in [A] only. Option C appears to combine terms prematurely: $-0.3[A] + 0.1(C - [A]) = -0.3[A] + 0.1C - 0.1[A] = -0.4[A] + 0.1C$ , not $-0.2[A] + 0.1C$ . Remember: in coupled reaction problems, always write the rate equation for each individual process first, then use conservation laws (like constant total concentration) to eliminate unwanted variables. Don't try to simplify too early—work systematically through each step.
An inverted conical tank with a height of 4 meters and a top radius of 2 meters is filled with water. Water leaks out through a small hole of area $a$ at the vertex. According to Torricelli's law, the rate of change of the volume of water, $V$ , is given by $dV/dt = -a\sqrt{2gh}$ , where $h$ is the height of the water and $g$ is the acceleration due to gravity. Which differential equation describes the water height $h(t)$ ?
Answer: $\frac{dh}{dt} = -\frac{4a\sqrt{2g}}{\pi}h^{-3/2}$ First, relate the volume $V$ of water in the cone to its height $h$ . Let $r$ be the radius of the water surface at height $h$ . By similar triangles, $r/h = R/H = 2/4 = 1/2$ , so $r = h/2$ . The volume is $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (h/2)^2 h = \frac{\pi}{12}h^3$ . Differentiate with respect to time $t$ using the chain rule: $dV/dt = \frac{d}{dh}(\frac{\pi}{12}h^3) \frac{dh}{dt} = \frac{3\pi}{12}h^2 \frac{dh}{dt} = \frac{\pi}{4}h^2 \frac{dh}{dt}$ . Now, set this equal to the expression from Torricelli's law: $\frac{\pi}{4}h^2 \frac{dh}{dt} = -a\sqrt{2gh}$ . Solving for $dh/dt$ gives $\frac{dh}{dt} = -\frac{4a\sqrt{2g}}{\pi} \frac{\sqrt{h}}{h^2} = -\frac{4a\sqrt{2g}}{\pi}h^{-3/2}$ .
A hot object with initial temperature 200°F is placed in a room maintained at 70°F. The object cools according to Newton's Law of Cooling with a cooling constant of $k = 0.1$ per minute. Simultaneously, an external heat source adds thermal energy at a rate equivalent to raising the object's temperature by 15°F per minute if no other thermal effects were present. What differential equation models the temperature $T(t)$ of the object?
Answer: $\frac{dT}{dt} = -0.1(T - 70) + 15$ Newton's Law gives cooling rate -k(T - T_ambient) = -0.1(T - 70). The external heating adds +15°F/min. The total rate is dT/dt = -0.1(T - 70) + 15. Choice B incorrectly shifts the ambient temperature by the heating rate. Choice C incorrectly expands Newton's Law as -0.1T + 0.1(70) = -0.1T + 7. Choice D incorrectly includes the heating term inside the temperature difference for Newton's Law.
According to Newton's law of cooling, the rate of change of an object's temperature is proportional to the difference between its temperature and the ambient temperature. An object is placed in an environment where the ambient temperature, $M(t)$ , is not constant but increases linearly from an initial temperature of $M_0$ at a rate of $\alpha$ degrees per hour. Let $T(t)$ be the object's temperature. The object cools when it is warmer than its surroundings. Let $k > 0$ be the constant of proportionality. Which differential equation models the temperature $T(t)$ ?
Answer: $\frac{dT}{dt} = -k(T - (M_0 + \alpha t))$ Newton's law of cooling states that $dT/dt = -k(T - M)$ , where $M$ is the ambient temperature and $k > 0$ . The problem states that the ambient temperature $M(t)$ increases linearly with time from an initial value of $M_0$ at a rate of $\alpha$ . This means $M(t) = M_0 + \alpha t$ . Substituting this into the cooling law gives $dT/dt = -k(T(t) - M(t))$ , which becomes $dT/dt = -k(T - (M_0 + \alpha t))$ .
An electric circuit consists of a resistor of resistance $R$ , an inductor of inductance $L$ , and a time-varying voltage source $E(t) = V_0 \sin(\omega t)$ , all connected in series. According to Kirchhoff's voltage law, the sum of voltage drops across the inductor ( $L(dI/dt)$ ) and the resistor ( $IR$ ) equals the source voltage. A student wants to model the total magnetic flux $\Phi(t)$ in the inductor, which is related to the current by $\Phi(t) = LI(t)$ . Which of the following is a correct differential equation for the magnetic flux $\Phi(t)$ ?
Answer: $\frac{d\Phi}{dt} + \frac{R}{L}\Phi = V_0 \sin(\omega t)$ Kirchhoff's voltage law gives the equation for the current $I(t)$ : $L \frac{dI}{dt} + RI = V_0 \sin(\omega t)$ . We are given the relation between magnetic flux $\Phi$ and current $I$ : $\Phi = LI$ . From this, we can express $I$ as $I = \Phi/L$ . We can also find the derivative of the current: $\frac{dI}{dt} = \frac{1}{L}\frac{d\Phi}{dt}$ . Now, substitute these expressions for $I$ and $dI/dt$ into Kirchhoff's law: $L \left(\frac{1}{L}\frac{d\Phi}{dt}\right) + R\left(\frac{\Phi}{L}\right) = V_0 \sin(\omega t)$ . Simplifying this gives the differential equation for the magnetic flux $\Phi(t)$ : $\frac{d\Phi}{dt} + \frac{R}{L}\Phi = V_0 \sin(\omega t)$ .

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