Differential Equations Quiz: Setting Up Des From Word Problems

What this quiz covers

This quiz focuses on Setting Up Des From Word Problems, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\frac{dS}{dt} = 6 - \frac{2S}{100 + t}$ (correct answer)
2. $\frac{dS}{dt} = 6 - \frac{2S}{100 + 2t}$
3. $\frac{dS}{dt} = 6 - \frac{3S}{100 + t}$
4. $\frac{dS}{dt} = 3 - \frac{2S}{100 + t}$

Explanation: 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.

Question 2

1. $\frac{dT}{dt} = -k(T - (M_0 + \alpha t))$ (correct answer)
2. $\frac{dT}{dt} = -k(T - M_0) + \alpha$
3. $\frac{dT}{dt} = -k(T - M_0)$
4. $\frac{dT}{dt} = -k(T - (M_0 + \alpha))$

Explanation: 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 $dT/dt = -k(T - (M_0 + \alpha t))$.

Question 3

1. $\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right) - h_0(1+\cos(\omega t))$
2. $\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right) - h_0(1+\cos(\omega t))P$ (correct answer)
3. $\frac{dP}{dt} = rP\left(1 - \frac{P}{K - h_0(1+\cos(\omega t))}\right)$
4. $\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right) + h_0(1+\cos(\omega t))P$

Explanation: 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: $dP/dt = rP(1 - P/K) - h_0(1 + \cos(\omega t))P$.

Question 4

1. $\frac{dI}{dt} = kI(N-I) - \gamma I$
2. $\frac{dI}{dt} = kI(N-I) - \gamma \sqrt{I}$ (correct answer)
3. $\frac{dI}{dt} = kI(N-I) + \gamma \sqrt{I}$
4. $\frac{dI}{dt} = kI(N-I) - \gamma I^2$

Explanation: 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: $dI/dt = kI(N-I) - \gamma\sqrt{I}$.

Question 5

1. $\frac{dV}{dt} = rV - k\sqrt{V} + D$
2. $\frac{dV}{dt} = rV + kV^2 - D$
3. $\frac{dV}{dt} = r - kV^2 + D$
4. $\frac{dV}{dt} = rV - kV^2 + D$ (correct answer)

Explanation: 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.

Question 6

1. $m\frac{dv}{dt} = mg - kv + T$
2. $m\frac{dv}{dt} = mg + kv + T$
3. $m\frac{dv}{dt} = mg - kv - T$ (correct answer)
4. $m\frac{dv}{dt} = -mg - kv - T$

Explanation: 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.

Question 7

1. $\frac{dh}{dt} = -kh^2$
2. $\frac{dh}{dt} = -k\sqrt{h}$
3. $\frac{dh}{dt} = -kh$ (correct answer)
4. $\frac{dh}{dt} = -k$

Explanation: 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.

Question 8

1. $\frac{dP}{dt} = k(1-P)$
2. $\frac{dP}{dt} = kP(P-1)$
3. $\frac{dP}{dt} = kP$
4. $\frac{dP}{dt} = kP(1-P)$ (correct answer)

Explanation: 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.

Question 9

1. $\frac{dA}{dt} = -k A - D_0 e^{-\alpha t}$
2. $\frac{dA}{dt} = -k \frac{A}{V} + D_0 e^{-\alpha t}$
3. $\frac{dA}{dt} = -k A + D_0 e^{-\alpha t}$ (correct answer)
4. $\frac{dA}{dt} = k A + D_0 e^{-\alpha t}$

Explanation: 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.

Question 10

1. $\frac{dx}{dt} = -k_1 x + k_2 (C-x)^2$ (correct answer)
2. $\frac{dx}{dt} = k_1 x - k_2 (C-x)^2$
3. $\frac{dx}{dt} = -k_1 x + k_2 (C-x)$
4. $\frac{dx}{dt} = -k_1 (C-x) + k_2 x^2$

Explanation: 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: $dx/dt = (\text{rate of formation}) - (\text{rate of consumption}) = k_2(C-x)^2 - k_1x$.

Question 11

1. $\frac{d\Phi}{dt} + \frac{R}{L}\Phi = V_0 \sin(\omega t)$ (correct answer)
2. $L\frac{d\Phi}{dt} + R\Phi = V_0 \sin(\omega t)$
3. $\frac{1}{L}\frac{d\Phi}{dt} + R\Phi = V_0 \sin(\omega t)$
4. $\frac{d\Phi}{dt} + R\Phi = V_0 \sin(\omega t)$

Explanation: 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 $\Phi(t)$: $\frac{d\Phi}{dt} + \frac{R}{L}\Phi = V_0 \sin(\omega t)$.

Question 12

1. $\frac{dh}{dt} = -\frac{a\sqrt{2g}}{\pi}h^{1/2}$
2. $\frac{dh}{dt} = -\frac{4a\sqrt{2g}}{\pi}h^{-3/2}$ (correct answer)
3. $\frac{dh}{dt} = -\frac{a\sqrt{2g}}{4\pi}h^{1/2}$
4. $\frac{dh}{dt} = -\frac{a\sqrt{2g}}{\pi}h^{-1/2}$

Explanation: 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 $\frac{dh}{dt} = -\frac{4a\sqrt{2g}}{\pi} \frac{\sqrt{h}}{h^2} = -\frac{4a\sqrt{2g}}{\pi}h^{-3/2}$.

Question 13

1. $\frac{dR}{dt} = 0.8R - 0.03FR, \quad \frac{dF}{dt} = -0.6F + 0.02FR(1 - R/200)$
2. $\frac{dR}{dt} = 0.8R - 0.03FR - R^2/200, \quad \frac{dF}{dt} = -0.6F + 0.02FR$
3. $\frac{dR}{dt} = 0.8R(1 - R/200) - 0.03FR, \quad \frac{dF}{dt} = -0.6F + 0.02FR$ (correct answer)
4. $\frac{dR}{dt} = 0.8R(1 - R/100) - 0.03FR, \quad \frac{dF}{dt} = -0.6F + 0.02FR$

Explanation: 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.

Question 14

1. $\frac{d[A]}{dt} = 0.3[A] - 0.1(C - [A])$
2. $\frac{d[A]}{dt} = -0.3[A] + 0.1[B]$
3. $\frac{d[A]}{dt} = -0.2[A] + 0.1C$
4. $\frac{d[A]}{dt} = -0.3[A] + 0.1(C - [A])$ (correct answer)

Explanation: 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.

Question 15

1. $\frac{d^2\theta}{dt^2} + 0.5\frac{d\theta}{dt} + 9.8\theta = 0.1\cos(2t)$
2. $\frac{d^2\theta}{dt^2} + 0.5\frac{d\theta}{dt} + 4.9\sin\theta = 0.1\cos(2t)$
3. $\frac{d^2\theta}{dt^2} + 0.5\frac{d\theta}{dt} + 4.9\theta = 0.1\cos(2t)$ (correct answer)
4. $\frac{d^2\theta}{dt^2} + 0.5\theta + 4.9\frac{d\theta}{dt} = 0.1\cos(2t)$

Explanation: 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.

Question 16

1. $\frac{d[D_f]}{dt} = 5 - 0.2[D_f] - 0.8[D_f] + 0.3[D_b] - 0.2[D_b]$
2. $\frac{d[D_f]}{dt} = 5 - 0.2[D_f] - 0.8[D_f] + 0.3[D_b]$ (correct answer)
3. $\frac{d[D_f]}{dt} = 5 - (0.2 + 0.8)[D_f] + 0.3[D_b]$
4. $\frac{d[D_f]}{dt} = 5 - 0.2([D_f] + [D_b]) - 0.8[D_f] + 0.3[D_b]$

Explanation: 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])

Question 17

1. $\frac{dT}{dt} = -0.1(T - 70) + 15$ (correct answer)
2. $\frac{dT}{dt} = -0.1(T - 85) + 15$
3. $\frac{dT}{dt} = -0.1T + 7 + 15$
4. $\frac{dT}{dt} = -0.1(T - 70 + 15)$

Explanation: 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.

Question 18

1. $\frac{dS}{dt} = -0.0001SI, \quad \frac{dI}{dt} = 0.0001SI - 0.05I - 50, \quad \frac{dR}{dt} = 0.05I + 50$
2. $\frac{dS}{dt} = -0.0001SI - 50, \quad \frac{dI}{dt} = 0.0001SI - 0.05I, \quad \frac{dR}{dt} = 0.05I + 50$ (correct answer)
3. $\frac{dS}{dt} = -0.0001SI - 50, \quad \frac{dI}{dt} = 0.0001SI - 0.05I + 50, \quad \frac{dR}{dt} = 0.05I$
4. $\frac{dS}{dt} = -0.0001SI - 50, \quad \frac{dI}{dt} = 0.0001SI - 0.05I, \quad \frac{dR}{dt} = 0.05I$

Explanation: 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.