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AP Calculus AB Quiz

AP Calculus AB Quiz: Modeling Situations With Differential Equations

Practice Modeling Situations With Differential Equations in AP Calculus AB with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 20

0 of 20 answered

The velocity v(t)v(t)v(t) of a falling object changes at a rate ggg minus a drag force proportional to v(t)v(t)v(t). Which differential equation models vvv?

Select an answer to continue

What this quiz covers

This quiz focuses on Modeling Situations With Differential Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.

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.

All questions

Question 1

The velocity v(t)v(t)v(t) of a falling object changes at a rate ggg minus a drag force proportional to v(t)v(t)v(t). Which differential equation models vvv?

  1. dvdt=g−kv\dfrac{dv}{dt}=g-kvdtdv​=g−kv (correct answer)
  2. dvdt=kv−g\dfrac{dv}{dt}=kv-gdtdv​=kv−g
  3. dvdt=g+kv\dfrac{dv}{dt}=g+kvdtdv​=g+kv
  4. dvdt=gkv\dfrac{dv}{dt}=\dfrac{g}{kv}dtdv​=kvg​
  5. dtdv=g−kv\dfrac{dt}{dv}=g-kvdvdt​=g−kv

Explanation: This problem models motion with gravity and air resistance, where velocity changes due to both downward gravitational acceleration and upward drag force. The gravitational acceleration contributes +g to the rate of change, while drag proportional to velocity contributes -kv, giving dv/dt = g - kv. The positive g term represents gravity accelerating the object downward, while the negative kv term represents air resistance opposing motion. Choice B has the wrong signs, representing upward acceleration and downward drag. Choice C would have both forces in the same direction, which is unphysical. When modeling falling objects with drag, gravity provides constant acceleration while drag opposes motion proportional to speed.

Question 2

A tank contains V(t)V(t)V(t) liters; it leaks at a rate proportional to the amount present. Which differential equation models VVV?

  1. dVdt=k\dfrac{dV}{dt}=kdtdV​=k
  2. dVdt=−kV\dfrac{dV}{dt}=-kVdtdV​=−kV (correct answer)
  3. dVdt=kV+t\dfrac{dV}{dt}=kV+tdtdV​=kV+t
  4. dVdt=−kV\dfrac{dV}{dt}=-\dfrac{k}{V}dtdV​=−Vk​
  5. dtdV=−kV\dfrac{dt}{dV}=-kVdVdt​=−kV

Explanation: This problem requires setting up a differential equation from a rate description involving proportional decay. The tank leaks at a rate proportional to the current amount V(t), which means the rate of change of volume is negative and directly related to the volume itself. Since "proportional to the amount present" means the leak rate equals k times V(t), we get dV/dt = -kV where k > 0. The negative sign is crucial because the volume is decreasing due to leaking. Choice A represents constant rate change, not proportional to volume, while choice D has the wrong form with V in the denominator. Always ensure the sign reflects whether the quantity is increasing or decreasing, and check that proportional relationships translate to multiplication by a constant.

Question 3

The rate of change of a tree’s height H(t)H(t)H(t) is proportional to 1H(t)\dfrac{1}{H(t)}H(t)1​. Which differential equation models HHH?

  1. dHdt=kH\dfrac{dH}{dt}=\dfrac{k}{H}dtdH​=Hk​ (correct answer)
  2. dHdt=kH\dfrac{dH}{dt}=kHdtdH​=kH
  3. dHdt=−kH\dfrac{dH}{dt}=-\dfrac{k}{H}dtdH​=−Hk​
  4. dHdt=Hk\dfrac{dH}{dt}=\dfrac{H}{k}dtdH​=kH​
  5. dtdH=kH\dfrac{dt}{dH}=\dfrac{k}{H}dHdt​=Hk​

Explanation: This problem establishes an inverse proportional relationship where the rate of change decreases as the height increases. The rate of change "is proportional to 1/H(t)," giving dHdt=kH\dfrac{dH}{dt} = \dfrac{k}{H}dtdH​=Hk​ where k>0k > 0k>0. This relationship might represent diminishing growth rates as trees get taller due to resource limitations or structural constraints. Choice C has the wrong sign representing decreasing height. Choice D incorrectly places H in the numerator instead of denominator. Inverse proportional relationships arise when growth becomes more difficult as the quantity increases.

Question 4

The number of cars N(t)N(t)N(t) in a parking lot decreases at a rate proportional to N(t)\sqrt{N(t)}N(t)​. Which differential equation models NNN?

  1. dNdt=−kN\dfrac{dN}{dt}=-k\sqrt{N}dtdN​=−kN​ (correct answer)
  2. dNdt=−kN\dfrac{dN}{dt}=-kNdtdN​=−kN
  3. dNdt=kN\dfrac{dN}{dt}=k\sqrt{N}dtdN​=kN​
  4. dNdt=−kN\dfrac{dN}{dt}=-\dfrac{k}{\sqrt{N}}dtdN​=−N​k​
  5. dtdN=−kN\dfrac{dt}{dN}=-k\sqrt{N}dNdt​=−kN​

Explanation: This problem describes a situation where the rate of decrease depends on the square root of the current amount rather than the amount itself. The cars leave "at a rate proportional to √N(t)," so dN/dt = -k√N where k > 0. The negative sign indicates decreasing number of cars, and the square root relationship might represent limited exit capacity or congestion effects. Choice B would be simple exponential decay. Choice C has the wrong sign representing increasing cars. Square root dependencies often arise in problems involving flow rates, diffusion, or situations where capacity constraints create nonlinear relationships.

Question 5

A lake’s pollutant concentration C(t)C(t)C(t) decreases at a rate proportional to C(t)C(t)C(t) due to natural cleanup. Which differential equation models CCC?

  1. dCdt=kC\dfrac{dC}{dt}=\dfrac{k}{C}dtdC​=Ck​
  2. dCdt=−kC\dfrac{dC}{dt}=-kCdtdC​=−kC (correct answer)
  3. dCdt=kC\dfrac{dC}{dt}=kCdtdC​=kC
  4. dtdC=−kC\dfrac{dt}{dC}=-kCdCdt​=−kC
  5. dCdt=−k(C−1)\dfrac{dC}{dt}=-k(C-1)dtdC​=−k(C−1)

Explanation: This problem models natural cleanup processes where pollutant concentration decreases at a rate proportional to the current concentration. The lake's "natural cleanup" causes the concentration to decrease proportionally to C(t), giving dC/dt = -kC where k > 0. The negative sign indicates decreasing concentration as the lake's natural processes remove pollutants. Choice C would represent increasing pollution rather than cleanup. Choice E introduces an unnecessary constant term that would represent a threshold concentration. Environmental cleanup problems typically follow exponential decay where the cleanup rate depends on how much pollution remains.

Question 6

A radioactive sample has mass m(t)m(t)m(t) that decays at a rate proportional to m(t)m(t)m(t). Which differential equation models mmm?

  1. dmdt=km\dfrac{dm}{dt}=kmdtdm​=km
  2. dmdt=−km\dfrac{dm}{dt}=-kmdtdm​=−km (correct answer)
  3. dmdt=−km\dfrac{dm}{dt}=-\dfrac{k}{m}dtdm​=−mk​
  4. dtdm=−km\dfrac{dt}{dm}=-kmdmdt​=−km
  5. dmdt=k−m\dfrac{dm}{dt}=k-mdtdm​=k−m

Explanation: This problem models radioactive decay where the mass decreases at a rate proportional to the current mass. Radioactive decay follows the law that "mass decays at a rate proportional to m(t)," giving dmdt=−km\frac{dm}{dt} = -kmdtdm​=−km where k>0k > 0k>0. The negative sign is crucial because the mass is decreasing due to radioactive particles leaving the sample. Choice A would represent exponential growth rather than decay. Choice C has an incorrect inverse relationship where the rate would increase as mass decreases. Radioactive decay is a classic example of exponential decay with rate proportional to the remaining amount.

Question 7

A sound’s intensity I(t)I(t)I(t) decreases at a rate proportional to I(t)I(t)I(t) plus background absorption bI(t)2bI(t)^2bI(t)2. Which differential equation models III?

  1. dIdt=−kI−bI2\dfrac{dI}{dt}=-kI-bI^2dtdI​=−kI−bI2 (correct answer)
  2. dIdt=kI+bI2\dfrac{dI}{dt}=kI+bI^2dtdI​=kI+bI2
  3. dIdt=−kI+b\dfrac{dI}{dt}=-kI+bdtdI​=−kI+b
  4. dIdt=−kI−bI2\dfrac{dI}{dt}=-\dfrac{k}{I}-bI^2dtdI​=−Ik​−bI2
  5. dtdI=−kI−bI2\dfrac{dt}{dI}=-kI-bI^2dIdt​=−kI−bI2

Explanation: This problem models sound intensity decrease with both linear and quadratic damping terms representing different absorption mechanisms. The intensity "decreases at a rate proportional to I(t)I(t)I(t) plus background absorption bI(t)2bI(t)^2bI(t)2," giving dIdt=−kI−bI2\dfrac{dI}{dt} = -kI - bI^2dtdI​=−kI−bI2 where k,b>0k, b > 0k,b>0. Both terms are negative because they represent different types of losses. The −kI-kI−kI term represents simple proportional loss while −bI2-bI^2−bI2 represents nonlinear absorption that increases with intensity. Choice B has wrong signs representing intensity increase. Sound absorption problems can involve multiple damping mechanisms with different dependencies on intensity.

Question 8

A car’s fuel amount F(t)F(t)F(t) decreases at a constant rate rrr liters per hour. Which differential equation models FFF?

  1. dFdt=−r\dfrac{dF}{dt}=-rdtdF​=−r (correct answer)
  2. dFdt=−rF\dfrac{dF}{dt}=-rFdtdF​=−rF
  3. dFdt=rF\dfrac{dF}{dt}=rFdtdF​=rF
  4. dtdF=−r\dfrac{dt}{dF}=-rdFdt​=−r
  5. dFdt=−rF\dfrac{dF}{dt}=\dfrac{-r}{F}dtdF​=F−r​

Explanation: This problem describes constant rate consumption where fuel decreases at a fixed rate regardless of the current fuel amount. Since the fuel "decreases at a constant rate r," the rate of change is simply dFdt=−r\dfrac{dF}{dt} = -rdtdF​=−r where r>0r > 0r>0. The negative sign indicates decreasing fuel, and the rate is constant (independent of F). Choice B would represent proportional consumption where more fuel leads to higher consumption rate. Choice E would represent inverse proportional consumption. Constant rate problems have the simplest differential equations with no dependence on the current amount of the quantity.

Question 9

A population P(t)P(t)P(t) grows at a rate proportional to both PPP and (5000−P)(5000-P)(5000−P). Which differential equation models PPP?

  1. dPdt=k(5000−P)\dfrac{dP}{dt}=k(5000-P)dtdP​=k(5000−P)
  2. dPdt=−kP(5000−P)\dfrac{dP}{dt}=-kP(5000-P)dtdP​=−kP(5000−P)
  3. dPdt=kP(5000−P)\dfrac{dP}{dt}=kP(5000-P)dtdP​=kP(5000−P) (correct answer)
  4. dPdt=k5000−PP\dfrac{dP}{dt}=k\dfrac{5000-P}{P}dtdP​=kP5000−P​
  5. dtdP=kP(5000−P)\dfrac{dt}{dP}=kP(5000-P)dPdt​=kP(5000−P)

Explanation: This problem involves modeling logistic growth with a differential equation. The key phrase "grows at a rate proportional to both P and (5000-P)" means we multiply these two factors: dP/dt = k·P·(5000-P). This creates the classic logistic growth model where growth slows as P approaches the carrying capacity of 5000. The positive k indicates growth, not decay. Choice B incorrectly includes a negative sign, which would model population decline. When modeling growth that depends on multiple factors, multiply those factors together in the rate expression.

Question 10

A spring’s displacement x(t)x(t)x(t) changes so that velocity is proportional to displacement and directed toward equilibrium. Which equation models xxx?

  1. dxdt=kx\dfrac{dx}{dt}=kxdtdx​=kx
  2. dxdt=−kx\dfrac{dx}{dt}=-kxdtdx​=−kx (correct answer)
  3. dxdt=−kt\dfrac{dx}{dt}=-k tdtdx​=−kt
  4. dtdx=−kx\dfrac{dt}{dx}=-kxdxdt​=−kx
  5. dxdt=−k+x\dfrac{dx}{dt}=-k+xdtdx​=−k+x

Explanation: This problem models simple harmonic motion using a differential equation. The velocity (dx/dt) being "proportional to displacement and directed toward equilibrium" means dx/dt = -kx, where the negative sign ensures motion toward x = 0. When x is positive, velocity is negative (moving left toward equilibrium), and when x is negative, velocity is positive (moving right toward equilibrium). Choice A (dx/dt = kx) would incorrectly model motion away from equilibrium, creating instability. For oscillatory systems, the restoring force (or velocity) opposes displacement, requiring a negative proportionality constant.

Question 11

A chemical amount A(t)A(t)A(t) decays at a rate proportional to the square root of AAA. Which differential equation models AAA?

  1. dAdt=kA\dfrac{dA}{dt}=k\sqrt{A}dtdA​=kA​
  2. dAdt=−kA2\dfrac{dA}{dt}=-kA^2dtdA​=−kA2
  3. dAdt=−kA\dfrac{dA}{dt}=-k\sqrt{A}dtdA​=−kA​ (correct answer)
  4. dtdA=−kA\dfrac{dt}{dA}=-k\sqrt{A}dAdt​=−kA​
  5. dAdt=−kA\dfrac{dA}{dt}=-kAdtdA​=−kA

Explanation: This problem models chemical decay with a differential equation involving a square root. The phrase "decays at a rate proportional to the square root of A" translates to dA/dt = -k√A, where the negative sign indicates decay (decreasing amount). This creates a slower decay than standard exponential decay since √A grows more slowly than A itself. Choice A (dA/dt = k√A) would incorrectly model growth rather than decay. When modeling decay with non-standard dependencies, maintain the negative sign while replacing the quantity with the specified function.

Question 12

A medication amount M(t)M(t)M(t) decreases at a rate proportional to MMM. Which differential equation models the elimination of the drug?

  1. dMdt=kM\dfrac{dM}{dt}=kMdtdM​=kM
  2. dMdt=−kM\dfrac{dM}{dt}=-kMdtdM​=−kM (correct answer)
  3. dtdM=−kM\dfrac{dt}{dM}=-kMdMdt​=−kM
  4. dMdt=−kt\dfrac{dM}{dt}=-k tdtdM​=−kt
  5. dMdt=−k+M\dfrac{dM}{dt}=-k+MdtdM​=−k+M

Explanation: This problem models exponential decay of medication using a differential equation. The phrase "decreases at a rate proportional to M" translates directly to dM/dt = -kM, where the negative sign indicates the medication amount is decreasing. This is the standard model for drug elimination following first-order kinetics. Choice A (dM/dt = kM) would incorrectly model growth of the medication amount rather than elimination. When a quantity decreases proportionally to itself, always include a negative sign in the differential equation to ensure the derivative is negative.

Question 13

A bacteria culture has N(t)N(t)N(t) cells; the growth rate is proportional to N2/3N^{2/3}N2/3. Which differential equation models NNN?

  1. dNdt=kN3/2\dfrac{dN}{dt}=kN^{3/2}dtdN​=kN3/2
  2. dNdt=kN2/3\dfrac{dN}{dt}=kN^{2/3}dtdN​=kN2/3 (correct answer)
  3. dNdt=−kN2/3\dfrac{dN}{dt}=-kN^{2/3}dtdN​=−kN2/3
  4. dtdN=kN2/3\dfrac{dt}{dN}=kN^{2/3}dNdt​=kN2/3
  5. dNdt=k(23)N\dfrac{dN}{dt}=k\left(\dfrac{2}{3}\right)NdtdN​=k(32​)N

Explanation: This problem requires modeling bacterial growth with a differential equation involving a power function. The phrase "growth rate is proportional to N^(2/3)" translates directly to dN/dt = kN^(2/3), where k is positive for growth. This non-standard growth model might represent surface-area-limited growth since N^(2/3) relates to surface area of a three-dimensional colony. Choice C incorrectly includes a negative sign, which would model decay rather than growth. When the rate depends on a power of the quantity, simply replace the quantity with that power in the standard proportional relationship.

Question 14

A savings account balance B(t)B(t)B(t) earns interest at a rate proportional to the current balance. Which differential equation models BBB?

  1. dBdt=kB\dfrac{dB}{dt}=kBdtdB​=kB (correct answer)
  2. dBdt=−kB\dfrac{dB}{dt}=-kBdtdB​=−kB
  3. dtdB=kB\dfrac{dt}{dB}=kBdBdt​=kB
  4. dBdt=kt\dfrac{dB}{dt}=k tdtdB​=kt
  5. dBdt=k+B\dfrac{dB}{dt}=k+BdtdB​=k+B

Explanation: This problem models compound interest growth with a differential equation. When interest is "proportional to the current balance," the rate of change equals k times the balance: dB/dt = kB. Since the balance is growing (earning interest), the derivative is positive, so we use +k, not -k. This creates exponential growth, the foundation of compound interest calculations. Choice B (dB/dt = -kB) would incorrectly model a decreasing balance. For financial growth problems, positive proportionality constants model appreciation while negative constants model depreciation.

Question 15

A rumor spreads so that the rate of new people informed is proportional to R(t)(1000−R(t))R(t)(1000-R(t))R(t)(1000−R(t)). Which equation models RRR?

  1. dRdt=kR(1000−R)\dfrac{dR}{dt}=kR(1000-R)dtdR​=kR(1000−R) (correct answer)
  2. dRdt=k(1000−R)\dfrac{dR}{dt}=k(1000-R)dtdR​=k(1000−R)
  3. dRdt=−kR(1000−R)\dfrac{dR}{dt}=-kR(1000-R)dtdR​=−kR(1000−R)
  4. dtdR=kR(1000−R)\dfrac{dt}{dR}=kR(1000-R)dRdt​=kR(1000−R)
  5. dRdt=kR1000−R\dfrac{dR}{dt}=k\dfrac{R}{1000-R}dtdR​=k1000−RR​

Explanation: This problem models rumor spreading using a logistic differential equation. The rate being "proportional to R(t)(1000-R(t))" translates to dR/dt = kR(1000-R), where R represents informed people and (1000-R) represents uninformed people. This models how rumors spread quickly at first (many uninformed people) but slow as more people become informed. Choice C incorrectly includes a negative sign, which would model the rumor dying out. For spreading phenomena involving interaction between two complementary groups, multiply the sizes of both groups in the rate expression.

Question 16

A hot metal cools so that dTdt\dfrac{dT}{dt}dtdT​ is proportional to 25−T25-T25−T; which differential equation models this?

  1. dTdt=k(T−25)\dfrac{dT}{dt}=k(T-25)dtdT​=k(T−25)
  2. dTdt=k(25−T)\dfrac{dT}{dt}=k(25-T)dtdT​=k(25−T) (correct answer)
  3. dTdt=kT(25−T)\dfrac{dT}{dt}=kT(25-T)dtdT​=kT(25−T)
  4. dtdT=k(25−T)\dfrac{dt}{dT}=k(25-T)dTdt​=k(25−T)
  5. dTdt=k(25−T)+25\dfrac{dT}{dt}=k(25-T)+25dtdT​=k(25−T)+25

Explanation: This problem involves modeling cooling with a differential equation. The metal's temperature change rate dT/dt is proportional to (25 - T), giving dT/dt = k(25 - T). Since the metal is hot (T > 25), the expression (25 - T) is negative, making dT/dt negative for cooling. This follows Newton's Law of Cooling with 25°C as the ambient temperature. Choice A incorrectly writes k(T - 25), which would be positive when T > 25, incorrectly modeling heating instead of cooling. When modeling temperature change, ensure the sign of the rate matches the physical process: negative for cooling, positive for heating.

Question 17

A savings balance S(t)S(t)S(t) earns interest at 4% per year, proportional to the current balance. Which equation models this?​​

  1. dSdt=0.04\dfrac{dS}{dt}=0.04dtdS​=0.04
  2. dSdt=0.04S\dfrac{dS}{dt}=\dfrac{0.04}{S}dtdS​=S0.04​
  3. dSdt=0.04S\dfrac{dS}{dt}=0.04SdtdS​=0.04S (correct answer)
  4. dtdS=0.04S\dfrac{dt}{dS}=0.04SdSdt​=0.04S
  5. dSdt=0.04t\dfrac{dS}{dt}=0.04tdtdS​=0.04t

Explanation: This problem models compound interest on a savings account. Interest at "4% per year, proportional to the current balance" translates to dS/dt = 0.04S, where 0.04 is the decimal form of 4%. The rate of change equals 0.04 times the current balance S. Choice A (dS/dt = 0.04) would add a constant $0.04 per year regardless of balance, which isn't how compound interest works. For percentage growth problems, convert the percentage to decimal and multiply by the current amount.

Question 18

A savings account balance B(t)B(t)B(t) earns interest at a rate proportional to B(t)B(t)B(t). Which differential equation models BBB?

  1. dBdt=kB\dfrac{dB}{dt}=\dfrac{k}{B}dtdB​=Bk​
  2. dBdt=kB\dfrac{dB}{dt}=kBdtdB​=kB (correct answer)
  3. dtdB=kB\dfrac{dt}{dB}=kBdBdt​=kB
  4. dBdt=k+B\dfrac{dB}{dt}=k+BdtdB​=k+B
  5. dBdt=−kB\dfrac{dB}{dt}=-kBdtdB​=−kB

Explanation: This problem models exponential growth where the savings account balance increases due to continuous compound interest. The phrase "earns interest at a rate proportional to B(t)" means the rate of increase equals k times the current balance, giving dB/dt = kB where k > 0. This represents exponential growth since more money earns more interest. Choice A would represent exponential decay with the wrong negative sign. Choice D represents linear growth by adding B rather than multiplying by a proportionality constant. Choice E also represents decay rather than growth. When modeling compound interest or exponential growth, the rate of change should be positive and directly proportional to the current amount.

Question 19

A hot object’s temperature T(t)T(t)T(t) cools at a rate proportional to T(t)−20T(t)-20T(t)−20. Which differential equation models TTT?

  1. dTdt=k(T−20)\dfrac{dT}{dt}=k(T-20)dtdT​=k(T−20)
  2. dTdt=−k(T−20)\dfrac{dT}{dt}=-k(T-20)dtdT​=−k(T−20) (correct answer)
  3. dTdt=−k(20−T)\dfrac{dT}{dt}=-k(20-T)dtdT​=−k(20−T)
  4. dTdt=−kT−20\dfrac{dT}{dt}=\dfrac{-k}{T-20}dtdT​=T−20−k​
  5. dtdT=−k(T−20)\dfrac{dt}{dT}=-k(T-20)dTdt​=−k(T−20)

Explanation: This problem applies Newton's Law of Cooling, where temperature change is proportional to the temperature difference from the ambient environment. The object cools, meaning its temperature T(t) decreases toward the ambient temperature of 20°. The rate of cooling is proportional to the temperature difference (T−20)(T-20)(T−20), so dV/dt=−k(T−20)dV/dt = -k(T-20)dV/dt=−k(T−20) where k > 0. The negative sign indicates cooling (decreasing temperature). Choice A lacks the negative sign, representing heating instead of cooling. Choice C uses (20−T)(20-T)(20−T) instead of (T−20)(T-20)(T−20), which would give the wrong sign behavior. When modeling temperature change, always identify the ambient temperature and use the difference (current−ambient)(current - ambient)(current−ambient) with appropriate sign for heating or cooling.

Question 20

A chemical reaction produces product P(t)P(t)P(t) at a rate proportional to the remaining reactant (50−P(t))(50-P(t))(50−P(t)). Which differential equation models PPP?

  1. dPdt=k(50−P)\dfrac{dP}{dt}=k(50-P)dtdP​=k(50−P) (correct answer)
  2. dPdt=−k(50−P)\dfrac{dP}{dt}=-k(50-P)dtdP​=−k(50−P)
  3. dPdt=kP(50−P)\dfrac{dP}{dt}=kP(50-P)dtdP​=kP(50−P)
  4. dPdt=k50−P\dfrac{dP}{dt}=\dfrac{k}{50-P}dtdP​=50−Pk​
  5. dtdP=k(50−P)\dfrac{dt}{dP}=k(50-P)dPdt​=k(50−P)

Explanation: This problem models a chemical reaction where the rate depends on the amount of unreacted starting material. The product P(t) forms "at a rate proportional to the remaining reactant (50-P(t))," giving dP/dt = k(50-P) where k > 0. The term (50-P) represents the amount of reactant still available for conversion to product. Choice B has the wrong sign, representing product disappearing. Choice C would represent autocatalytic behavior where product formation accelerates with existing product. Chemical reaction rates typically depend on concentrations of reactants, making the rate proportional to remaining unreacted material.