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AP Calculus BC Quiz

AP Calculus BC Quiz: Modeling Situations With Differential Equations

Practice Modeling Situations With Differential Equations in AP Calculus BC 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

A medication amount AAA in the bloodstream decreases at a rate proportional to the square root of the amount present. Which models A(t)A(t)A(t)?

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 BC.

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

A medication amount AAA in the bloodstream decreases at a rate proportional to the square root of the amount present. Which models A(t)A(t)A(t)?

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

Explanation: This problem requires modeling medication decay with a differential equation. The phrase "decreases at a rate proportional to the square root" means dA/dt = -k√A, where k > 0. The negative sign indicates decrease, and the square root appears directly in the rate expression. Choice E would model standard exponential decay rather than the specified square-root relationship. When the rate depends on a function of the quantity, that function appears directly in the differential equation.

Question 2

A cup of coffee at temperature TTT cools in a 20∘20^\circ20∘C room; the cooling rate is proportional to T−20T-20T−20. Which equation models T(t)T(t)T(t)?

  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=kT−20\dfrac{dT}{dt}=\dfrac{k}{T-20}dtdT​=T−20k​
  4. dTdt=−kT+20\dfrac{dT}{dt}=-kT+20dtdT​=−kT+20
  5. dTdt=k−(T−20)\dfrac{dT}{dt}=k- (T-20)dtdT​=k−(T−20)

Explanation: This problem involves modeling Newton's law of cooling with a differential equation. The cooling rate being "proportional to T-20" means dTdt=−k(T−20)\dfrac{dT}{dt} = -k(T-20)dtdT​=−k(T−20), where k>0k > 0k>0. The negative sign is essential because when T>20T > 20T>20 (coffee hotter than room), the temperature decreases, making dTdt\dfrac{dT}{dt}dtdT​ negative. Choice A would incorrectly predict the coffee heating up when above room temperature. For temperature problems, always verify your equation predicts cooling when T>T >T> ambient and heating when T<T <T< ambient.

Question 3

A savings account balance BBB earns interest at 3% continuously and receives deposits of 200peryear200 per year200peryear. Which models B(t)B(t)B(t)?

  1. dBdt=0.03B−200\dfrac{dB}{dt}=0.03B-200dtdB​=0.03B−200
  2. dBdt=0.03B+200\dfrac{dB}{dt}=0.03B+200dtdB​=0.03B+200 (correct answer)
  3. B=0.03dBdt+200B=0.03\dfrac{dB}{dt}+200B=0.03dtdB​+200
  4. dBdt=0.03(B+200)\dfrac{dB}{dt}=0.03(B+200)dtdB​=0.03(B+200)
  5. dBdt=0.03B+200\dfrac{dB}{dt}=\dfrac{0.03}{B}+200dtdB​=B0.03​+200

Explanation: This problem models a savings account with both interest and deposits using a differential equation. The account grows from two sources: continuous interest at 3% means 0.03B contribution, and deposits of $200 per year add a constant +200 term. Combining these gives dB/dt = 0.03B + 200. The interest term 0.03B is proportional to the balance while deposits add at a constant rate. Choice A incorrectly subtracts the deposits, which would model withdrawals instead. When modeling multiple effects, add terms for increases and subtract terms for decreases.

Question 4

A bacteria culture grows at a rate proportional to its current population PPP. Which differential equation models P(t)P(t)P(t)?

  1. dPdt=kP\dfrac{dP}{dt}=kPdtdP​=kP (correct answer)
  2. P=kdPdtP=k\dfrac{dP}{dt}P=kdtdP​
  3. dPdt=k1P\dfrac{dP}{dt}=k\dfrac{1}{P}dtdP​=kP1​
  4. dPdt=k(P−1)\dfrac{dP}{dt}=k(P-1)dtdP​=k(P−1)
  5. dPdt=k−P\dfrac{dP}{dt}=k-PdtdP​=k−P

Explanation: This problem involves modeling exponential growth with a differential equation. The phrase "grows at a rate proportional to its current population P" directly translates to dP/dt = kP, where k > 0 is the growth constant. This is the standard form for exponential growth, where the rate of change equals a constant times the current value. Choice B incorrectly reverses the relationship by making P proportional to the rate rather than the rate proportional to P. When you see "rate proportional to," immediately write dP/dt = k × (whatever follows).

Question 5

A water tank contains WWW liters and leaks at a rate proportional to W\sqrt{W}W​. Which differential equation models W(t)W(t)W(t)?

  1. dWdt=kW\dfrac{dW}{dt}=k\sqrt{W}dtdW​=kW​
  2. dWdt=−kW\dfrac{dW}{dt}=-k\sqrt{W}dtdW​=−kW​ (correct answer)
  3. W=−kWW=-k\sqrt{W}W=−kW​
  4. dWdt=−kW2\dfrac{dW}{dt}=-kW^2dtdW​=−kW2
  5. dWdt=−k/W\dfrac{dW}{dt}=-k/\sqrt{W}dtdW​=−k/W​

Explanation: This problem requires modeling a leaking tank with a differential equation. The phrase "leaks at a rate proportional to √W" means the rate of water loss is k√W for some positive constant k. Since water is leaving the tank, dW/dt must be negative, giving us dW/dt = -k√W. This models situations where flow rate depends on pressure, which varies with the square root of water height. Choice A would incorrectly model water entering rather than leaving the tank. Remember that "rate of decrease" always requires a negative sign in the differential equation.

Question 6

A hot metal rod at temperature TTT warms toward 100°C so that change rate is proportional to 100−T100-T100−T. Which models T(t)T(t)T(t)?

  1. dTdt=k(T−100)\dfrac{dT}{dt}=k(T-100)dtdT​=k(T−100)
  2. dTdt=k(100−T)\dfrac{dT}{dt}=k(100-T)dtdT​=k(100−T) (correct answer)
  3. T=k(100−T)T=k(100-T)T=k(100−T)
  4. dTdt=kT−100\dfrac{dT}{dt}=kT-100dtdT​=kT−100
  5. dTdt=k/(100−T)\dfrac{dT}{dt}=k/(100-T)dtdT​=k/(100−T)

Explanation: This problem models Newton's law of heating using a differential equation. The phrase "change rate is proportional to 100-T" means dT/dt = k(100-T) for some constant k. Since the rod is warming toward 100°C from a lower temperature, T < 100 initially, making 100-T positive. For warming to occur, dT/dt must be positive, which is satisfied when k > 0. Choice A would incorrectly model cooling away from 100°C rather than warming toward it. When modeling temperature change, the sign of (Target - Current) determines whether heating or cooling occurs.

Question 7

A population PPP follows logistic growth with carrying capacity 5000 and growth constant kkk; which differential equation models P(t)P(t)P(t)?

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

Explanation: This problem involves modeling situations with differential equations, a key skill in AP Calculus BC. The population follows logistic growth with carrying capacity 5000, meaning growth is proportional to both current PPP and remaining capacity (5000−P5000 - P5000−P). This gives dPdt=kP(5000−P)\dfrac{dP}{dt} = kP(5000 - P)dtdP​=kP(5000−P), where k>0k > 0k>0 adjusts the growth rate, capturing initial exponential-like growth that tapers off. The product form ensures the rate is zero at P=0P = 0P=0 or P=5000P = 5000P=5000. A tempting distractor like choice B, dPdt=k(5000−P)\dfrac{dP}{dt} = k(5000 - P)dtdP​=k(5000−P), fails because it lacks the P factor, implying constant growth regardless of current population size. A transferable modeling strategy is to use product terms in logistic models to incorporate both growth potential and environmental limits.

Question 8

Water drains so the volume’s decrease rate is proportional to V\sqrt{V}V​; which differential equation models V(t)V(t)V(t)?

  1. dVdt=kV\dfrac{dV}{dt}=k\sqrt{V}dtdV​=kV​
  2. dVdt=−kV\dfrac{dV}{dt}=-k\sqrt{V}dtdV​=−kV​ (correct answer)
  3. dVdt=−kV\dfrac{dV}{dt}=-\dfrac{k}{\sqrt{V}}dtdV​=−V​k​
  4. V=−kVV=-k\sqrt{V}V=−kV​
  5. V′=−kV\sqrt{V'}=-kVV′​=−kV

Explanation: This problem involves modeling situations with differential equations, a key skill in AP Calculus BC. Water drains such that the volume decrease rate is proportional to the square root of VVV, often due to Torricelli's law relating to height. This is dVdt=−kV\dfrac{dV}{dt} = -k \sqrt{V}dtdV​=−kV​, with k>0k > 0k>0, ensuring a negative rate as volume decreases. The square root reflects dependence on the surface level or pressure. A tempting distractor like choice A, dVdt=kV\dfrac{dV}{dt} = k \sqrt{V}dtdV​=kV​, fails because it would predict increasing volume, not drainage. A transferable modeling strategy is to incorporate nonlinear terms like square roots when rates depend on physical properties such as height or pressure, and add negatives for decreases.

Question 9

A car’s speed vvv decreases at a rate proportional to its speed due to drag; which differential equation models v(t)v(t)v(t)?

  1. dvdt=kv\dfrac{dv}{dt}=kvdtdv​=kv
  2. dvdt=k−v\dfrac{dv}{dt}=k-vdtdv​=k−v
  3. dvdt=−kv\dfrac{dv}{dt}=-kvdtdv​=−kv (correct answer)
  4. v=−kvv=-kvv=−kv
  5. dvdt=−kv\dfrac{dv}{dt}=-\dfrac{k}{v}dtdv​=−vk​

Explanation: This problem involves modeling situations with differential equations, a key skill in AP Calculus BC. The car's speed vvv decreases at a rate proportional to vvv itself due to drag, meaning slowdown intensifies with higher speed. This is modeled as dvdt=−kv\dfrac{dv}{dt} = -kvdtdv​=−kv, with k>0k > 0k>0 to ensure a negative rate for positive vvv. The negative sign captures the deceleration. A tempting distractor like choice A, dvdt=kv\dfrac{dv}{dt} = kvdtdv​=kv, fails because it would predict increasing speed, contradicting the drag-induced decrease. A transferable modeling strategy is to include a negative sign in proportionality for decay or reduction processes to match the direction of change.

Question 10

A cup of coffee has temperature T(t)T(t)T(t) in a 20∘20^\circ20∘C room; cooling rate is proportional to T(t)−20T(t)-20T(t)−20. Which differential equation models T(t)T(t)T(t)?​

  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=−kT−20\dfrac{dT}{dt}=-kT-20dtdT​=−kT−20
  4. T(t)=−k (T−20)T(t)=-k\,(T-20)T(t)=−k(T−20)
  5. dTdt=−k (20−T)2\dfrac{dT}{dt}=-k\,(20-T)^2dtdT​=−k(20−T)2

Explanation: This problem models Newton's Law of Cooling using a differential equation. The cooling rate is proportional to the temperature difference T(t) - 20, where 20°C is room temperature. Since the coffee is cooling (temperature decreasing), we need a negative rate: dT/dt = -k(T - 20) where k > 0. This ensures that when T > 20, the derivative is negative and temperature decreases. Choice C incorrectly writes -kT - 20, which doesn't represent the temperature difference and would incorrectly suggest cooling even at absolute zero. When modeling temperature change, the rate depends on the difference from ambient temperature, not the absolute temperature.

Question 11

A tank contains yyy liters of brine; it is drained at a rate proportional to the amount present. Which differential equation models y(t)y(t)y(t)?​

  1. dydt=ky\dfrac{dy}{dt}=kydtdy​=ky
  2. dydt=ky\dfrac{dy}{dt}=\dfrac{k}{y}dtdy​=yk​
  3. dydt=−ky\dfrac{dy}{dt}=-kydtdy​=−ky (correct answer)
  4. dydt=−ky\dfrac{dy}{dt}=-\dfrac{k}{y}dtdy​=−yk​
  5. dydt=k−y\dfrac{dy}{dt}=k- ydtdy​=k−y

Explanation: This problem requires modeling a draining tank with a differential equation. The phrase "drained at a rate proportional to the amount present" means the rate of change dy/dt equals -k times y, where k > 0 is the proportionality constant. The negative sign is crucial because draining decreases the amount, so dy/dt < 0 when y > 0. Choice A would incorrectly model growth rather than drainage. When translating verbal descriptions to differential equations, always check whether the quantity increases or decreases to determine the sign.

Question 12

A spring-mass system has displacement xxx where acceleration is proportional to −x-x−x and velocity contributes no damping. Which differential equation models x(t)x(t)x(t)?

  1. d2xdt2=kx\dfrac{d^2x}{dt^2}=kxdt2d2x​=kx
  2. d2xdt2=−kx\dfrac{d^2x}{dt^2}=-kxdt2d2x​=−kx (correct answer)
  3. dxdt=−kx\dfrac{dx}{dt}=-kxdtdx​=−kx
  4. d2xdt2=−kx\dfrac{d^2x}{dt^2}=-\dfrac{k}{x}dt2d2x​=−xk​
  5. dxdt=k−x\dfrac{dx}{dt}=k-xdtdx​=k−x

Explanation: This problem models simple harmonic motion with a differential equation. A spring force proportional to -x creates acceleration (second derivative) proportional to -x, giving d²x/dt² = -kx. This is a second-order equation because acceleration involves the second derivative. Choice C incorrectly uses the first derivative, which would model exponential decay rather than oscillation. Spring-mass systems always produce second-order equations relating acceleration to position.

Question 13

A cup of coffee at temperature T(t)T(t)T(t) cools in a 20∘20^\circ20∘C room; its cooling rate is proportional to T−20T-20T−20. Which differential equation models this?

  1. dTdt=k(T−20)\dfrac{dT}{dt}=k(T-20)dtdT​=k(T−20)
  2. dTdt=k(20−T)\dfrac{dT}{dt}=k(20-T)dtdT​=k(20−T) (correct answer)
  3. dTdt=kT\dfrac{dT}{dt}=kTdtdT​=kT
  4. T=k(T−20)T=k(T-20)T=k(T−20)
  5. dTdt=k1T−20\dfrac{dT}{dt}=k\dfrac{1}{T-20}dtdT​=kT−201​

Explanation: This question tests the skill of modeling situations with differential equations. The verbal description indicates that the coffee is cooling in a 20°C room, with the cooling rate proportional to the temperature difference T - 20. Since the coffee is hotter than the room, the rate of change dT/dt should be negative to reflect cooling, leading to the form dT/dt = k(20 - T) where k is positive. This matches Newton's law of cooling, where the temperature approaches the ambient temperature over time. A tempting distractor is choice A, which has the sign reversed and would imply heating instead of cooling when T > 20. Always ensure the sign of the proportionality reflects the direction of change, such as negative for cooling when the object is hotter than the surroundings.

Question 14

A tank contains y(t)y(t)y(t) liters of salt; brine enters adding salt at 333 g/min and salt leaves at 0.1y0.1y0.1y g/min. Which differential equation models yyy?

  1. dydt=0.1y−3\dfrac{dy}{dt}=0.1y-3dtdy​=0.1y−3
  2. dydt=3−0.1y\dfrac{dy}{dt}=3-0.1ydtdy​=3−0.1y (correct answer)
  3. y=3−0.1yy=3-0.1yy=3−0.1y
  4. dydt=0.1(3−y)\dfrac{dy}{dt}=0.1(3-y)dtdy​=0.1(3−y)
  5. dydt=30.1y\dfrac{dy}{dt}=\dfrac{3}{0.1y}dtdy​=0.1y3​

Explanation: This question tests the skill of modeling situations with differential equations. The verbal description involves a tank where salt is added at a constant rate of 3 g/min and removed at a rate of 0.1y g/min, with y(t) representing the amount of salt. The net rate of change dy/dt is therefore the inflow minus the outflow, giving dydt=3−0.1y\dfrac{dy}{dt} = 3 - 0.1ydtdy​=3−0.1y. This is a linear differential equation capturing the balance between constant addition and concentration-dependent removal. A tempting distractor is choice A, which reverses the signs and would imply salt decreases when y is small, contradicting the net addition. In mixture problems, always set up the equation as rate in minus rate out for the accumulating quantity.

Question 15

A spring-mass system has displacement x(t)x(t)x(t) and velocity v(t)=dxdtv(t)=\dfrac{dx}{dt}v(t)=dtdx​; acceleration satisfies dvdt=−4x\dfrac{dv}{dt}=-4xdtdv​=−4x. Which differential equation for xxx models motion?

  1. d2xdt2=−4x\dfrac{d^2x}{dt^2}=-4xdt2d2x​=−4x (correct answer)
  2. dxdt=−4x\dfrac{dx}{dt}=-4xdtdx​=−4x
  3. d2xdt2=−4dxdt\dfrac{d^2x}{dt^2}=-4\dfrac{dx}{dt}dt2d2x​=−4dtdx​
  4. x=−4d2xdt2x=-4\dfrac{d^2x}{dt^2}x=−4dt2d2x​
  5. d2xdt2=−4x\dfrac{d^2x}{dt^2}=\dfrac{-4}{x}dt2d2x​=x−4​

Explanation: This question tests the skill of modeling situations with differential equations. The verbal description provides that acceleration dvdt=−4x\dfrac{dv}{dt} = -4xdtdv​=−4x, and since v=dxdtv = \dfrac{dx}{dt}v=dtdx​, this implies the second derivative d2xdt2=−4x\dfrac{d^2 x}{dt^2} = -4xdt2d2x​=−4x. This equation models simple harmonic motion in a spring-mass system, where acceleration is proportional to displacement but opposite in direction. The negative sign ensures oscillatory behavior around equilibrium. A tempting distractor is choice B, which is first-order and would imply exponential decay, not oscillation. When dealing with higher-order dynamics like acceleration, express the model using second derivatives for position-based forces.

Question 16

A bacteria culture has population P(t)P(t)P(t); its growth rate is proportional to the current population. Which differential equation models this situation?

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

Explanation: This question tests the skill of modeling situations with differential equations. The verbal description states that the bacteria population's growth rate is proportional to the current population P(t)P(t)P(t), meaning the rate of increase dPdt\frac{dP}{dt}dtdP​ equals kPkPkP for some positive constant k. This setup describes exponential growth, common in unrestricted population models where more individuals lead to faster reproduction. No limiting factors are mentioned, so the equation remains simple without additional terms. A tempting distractor is choice C, which includes (P−1)(P-1)(P−1) and might confuse with logistic models, but it incorrectly alters the proportionality to the population itself. When modeling proportional growth, express the derivative directly as a constant times the function for exponential behavior.

Question 17

An investment balance B(t)B(t)B(t) earns continuous interest at 5%5\%5% per year and has withdrawals of 200020002000 dollars per year. Which differential equation models BBB?

  1. dBdt=0.05B−2000\dfrac{dB}{dt}=0.05B-2000dtdB​=0.05B−2000 (correct answer)
  2. dBdt=0.05B+2000\dfrac{dB}{dt}=0.05B+2000dtdB​=0.05B+2000
  3. dBdt=0.05−2000B\dfrac{dB}{dt}=0.05-2000BdtdB​=0.05−2000B
  4. B=0.05B−2000B=0.05B-2000B=0.05B−2000
  5. dBdt=0.05B−2000\dfrac{dB}{dt}=\dfrac{0.05}{B}-2000dtdB​=B0.05​−2000

Explanation: This question tests the skill of modeling situations with differential equations. The verbal description involves an investment balance B(t) earning 5% continuous interest, adding 0.05B per year, minus constant withdrawals of 2000 per year, yielding dB/dt = 0.05B - 2000. This balances exponential growth from interest with a fixed subtraction for outflows. The equation can lead to growth or decline depending on the initial balance and rates. A tempting distractor is choice B, which adds instead of subtracts the 2000, incorrectly modeling deposits rather than withdrawals. In financial models, add terms for income or interest and subtract for expenses or withdrawals in the rate of change.

Question 18

A drug amount AAA in blood decreases at a rate proportional to AAA. Which differential equation models A(t)A(t)A(t)?

  1. dAdt=kA\dfrac{dA}{dt}=kAdtdA​=kA
  2. A=kdAdtA=k\dfrac{dA}{dt}A=kdtdA​
  3. dAdt=−kA\dfrac{dA}{dt}=-kAdtdA​=−kA (correct answer)
  4. dAdt=−k+A\dfrac{dA}{dt}=-k+AdtdA​=−k+A
  5. dAdt=−k/A\dfrac{dA}{dt}=-k/AdtdA​=−k/A

Explanation: This problem models drug elimination using a differential equation. The phrase "decreases at a rate proportional to A" means the rate of decrease is kA for some positive constant k. Since the amount is decreasing, dA/dt must be negative, giving us dA/dt = -kA. This is the standard model for first-order elimination, common in pharmacokinetics. Choice A would incorrectly model drug accumulation rather than elimination. When modeling decay or decrease, always include a negative sign to ensure the quantity decreases over time.

Question 19

A spring-mass system has acceleration proportional to negative displacement xxx from equilibrium; which differential equation models x(t)x(t)x(t)?

  1. d2xdt2=kx\dfrac{d^2x}{dt^2}=kxdt2d2x​=kx
  2. d2xdt2=−kx\dfrac{d^2x}{dt^2}=-kxdt2d2x​=−kx (correct answer)
  3. dxdt=−kx\dfrac{dx}{dt}=-kxdtdx​=−kx
  4. x′′=k1xx''=k\dfrac{1}{x}x′′=kx1​
  5. x=−kx′′x=-kx''x=−kx′′

Explanation: This problem involves modeling situations with differential equations, a key skill in AP Calculus BC. In a spring-mass system, acceleration is proportional to the negative displacement x from equilibrium, per Hooke's law, pulling back toward the center. This second-order equation is d2xdt2=−kx\dfrac{d^2x}{dt^2} = -kxdt2d2x​=−kx, with k>0k > 0k>0, leading to oscillatory motion. The negative sign ensures restoring force opposes displacement. A tempting distractor like choice A, d2xdt2=kx\dfrac{d^2x}{dt^2} = kxdt2d2x​=kx, fails because it would predict unstable motion away from equilibrium, not oscillation. A transferable modeling strategy is to use second-order equations for acceleration-based systems and incorporate negative signs for restoring forces.

Question 20

A population P(t)P(t)P(t) follows logistic growth with carrying capacity 500500500 and growth constant kkk. Which differential equation models this?

  1. dPdt=kP(500−P)\dfrac{dP}{dt}=kP(500-P)dtdP​=kP(500−P) (correct answer)
  2. dPdt=k(500−P)\dfrac{dP}{dt}=k(500-P)dtdP​=k(500−P)
  3. dPdt=kP(500+P)\dfrac{dP}{dt}=kP(500+P)dtdP​=kP(500+P)
  4. P=kP(500−P)P=kP(500-P)P=kP(500−P)
  5. dPdt=kP(500−P)\dfrac{dP}{dt}=\dfrac{k}{P}(500-P)dtdP​=Pk​(500−P)

Explanation: This question tests the skill of modeling situations with differential equations. The verbal description specifies logistic growth for population P(t) with carrying capacity 500 and growth constant k, leading to dP/dt = kP(500 - P). This form incorporates exponential growth at low populations via kP and slowing as P approaches 500 due to the (500 - P) term. The product ensures the rate is positive below the carrying capacity and zero at equilibrium. A tempting distractor is choice B, which lacks the P factor and would imply linear decay toward 500, not density-dependent growth. In logistic models, multiply the growth rate by both the population and the remaining capacity factor for realistic limits.