Modeling Situations with Differential Equations - AP Calculus BC

A spring-mass system has displacement $x$ where acceleration is proportional to $-x$ and velocity contributes no damping. Which differential equation models $x(t)$ ?
Answer: $\dfrac{d^2x}{dt^2}=-kx$ 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.
A population $P(t)$ grows at a rate proportional to its current size. Which differential equation models this situation?
Answer: $\dfrac{dP}{dt}=kP$ This problem involves modeling exponential growth with a differential equation. When a population grows at a rate proportional to its current size, we write the rate of change dP/dt as k times the population P, giving dP/dt = kP. This is the fundamental equation for exponential growth, where k is the growth constant. Choice C incorrectly suggests constant growth rather than proportional growth, which would result in linear rather than exponential behavior. The general principle is that "rate proportional to quantity" translates directly to dQ/dt = kQ in differential equation form.
A car's velocity $v$ decreases due to drag with magnitude proportional to $v^2$ , opposite the motion. Which differential equation models $v(t)$ ?
Answer: $\dfrac{dv}{dt}=-kv^2$ This problem models velocity decrease due to drag using a differential equation. Drag "proportional to v²" that opposes motion means dv/dt = -kv², where k > 0. The negative sign is crucial because drag always opposes velocity, causing deceleration. Choice A would incorrectly predict acceleration from drag, violating physics. For force-based problems, always verify the sign matches the physical direction of the effect.
A cup of coffee at temperature $T(t)$ cools in a $20^\circ$ C room; its cooling rate is proportional to $T-20$ . Which differential equation models this?
Answer: $\dfrac{dT}{dt}=k(20-T)$ 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.
A capacitor's charge $q(t)$ decreases so its rate of change is proportional to the negative of the charge. Which equation models $q(t)$ ?
Answer: $\dfrac{dq}{dt}=-kq$ This problem involves modeling capacitor discharge with a differential equation. The phrase "rate of change is proportional to the negative of the charge" directly translates to dq/dt = -kq, where k is a positive constant. This describes exponential decay of charge over time, typical in RC circuits. Choice A would incorrectly model charging rather than discharging, causing charge to grow exponentially. The key insight is to parse the wording carefully: "proportional to the negative of" means we multiply by -k, not that the rate itself is already negative.
A population $P$ follows logistic growth with carrying capacity 5000 and growth constant $k$ ; which differential equation models $P(t)$ ?
Answer: $\dfrac{dP}{dt}=kP(5000-P)$ 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 $P$ and remaining capacity ( $5000 - P$ ). This gives $\dfrac{dP}{dt} = kP(5000 - P)$ , where $k > 0$ adjusts the growth rate, capturing initial exponential-like growth that tapers off. The product form ensures the rate is zero at $P = 0$ or $P = 5000$ . A tempting distractor like choice B, $\dfrac{dP}{dt} = k(5000 - P)$ , 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.
An investment balance $B(t)$ earns continuous interest at $5\%$ per year and has withdrawals of $2000$ dollars per year. Which differential equation models $B$ ?
Answer: $\dfrac{dB}{dt}=0.05B-2000$ 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.
A drug amount $A$ in blood decreases at a rate proportional to $A$ . Which differential equation models $A(t)$ ?
Answer: $\dfrac{dA}{dt}=-kA$ 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.
A cup of coffee has temperature $T(t)$ in a $20^\circ$ C room; cooling rate is proportional to $T(t)-20$ . Which differential equation models $T(t)$ ?
Answer: $\dfrac{dT}{dt}=-k\,(T-20)$ 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.
A car's speed $v$ decreases at a rate proportional to its speed due to drag; which differential equation models $v(t)$ ?
Answer: $\dfrac{dv}{dt}=-kv$ This problem involves modeling situations with differential equations, a key skill in AP Calculus BC. The car's speed $v$ decreases at a rate proportional to $v$ itself due to drag, meaning slowdown intensifies with higher speed. This is modeled as $\dfrac{dv}{dt} = -kv$ , with $k > 0$ to ensure a negative rate for positive $v$ . The negative sign captures the deceleration. A tempting distractor like choice A, $\dfrac{dv}{dt} = kv$ , 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.
A bacteria culture has carrying capacity $5000$ ; growth rate is proportional to $P$ and to $5000-P$ . Which models $P(t)$ ?
Answer: $\dfrac{dP}{dt}=kP(5000-P)$ This problem models logistic growth with a differential equation. Growth rate "proportional to P and to 5000-P" translates directly to dP/dt = kP(5000-P). This captures both exponential growth at small P and saturation as P approaches the carrying capacity 5000. Choice C incorrectly includes a negative sign, which would predict population decline. Logistic models always have positive k to ensure growth when P is between 0 and carrying capacity.
A hot metal rod has temperature $T(t)$ in a $100^\circ$ C furnace; its heating rate is proportional to $100-T$ . Which differential equation models this?
Answer: $\dfrac{dT}{dt}=k(100-T)$ This question tests the skill of modeling situations with differential equations. The verbal description states that a hot metal rod in a 100°C furnace has a heating rate proportional to 100 - T, meaning dT/dt = k(100 - T) for positive k when T < 100. This ensures the temperature increases toward the furnace temperature, aligning with Newton's law adapted for heating. The difference drives the change, approaching equilibrium as it diminishes. A tempting distractor is choice A, which reverses the difference and would cause cooling instead of heating for T < 100. When modeling temperature changes, orient the proportionality to make the derivative positive for heating or negative for cooling as appropriate.
A spring-mass system has displacement $x(t)$ and velocity $v(t)=\dfrac{dx}{dt}$ ; acceleration satisfies $\dfrac{dv}{dt}=-4x$ . Which differential equation for $x$ models motion?
Answer: $\dfrac{d^2x}{dt^2}=-4x$ This question tests the skill of modeling situations with differential equations. The verbal description provides that acceleration $\dfrac{dv}{dt} = -4x$ , and since $v = \dfrac{dx}{dt}$ , this implies the second derivative $\dfrac{d^2 x}{dt^2} = -4x$ . 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.
A tank contains $A(t)$ liters of acid; it is drained so the outflow rate is proportional to $\sqrt{A}$ . Which differential equation models $A(t)$ ?
Answer: $\dfrac{dA}{dt}=-k\sqrt{A}$ This problem requires modeling a draining process with a differential equation. The tank is being drained, so the amount A(t) is decreasing, meaning dA/dt must be negative. Since the outflow rate is proportional to √A, we have dA/dt = -k√A, where k is a positive constant. Choice A would incorrectly model filling rather than draining, as it would make dA/dt positive. The key insight is that "outflow" or "drainage" always corresponds to a negative rate of change, requiring a negative sign in the differential equation.
A rumor spreads so that the rate of new people hearing it is proportional to those who have heard it times those who have not. Which models $R(t)$ ?
Answer: $\dfrac{dR}{dt}=kR(N-R)$ This problem models rumor spreading with a differential equation. The rate depends on "those who have heard it times those who have not," which translates to R(N-R), where N is the total population. Thus dR/dt = kR(N-R), representing logistic growth. Choice B incorrectly suggests the rate depends only on those who haven't heard it. This classic model captures how spreading slows as fewer people remain uninformed, making it useful for modeling epidemics and information diffusion.
A hot cup cools so that its temperature change rate is proportional to $T-20$ (°C); which differential equation models $T(t)$ ?
Answer: $\dfrac{dT}{dt}=k(20-T)$ This problem involves modeling situations with differential equations, a key skill in AP Calculus BC. The verbal description states that the rate of temperature change for a hot cup is proportional to $T - 20$ , where 20°C is the ambient temperature, indicating Newton's Law of Cooling. Since the cup is cooling, the rate $dT/dt$ should be negative when $T > 20$ , so we express it as $dT/dt = k(20 - T)$ with k > 0 to ensure the temperature decreases toward the ambient. This setup captures the proportionality while accounting for the direction of change. A tempting distractor like choice B, $dT/dt = k(T - 20)$ , fails because it would predict heating when $T > 20$ , which contradicts the cooling scenario. A transferable modeling strategy is to always consider the direction of change and adjust the sign of the proportionality constant accordingly.
A bacteria culture has population $P(t)$ ; its growth rate is proportional to the current population. Which differential equation models this situation?
Answer: $\dfrac{dP}{dt}=kP$ 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 $P(t)$ , meaning the rate of increase $\frac{dP}{dt}$ equals $kP$ 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 $(P-1)$ 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.
A cup of coffee at $90^\circ$ C cools in a $20^\circ$ C room; its cooling rate is proportional to $T-20$ . Which differential equation models $T(t)$ ?
Answer: $\dfrac{dT}{dt}=k(20-T)$ This problem requires modeling Newton's Law of Cooling using a differential equation. The cooling rate (dT/dt) is proportional to the temperature difference between the coffee and room temperature, which is (T - 20). Since the coffee is cooling from 90°C toward 20°C, the temperature is decreasing, so dT/dt must be negative. This gives us dT/dt = -k(T - 20) = k(20 - T), where k is a positive constant. Choice A would incorrectly model heating rather than cooling, as it would make dT/dt positive when T > 20. The key strategy is to check the sign: when the object is warmer than the environment, the rate of change should be negative.
A car's speed $v$ decreases due to drag proportional to $v^2$ . Which differential equation models $v(t)$ while coasting?
Answer: $\dfrac{dv}{dt}=-kv^2$ This problem models velocity decrease due to drag using a differential equation. The phrase "drag proportional to v²" means the deceleration force is kv² for some positive constant k. Since drag opposes motion and decreases speed, dv/dt must be negative, giving us dv/dt = -kv². This quadratic drag model applies at higher speeds where air resistance dominates. Choice A would incorrectly model acceleration rather than deceleration. Remember that resistive forces always oppose motion, requiring a negative sign in the differential equation.
A tank contains $y$ liters of brine; it is drained at a rate proportional to the amount present. Which differential equation models $y(t)$ ?
Answer: $\dfrac{dy}{dt}=-ky$ 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.

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