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.

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

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.

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).

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.

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.

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.

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 $V$, often due to Torricelli's law relating to height. This is $\dfrac{dV}{dt} = -k \sqrt{V}$, with $k > 0$, ensuring a negative rate as volume decreases. The square root reflects dependence on the surface level or pressure. A tempting distractor like choice A, $\dfrac{dV}{dt} = k \sqrt{V}$, 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.

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.

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.