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.

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.

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 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.

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 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 $\dfrac{dy}{dt} = 3 - 0.1y$. 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.

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.

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.

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.

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.