Modeling Situations with Differential Equations

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AP Calculus AB › Modeling Situations with Differential Equations

Questions 1 - 10
1

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

$\dfrac{dv}{dt}=g-kv$

$\dfrac{dv}{dt}=g+kv$

$\dfrac{dv}{dt}=\dfrac{g}{kv}$

$\dfrac{dv}{dt}=kv-g$

$\dfrac{dt}{dv}=g-kv$

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.

2

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

$\dfrac{dt}{dV}=-kV$

$\dfrac{dV}{dt}=-\dfrac{k}{V}$

$\dfrac{dV}{dt}=-kV$

$\dfrac{dV}{dt}=kV+t$

$\dfrac{dV}{dt}=k$

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.

3

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

$\dfrac{dH}{dt}=kH$

$\dfrac{dH}{dt}=-\dfrac{k}{H}$

$\dfrac{dt}{dH}=\dfrac{k}{H}$

$\dfrac{dH}{dt}=\dfrac{H}{k}$

$\dfrac{dH}{dt}=\dfrac{k}{H}$

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 $\dfrac{dH}{dt} = \dfrac{k}{H}$ where $k > 0$. 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.

4

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

$\dfrac{dN}{dt}=-\dfrac{k}{\sqrt{N}}$

$\dfrac{dN}{dt}=-k\sqrt{N}$

$\dfrac{dN}{dt}=k\sqrt{N}$

$\dfrac{dN}{dt}=-kN$

$\dfrac{dt}{dN}=-k\sqrt{N}$

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.

5

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

$\dfrac{dC}{dt}=-k(C-1)$

$\dfrac{dC}{dt}=\dfrac{k}{C}$

$\dfrac{dC}{dt}=kC$

$\dfrac{dt}{dC}=-kC$

$\dfrac{dC}{dt}=-kC$

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.

6

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

$\dfrac{dm}{dt}=km$

$\dfrac{dm}{dt}=-km$

$\dfrac{dm}{dt}=k-m$

$\dfrac{dm}{dt}=-\dfrac{k}{m}$

$\dfrac{dt}{dm}=-km$

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 $\frac{dm}{dt} = -km$ where $k > 0$. 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.

7

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

$\dfrac{dI}{dt}=kI+bI^2$

$\dfrac{dI}{dt}=-kI-bI^2$

$\dfrac{dI}{dt}=-\dfrac{k}{I}-bI^2$

$\dfrac{dI}{dt}=-kI+b$

$\dfrac{dt}{dI}=-kI-bI^2$

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 $I(t)$ plus background absorption $bI(t)^2$," giving $\dfrac{dI}{dt} = -kI - bI^2$ where $k, b > 0$. Both terms are negative because they represent different types of losses. The $-kI$ term represents simple proportional loss while $-bI^2$ 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.

8

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

$\dfrac{dF}{dt}=\dfrac{-r}{F}$

$\dfrac{dF}{dt}=-rF$

$\dfrac{dt}{dF}=-r$

$\dfrac{dF}{dt}=-r$

$\dfrac{dF}{dt}=rF$

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 $\dfrac{dF}{dt} = -r$ where $r > 0$. 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.

9

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

$\dfrac{dP}{dt}=-kP(5000-P)$

$\dfrac{dP}{dt}=k(5000-P)$

$\dfrac{dP}{dt}=kP(5000-P)$

$\dfrac{dP}{dt}=k\dfrac{5000-P}{P}$

$\dfrac{dt}{dP}=kP(5000-P)$

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.

10

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

$\dfrac{dt}{dx}=-kx$

$\dfrac{dx}{dt}=-k t$

$\dfrac{dx}{dt}=kx$

$\dfrac{dx}{dt}=-k+x$

$\dfrac{dx}{dt}=-kx$

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.

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