Modeling Situations with Differential Equations - AP Calculus AB

The amount of medicine $M(t)$ decreases at a rate proportional to $M(t)$ . Which differential equation models $M$ ?
Answer: $\dfrac{dM}{dt}=-kM$ This problem models exponential decay where the medicine amount decreases at a rate proportional to the current amount. Since the medicine "decreases at a rate proportional to $M(t)$ ," we need $\dfrac{dM}{dt} = -kM$ where $k > 0$ . The negative sign is essential because the amount is decreasing over time due to the body processing the medicine. Choice B would represent growth rather than decay by missing the negative sign. Choice C has an incorrect inverse relationship with $M$ in the denominator rather than as a factor. Choice E incorrectly adds a time-dependent term. For exponential decay problems, always include the negative sign to represent the decreasing quantity and ensure the rate is directly proportional to the current amount.
A balloon's radius $r(t)$ increases so that its surface area $4\pi r^2$ grows at a constant rate $c$ . Which differential equation models $r$ ?
Answer: $\dfrac{dr}{dt}=\dfrac{c}{8\pi r}$ This problem requires relating the rate of change of radius to the rate of change of surface area through the chain rule. The surface area S = 4πr² increases at rate c, so dS/dt = c. Using the chain rule: dS/dt = dS/dr · dr/dt = 8πr · dr/dt = c, which gives dr/dt = c/(8πr). The factor 8πr comes from differentiating 4πr² with respect to r. Choice C incorrectly uses the derivative of volume (4πr²) instead of surface area. Choice A ignores the geometric relationship entirely. When relating rates of change for geometric quantities, always use the chain rule with the appropriate derivative of the geometric formula.
A radioactive sample has mass $m(t)$ that decays at a rate proportional to $m(t)$ . Which differential equation models $m$ ?
Answer: $\dfrac{dm}{dt}=-km$ 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.
A tank has $S(t)$ grams of salt; brine enters adding salt at constant $4$ g/min and no salt leaves. Which equation models $S$ ?
Answer: $\dfrac{dS}{dt}=4$ This problem models a constant input rate with a differential equation. When salt enters at a "constant 4 g/min" with no removal, the rate of change is simply dS/dt = 4, independent of the current amount S. This creates linear growth rather than exponential growth. Choice B (dS/dt = 4S) incorrectly makes the rate proportional to the amount present, which would model exponential growth. When a quantity changes at a constant rate (not proportional to anything), the differential equation contains only constants on the right side.
A company's profit $P(t)$ increases at constant rate $c$ but decreases at a rate proportional to current profit. Which differential equation models $P$ ?
Answer: $\dfrac{dP}{dt}=c-kP$ This problem combines constant profit generation with proportional profit reduction, modeling a business with both income and expenses. The profit "increases at constant rate c but decreases at a rate proportional to current profit," giving dP/dt = c - kP where c,k > 0. The +c term represents steady income while -kP represents expenses that scale with profit size. Choice B has the terms switched, which would create unstable growth. Choice C has both terms positive, missing the expense component. Business profit models often balance constant income against proportional expenses or taxes.
A ball rolling on grass has speed $v(t)$ decreasing at a rate proportional to $v(t)^3$ . Which differential equation models $v$ ?
Answer: $\dfrac{dv}{dt}=-kv^3$ This problem describes friction with a cubic dependency on speed, representing high-speed drag or rolling resistance. The ball's speed "decreases at a rate proportional to v(t)³," giving dv/dt = -kv³ where k > 0. The negative sign indicates decreasing speed due to friction, and the cubic relationship represents nonlinear drag that increases rapidly with speed. Choice B would be linear friction. Choice C has the wrong sign representing acceleration. Cubic friction terms often appear in high-speed scenarios where air resistance or other nonlinear effects dominate.
A rumor spreads so that $R(t)$ increases at a rate proportional to $R(t)$ and to $(500-R(t))$ . Which differential equation models $R$ ?
Answer: $\dfrac{dR}{dt}=kR(500-R)$ This problem describes rumor spreading following a logistic model where the spread rate depends on both current believers and potential new believers. The rate is proportional to R(t) (current believers who can spread the rumor) and to (500-R(t)) (people who haven't heard it yet), giving dR/dt = kR(500-R). This creates the characteristic S-curve of information spread in populations. Choice B ignores the dependence on current rumor bearers. Choice C uses division instead of multiplication, representing a different relationship. Rumor and information spread models typically follow logistic growth patterns with the product of informed and uninformed populations.
A hot drink has temperature $T(t)$ in a $20^\circ$ C room; cooling rate is proportional to $T-20$ . Which equation models $T$ ?
Answer: $\dfrac{dT}{dt}=-k(T-20)$ This is a Newton's Law of Cooling problem requiring differential equation modeling. The cooling rate being "proportional to T-20" means dT/dt = -k(T-20), where the negative sign indicates cooling (temperature decreasing). Since the drink starts hot and cools toward room temperature (20°C), T > 20 and T-20 is positive, so we need the negative sign to make dT/dt negative. Choice B (dT/dt = k(T-20)) would incorrectly model heating rather than cooling. For temperature problems, the rate of change is proportional to the difference between object temperature and ambient temperature, with sign indicating heating or cooling direction.
A lake's water level $L(t)$ rises at rate proportional to rainfall $R(t)$ and $R(t) = R_0$ is constant. Which differential equation models $L$ ?
Answer: $\dfrac{dL}{dt}=kR_0$ This problem models water level change due to constant rainfall, creating a simple linear relationship. The lake's water level "rises at rate proportional to rainfall $R(t)$ and $R(t)=R_0$ is constant," giving $\dfrac{dL}{dt} = k R_0$ where $k > 0$ . Since rainfall is constant at $R_0$ , the proportional rate becomes a constant $k R_0$ . Choice B would represent exponential growth in lake level. Choice D incorrectly multiplies by both k, $R_0$ , and $L$ . When the driving factor in a proportional relationship is constant, the differential equation becomes a simple constant rate equation.
A bacteria culture has $N(t)$ cells; the growth rate is proportional to $N^{2/3}$ . Which differential equation models $N$ ?
Answer: $\dfrac{dN}{dt}=kN^{2/3}$ This problem requires modeling bacterial growth with a differential equation involving a power function. The phrase "growth rate is proportional to $N^{2/3}$ " translates directly to dN/dt = $kN^{2/3}$ , where k is positive for growth. This non-standard growth model might represent surface-area-limited growth since $N^{2/3}$ relates to surface area of a three-dimensional colony. Choice C incorrectly includes a negative sign, which would model decay rather than growth. When the rate depends on a power of the quantity, simply replace the quantity with that power in the standard proportional relationship.
A savings account balance $B(t)$ earns interest at a rate proportional to $B(t)$ . Which differential equation models $B$ ?
Answer: $\dfrac{dB}{dt}=kB$ This problem models exponential growth where the savings account balance increases due to continuous compound interest. The phrase "earns interest at a rate proportional to B(t)" means the rate of increase equals k times the current balance, giving dB/dt = kB where k > 0. This represents exponential growth since more money earns more interest. Choice A would represent exponential decay with the wrong negative sign. Choice D represents linear growth by adding B rather than multiplying by a proportionality constant. Choice E also represents decay rather than growth. When modeling compound interest or exponential growth, the rate of change should be positive and directly proportional to the current amount.
A car's speed $v(t)$ increases at a constant rate of $3$ m/s $^2$ . Which differential equation models this?
Answer: $\dfrac{dv}{dt}=3$ This problem models constant acceleration in physics. A "constant rate of 3 m/s²" means the velocity's rate of change is constant, so dv/dt = 3. This represents uniform acceleration where velocity increases by 3 m/s every second. Choice B (dv/dt = 3v) would model exponential growth of velocity, not constant acceleration. When the rate of change is described as constant, the differential equation has a constant on the right side, not a variable-dependent term.
A hot object's temperature $T(t)$ cools at a rate proportional to $T(t)-20$ . Which differential equation models $T$ ?
Answer: $\dfrac{dT}{dt}=-k(T-20)$ This problem applies Newton's Law of Cooling, where temperature change is proportional to the temperature difference from the ambient environment. The object cools, meaning its temperature T(t) decreases toward the ambient temperature of 20°. The rate of cooling is proportional to the temperature difference $(T-20)$ , so $dV/dt = -k(T-20)$ where k > 0. The negative sign indicates cooling (decreasing temperature). Choice A lacks the negative sign, representing heating instead of cooling. Choice C uses $(20-T)$ instead of $(T-20)$ , which would give the wrong sign behavior. When modeling temperature change, always identify the ambient temperature and use the difference $(current - ambient)$ with appropriate sign for heating or cooling.
A tank has $y(t)$ liters; it is filled at constant rate $a$ and drains at rate proportional to $y(t)$ . Which differential equation models $y$ ?
Answer: $\dfrac{dy}{dt}=a-ky$ This problem combines constant filling with proportional draining, requiring addition of a constant rate term and a proportional rate term. The tank "is filled at constant rate a and drains at rate proportional to y(t)," giving $\frac{dy}{dt} = a - ky$ where $k > 0$ . The + $a$ term represents constant filling while - $ky$ represents proportional draining. Choice B has the signs reversed, representing filling that depends on current volume. Choice C incorrectly places k inside the drainage term. Tank problems with constant inflow and proportional outflow always combine constant and proportional terms with opposite signs.
A virus count $V(t)$ decreases due to treatment at a rate proportional to $V(t)$ and to dosage $d$ (constant). Which differential equation models $V$ ?
Answer: $\dfrac{dV}{dt}=-kdV$ This problem models treatment effectiveness where the rate of virus reduction depends on both current virus count and treatment dosage. The virus count "decreases due to treatment at a rate proportional to V(t) and to dosage d," giving dV/dt = -kdV where k > 0. The negative sign indicates decreasing virus count, and the rate depends on both the current viral load and the constant dosage strength. Choice B has an incorrect form with V and d in different relationships. Medical treatment models often show effectiveness proportional to both pathogen load and treatment intensity.
A candle's height $h(t)$ decreases at a rate proportional to its current height squared. Which differential equation models $h$ ?
Answer: $\dfrac{dh}{dt}=-kh^2$ This problem describes burning where the rate depends on the square of the current height, possibly representing increased surface area or burning intensity. The candle "decreases at a rate proportional to its current height squared," giving dh/dt = -kh² where k > 0. The negative sign indicates decreasing height, and the h² dependence suggests the burning rate accelerates as conditions change. Choice B would be simple exponential decay. Choice C has the wrong sign representing growth rather than burning. Quadratic dependencies in rate problems often arise from area effects or accelerating processes.
The rate of change of pressure $p(t)$ is proportional to the square root of $p(t)$ . Which differential equation models $p$ ?
Answer: $\dfrac{dp}{dt}=k\sqrt{p}$ This problem establishes a direct proportional relationship between the rate of change and the square root of the current pressure. The rate of change "is proportional to the square root of p(t)," giving dp/dt = k√p where k > 0. The positive sign indicates increasing pressure, and the square root relationship might represent physical processes like diffusion or flow through porous media. Choice B would be exponential growth. Choice C has an inverse relationship that would slow down as pressure increases. Square root dependencies often appear in physical processes involving gradients or flow rates.
A bank account balance $B(t)$ changes due to continuous interest proportional to $B(t)$ and constant withdrawals $w$ . Which differential equation models $B$ ?
Answer: $\dfrac{dB}{dt}=kB-w$ This problem combines continuous compound interest with constant withdrawals, requiring both proportional and constant rate terms. The balance increases due to interest proportional to B(t) and decreases due to constant withdrawals w, giving dB/dt = kB - w. The kB term represents interest earnings while -w represents steady withdrawals. Choice B incorrectly factors the terms together rather than keeping them separate. Choice E has both terms negative, representing interest as a loss. Bank account problems with interest and withdrawals always combine proportional growth with constant decrease terms additively.
The rate of change of a rumor's spread $R(t)$ is proportional to $R(1-R)$ . Which differential equation models this?
Answer: $\dfrac{dR}{dt}=kR(1-R)$ This problem models rumor spread in a population using logistic dynamics. The rate "proportional to R(1-R)" translates directly to dR/dt = kR(1-R), where R represents the fraction who know the rumor. The R factor represents those who can spread it, while (1-R) represents those who can still learn it. Choice B (dR/dt = k(1-R)) misses the R factor, failing to account for the spreading population. For interaction models, multiply factors representing both interacting groups.
A bacteria culture has population $P(t)$ and grows at a rate proportional to $P$ ; which differential equation models this?
Answer: $\dfrac{dP}{dt}=kP$ This problem requires modeling a situation with a differential equation. The bacteria population P(t) grows at a rate proportional to P, which means the rate of change dP/dt equals some constant k times the current population P. This translates directly to the differential equation dP/dt = kP, where k is the proportionality constant. The key phrase "proportional to P" tells us to multiply P by a constant, not add it or use other operations. Choice D incorrectly inverts the derivative notation, writing dt/dP instead of dP/dt, which would represent the reciprocal rate. When modeling growth or decay, always identify what quantity changes (P), what it changes with respect to (t), and what the rate is proportional to (P itself).

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