Calculus 2 Quiz: Modeling With Differential Equations
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Modeling With Differential EquationsQuestion 1 of 20

A tank initially contains 100 gallons of pure water. Brine containing 2 pounds of salt per gallon flows into the tank at 3 gallons per minute, and the well-mixed solution flows out at 2 gallons per minute. Which differential equation correctly models the amount of salt S(t)S(t) in the tank at time tt minutes?

dSdt=6−2S100+t\frac{dS}{dt} = 6 - \frac{2S}{100 + t}
dSdt=6−3S100+t\frac{dS}{dt} = 6 - \frac{3S}{100 + t}
dSdt=6−2S100\frac{dS}{dt} = 6 - \frac{2S}{100}
dSdt=4−2S100+t\frac{dS}{dt} = 4 - \frac{2S}{100 + t}
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Calculus 2 Quiz

Calculus 2 Quiz: Modeling With Differential Equations

Practice Modeling With Differential Equations in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Modeling With Differential Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A tank initially contains 100 gallons of pure water. Brine containing 2 pounds of salt per gallon flows into the tank at 3 gallons per minute, and the well-mixed solution flows out at 2 gallons per minute. Which differential equation correctly models the amount of salt S(t)S(t) in the tank at time tt minutes?

  1. dSdt=6−2S100+t\frac{dS}{dt} = 6 - \frac{2S}{100 + t} (correct answer)
  2. dSdt=6−3S100+t\frac{dS}{dt} = 6 - \frac{3S}{100 + t}
  3. dSdt=6−2S100\frac{dS}{dt} = 6 - \frac{2S}{100}
  4. dSdt=4−2S100+t\frac{dS}{dt} = 4 - \frac{2S}{100 + t}
Explanation: Salt enters at rate = (concentration in)(flow rate in) = 2 lb/gal × 3 gal/min = 6 lb/min. The tank volume changes: V(t) = 100 + (3-2)t = 100 + t gallons. Salt leaves at rate = (concentration out)(flow rate out) = S(t)100+t\frac{S(t)}{100+t} × 2 = 2S100+t\frac{2S}{100+t} lb/min. Therefore dSdt=6−2S100+t\frac{dS}{dt} = 6 - \frac{2S}{100+t}. Choice B uses wrong outflow coefficient, C ignores volume change, D uses wrong inflow rate.

Question 2

In a fixed population of NN individuals, the rate at which a rumor spreads is jointly proportional to the number of people who have heard the rumor, P(t)P(t), and the number of people who have not yet heard it. Which differential equation describes the spread of the rumor?

  1. dPdt=kP\frac{dP}{dt} = kP
  2. dPdt=k(N−P)\frac{dP}{dt} = k(N-P)
  3. dPdt=kP(N−P)\frac{dP}{dt} = kP(N-P) (correct answer)
  4. dPdt=kPN−P\frac{dP}{dt} = k\frac{P}{N-P}
Explanation: The number of people who have heard the rumor is PP. The number who have not heard it is N−PN-P. The phrase 'jointly proportional' means the rate, dPdt\frac{dP}{dt}, is proportional to the product of these two quantities. Therefore, dPdt=kP(N−P)\frac{dP}{dt} = kP(N-P) for some proportionality constant kk. This is the standard logistic growth model, often used for such scenarios.

Question 3

A spherical raindrop falls through air and grows by condensation. Its radius increases at a rate proportional to its surface area, and it falls with velocity proportional to its radius. If r(t)r(t) is the radius at time tt, which system of differential equations best models this situation?

  1. drdt=kr2\frac{dr}{dt} = kr^2, dvdt=gr−bv2\frac{dv}{dt} = gr - bv^2
  2. drdt=kr2\frac{dr}{dt} = kr^2, v=crv = cr (correct answer)
  3. drdt=kr\frac{dr}{dt} = kr, dvdt=gr−bv\frac{dv}{dt} = gr - bv
  4. drdt=k\frac{dr}{dt} = k, v=cr2v = cr^2
Explanation: The radius increases at a rate proportional to surface area: drdt=k⋅4πr2=kr2\frac{dr}{dt} = k \cdot 4\pi r^2 = kr^2 (absorbing 4π4\pi into kk). The velocity is proportional to radius: v=crv = cr. This is not a rate equation but a direct relationship. Choice A incorrectly models velocity with acceleration. Choice C uses wrong proportionality for radius growth. Choice D assumes constant radius growth rate.

Question 4

A tank initially contains 100 L of pure water. A brine solution with a salt concentration of 50 g/L is pumped into the tank at a rate of 2 L/min. The well-mixed solution is pumped out at a rate of 1 L/min. Let A(t)A(t) be the amount of salt in the tank in grams at time tt in minutes. Which differential equation models this scenario?

  1. dAdt=100−A100\frac{dA}{dt} = 100 - \frac{A}{100}
  2. dAdt=100−A100+t\frac{dA}{dt} = 100 - \frac{A}{100+t} (correct answer)
  3. dAdt=50−A100+t\frac{dA}{dt} = 50 - \frac{A}{100+t}
  4. dAdt=100−A100−t\frac{dA}{dt} = 100 - \frac{A}{100-t}
Explanation: The rate of change of salt, dAdt\frac{dA}{dt}, is the rate of salt in minus the rate of salt out. Rate in = (concentration in) × (flow rate in) = (50 g/L) × (2 L/min) = 100 g/min. The volume of the solution in the tank at time tt is V(t)=100+(2−1)t=100+tV(t) = 100 + (2-1)t = 100+t L. The concentration of salt in the tank at time tt is C(t)=A(t)V(t)=A100+tC(t) = \frac{A(t)}{V(t)} = \frac{A}{100+t} g/L. Rate out = (concentration out) × (flow rate out) = (A100+t)×(1 L/min)=A100+t\left(\frac{A}{100+t}\right) \times (1\text{ L/min}) = \frac{A}{100+t} g/min. Therefore, the differential equation is dAdt=100−A100+t\frac{dA}{dt} = 100 - \frac{A}{100+t}.

Question 5

A fish population, P(t)P(t), in a lake grows according to the logistic model with a carrying capacity of MM and intrinsic growth rate rr. Fishermen are allowed to harvest the fish at a constant rate of HH fish per year. Which differential equation correctly models the fish population?

  1. dPdt=rP(1−PM)−HP\frac{dP}{dt} = rP\left(1 - \frac{P}{M}\right) - HP
  2. dPdt=rP(1−PM)+H\frac{dP}{dt} = rP\left(1 - \frac{P}{M}\right) + H
  3. dPdt=rP−H\frac{dP}{dt} = rP - H
  4. dPdt=rP(1−PM)−H\frac{dP}{dt} = rP\left(1 - \frac{P}{M}\right) - H (correct answer)
Explanation: The logistic growth model is given by dPdt=rP(1−PM)\frac{dP}{dt} = rP\left(1 - \frac{P}{M}\right). Harvesting removes fish from the population, so it is a negative term in the rate of change. Since the harvesting occurs at a constant rate HH, this term is simply −H-H. Combining these gives the model dPdt=rP(1−PM)−H\frac{dP}{dt} = rP\left(1 - \frac{P}{M}\right) - H.

Question 6

A metal object's temperature T(t)T(t) changes over time. According to Newton's Law of Cooling, its rate of change is proportional to the difference between its temperature and the ambient temperature. The object is placed in a special chamber where the ambient temperature, A(t)A(t), is not constant but increases linearly with time according to A(t)=20+0.5tA(t) = 20 + 0.5t, where tt is in minutes. If the object is cooling and kk is a positive constant of proportionality, which equation models the object's temperature?

  1. dTdt=−k(T−(20+0.5t))\frac{dT}{dt} = -k(T - (20 + 0.5t)) (correct answer)
  2. dTdt=−k(T−20)\frac{dT}{dt} = -k(T - 20)
  3. dTdt=k(T−(20+0.5t))\frac{dT}{dt} = k(T - (20 + 0.5t))
  4. dTdt=−k(T−20)−0.5\frac{dT}{dt} = -k(T - 20) - 0.5
Explanation: Newton's Law of Cooling states dTdt\frac{dT}{dt} is proportional to T−AT - A. So, dTdt=C(T−A(t))\frac{dT}{dt} = C(T - A(t)). Since the object is cooling, T>A(t)T > A(t) and dTdt<0\frac{dT}{dt} < 0. This means the constant of proportionality CC must be negative. Letting C=−kC = -k where kk is a positive constant gives dTdt=−k(T−A(t))\frac{dT}{dt} = -k(T - A(t)). Substituting A(t)=20+0.5tA(t) = 20 + 0.5t gives dTdt=−k(T−(20+0.5t))\frac{dT}{dt} = -k(T - (20 + 0.5t)).

Question 7

Two chemicals, X and Y, react to form a substance Z. For every 1 gram of Z produced, 0.4 g of X and 0.6 g of Y are consumed. The rate of formation of Z is proportional to the product of the amounts of X and Y remaining. Initially, there are 100 g of X and 150 g of Y. Let z(t)z(t) be the amount of Z in grams at time tt. Which equation models this reaction?

  1. dzdt=k(100−z)(150−z)\frac{dz}{dt} = k(100 - z)(150 - z)
  2. dzdt=k(100−0.4z)(150−0.6z)\frac{dz}{dt} = k(100 - 0.4z)(150 - 0.6z) (correct answer)
  3. dzdt=k(0.4z)(0.6z)\frac{dz}{dt} = k(0.4z)(0.6z)
  4. dzdt=k((100−0.4z)+(150−0.6z))\frac{dz}{dt} = k((100 - 0.4z) + (150 - 0.6z))
Explanation: Let z(t)z(t) be the amount of Z. The amount of X consumed is 0.4z0.4z, and the amount of Y consumed is 0.6z0.6z. The amounts remaining are X(t)=100−0.4zX(t) = 100 - 0.4z and Y(t)=150−0.6zY(t) = 150 - 0.6z. The rate of formation of Z, dzdt\frac{dz}{dt}, is proportional to the product of the remaining amounts: dzdt=k(X(t))(Y(t))=k(100−0.4z)(150−0.6z)\frac{dz}{dt} = k(X(t))(Y(t)) = k(100 - 0.4z)(150 - 0.6z).

Question 8

In a simple predator-prey model, let R(t)R(t) be the population of rabbits and W(t)W(t) be the population of wolves. In the absence of wolves, the rabbit population grows at a rate proportional to its current size. The presence of wolves decreases the rabbit population at a rate jointly proportional to the number of rabbits and the number of wolves. Using aa and bb as positive constants, which equation models the rabbit population?

  1. dRdt=aR+bRW\frac{dR}{dt} = aR + bRW
  2. dRdt=aR(1−R/K)−bRW\frac{dR}{dt} = aR(1 - R/K) - bRW
  3. dRdt=aR−bW\frac{dR}{dt} = aR - bW
  4. dRdt=aR−bRW\frac{dR}{dt} = aR - bRW (correct answer)
Explanation: The natural growth of the rabbit population is proportional to its size, giving the term aRaR, where a>0a > 0. The predation by wolves decreases the rabbit population. This interaction rate is 'jointly proportional' to both populations, meaning it is proportional to their product, RWRW. Since this decreases the rabbit population, the term is −bRW-bRW, where b>0b > 0. The complete model is dRdt=aR−bRW\frac{dR}{dt} = aR - bRW.

Question 9

A patient receives a drug intravenously at a constant rate of RR mg/hour. The body eliminates the drug at a rate proportional to the amount, A(t)A(t), present in the bloodstream. Let k>0k > 0 be the elimination rate constant. Which differential equation models the amount of the drug in the bloodstream?

  1. dAdt=R−kA\frac{dA}{dt} = R - kA (correct answer)
  2. dAdt=−kA\frac{dA}{dt} = -kA
  3. dAdt=R+kA\frac{dA}{dt} = R + kA
  4. dAdt=R−kA2\frac{dA}{dt} = R - kA^2
Explanation: The rate of change, dAdt\frac{dA}{dt}, is the rate of administration minus the rate of elimination. The drug is administered at a constant rate RR. The body eliminates the drug at a rate proportional to the amount present, which means the elimination rate is kAkA. Since this process removes the drug, it is a negative term. Therefore, the overall model is dAdt=R−kA\frac{dA}{dt} = R - kA.

Question 10

The temperature T(t)T(t) of a chemical reaction, in degrees Celsius, is modeled by the equation dTdt=0.2(120−T)+5e−0.1t\frac{dT}{dt} = 0.2(120 - T) + 5e^{-0.1t}. Which statement best describes the physical situation being modeled?

  1. The reaction cools towards an ambient temperature of 120°C while an external cooling source is applied.
  2. The reaction exchanges heat with a 120°C environment and is also heated by an external source that decays over time. (correct answer)
  3. The ambient temperature is decaying exponentially, and the reaction is simultaneously generating heat.
  4. The reaction heats towards 120°C and is also cooled by a system whose effect diminishes over time.
Explanation: The term 0.2(120−T)0.2(120 - T) represents Newton's Law of Cooling/Heating, where the system's temperature TT tends toward an ambient temperature of 120°C. The term +5e−0.1t+5e^{-0.1t} is a positive, time-dependent term that adds to the rate of change. This represents an additional heat source. Since the term decreases as tt increases, it is a heat source that decays exponentially over time.

Question 11

In an SIR model of an epidemic, S(t)S(t) is the number of susceptible people, I(t)I(t) is the number of infected people, and R(t)R(t) is the number of recovered people. The rate of new infections is proportional to the product of SS and II. Infected people recover at a rate proportional to the number of infected people. Let β\beta be the infection rate constant and γ\gamma be the recovery rate constant. Which equation models the rate of change of the infected population, dIdt\frac{dI}{dt}?

  1. dIdt=−βSI\frac{dI}{dt} = -\beta SI
  2. dIdt=βSI−γI\frac{dI}{dt} = \beta SI - \gamma I (correct answer)
  3. dIdt=βSI+γI\frac{dI}{dt} = \beta SI + \gamma I
  4. dIdt=βSI−γ\frac{dI}{dt} = \beta SI - \gamma
Explanation: The infected population II increases as susceptible people get infected. This happens at a rate of βSI\beta SI. The infected population decreases as people recover. This happens at a rate proportional to the number of infected people, γI\gamma I. The net rate of change for the infected population is the rate of new infections minus the rate of recovery, which gives dIdt=βSI−γI\frac{dI}{dt} = \beta SI - \gamma I.

Question 12

A spherical raindrop evaporates at a rate proportional to its surface area. Let V(t)V(t) be the volume and r(t)r(t) be the radius of the raindrop. Which differential equation describes the rate of change of the radius, drdt\frac{dr}{dt}?

  1. drdt=−kr2\frac{dr}{dt} = -kr^2
  2. drdt=−kr\frac{dr}{dt} = -kr
  3. drdt=−k\frac{dr}{dt} = -k (correct answer)
  4. drdt=−kr\frac{dr}{dt} = -\frac{k}{r}
Explanation: The rate of evaporation is proportional to the surface area A=4πr2A = 4\pi r^2. So, dVdt=−cA=−c(4πr2)\frac{dV}{dt} = -c A = -c(4\pi r^2) for some positive constant cc. The volume of a sphere is V=43πr3V = \frac{4}{3}\pi r^3. Differentiating with respect to tt gives dVdt=ddt(43πr3)=4πr2drdt\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) = 4\pi r^2 \frac{dr}{dt}. Equating the two expressions for dVdt\frac{dV}{dt} gives 4πr2drdt=−c(4πr2)4\pi r^2 \frac{dr}{dt} = -c(4\pi r^2). Dividing both sides by 4πr24\pi r^2 results in drdt=−c\frac{dr}{dt} = -c. Letting k=ck=c, we find drdt=−k\frac{dr}{dt} = -k, a constant.

Question 13

The population P(t)P(t) of a city experiences a natural growth rate proportional to its size, with constant k1>0k_1 > 0. In addition, there is a net migration into the city at a rate that is also proportional to the city's current population size, with constant k2>0k_2 > 0. Which differential equation models this city's population growth?

  1. dPdt=(k1+k2)P\frac{dP}{dt} = (k_1 + k_2)P (correct answer)
  2. dPdt=k1P+k2\frac{dP}{dt} = k_1P + k_2
  3. dPdt=k1P(M−P)+k2P\frac{dP}{dt} = k_1P(M-P) + k_2P
  4. dPdt=k1k2P2\frac{dP}{dt} = k_1 k_2 P^2
Explanation: The rate of change of the population has two components. The natural growth is k1Pk_1P. The net migration is also proportional to the population, so it is k2Pk_2P. Both terms contribute positively to the growth. Therefore, the total rate of change is the sum of these two rates: dPdt=k1P+k2P\frac{dP}{dt} = k_1P + k_2P. This can be factored as dPdt=(k1+k2)P\frac{dP}{dt} = (k_1 + k_2)P.

Question 14

The rate at which a student learns a new set of vocabulary words is proportional to the difference between the maximum number of words they can learn, MM, and the number of words they have already learned, W(t)W(t). Let k>0k > 0 be the constant of proportionality. Which differential equation models this learning process?

  1. dWdt=kW\frac{dW}{dt} = kW
  2. dWdt=k(M−W)\frac{dW}{dt} = k(M - W) (correct answer)
  3. dWdt=kW(M−W)\frac{dW}{dt} = kW(M - W)
  4. dWdt=k(W−M)\frac{dW}{dt} = k(W - M)
Explanation: The quantity M−WM-W represents the number of words yet to be learned. The problem states that the learning rate, dWdt\frac{dW}{dt}, is proportional to this quantity. Therefore, we can write the relationship as dWdt=k(M−W)\frac{dW}{dt} = k(M - W) for some constant kk. Since W<MW < M and WW is increasing, dWdt>0\frac{dW}{dt} > 0, which is consistent with k>0k > 0.

Question 15

According to Kirchhoff's Voltage Law, the sum of voltage drops around a closed circuit equals the source voltage. In a series circuit containing a resistor with resistance RR, an inductor with inductance LL, and a constant voltage source EE, the voltage drop across the resistor is RI(t)RI(t) and across the inductor is LdIdtL\frac{dI}{dt}, where I(t)I(t) is the current. Which differential equation models the current in this RL circuit?

  1. LdIdt−RI=EL\frac{dI}{dt} - RI = E
  2. dIdt=E+RIL\frac{dI}{dt} = \frac{E+RI}{L}
  3. LdIdt+RI=EL\frac{dI}{dt} + RI = E (correct answer)
  4. I(t)=ER+LdIdtI(t) = \frac{E}{R} + L\frac{dI}{dt}
Explanation: Kirchhoff's Voltage Law states that the sum of the voltage drops across the components equals the voltage of the source. The voltage drop across the resistor is VR=RIV_R = RI. The voltage drop across the inductor is VL=LdIdtV_L = L\frac{dI}{dt}. Therefore, VR+VL=EV_R + V_L = E, which gives the equation RI+LdIdt=ERI + L\frac{dI}{dt} = E, commonly written as LdIdt+RI=EL\frac{dI}{dt} + RI = E.

Question 16

A 500-liter tank is full of pure water. Brine with a concentration of 20 g/L is pumped in at a rate of 10 L/min. The well-mixed solution is pumped out at the same rate. Let c(t)c(t) be the concentration of salt in the tank (in g/L) at time tt. Which differential equation models the rate of change of concentration, dcdt\frac{dc}{dt}?

  1. dcdt=200−10c\frac{dc}{dt} = 200 - 10c
  2. dcdt=150(20−c)\frac{dc}{dt} = \frac{1}{50}(20 - c) (correct answer)
  3. dcdt=10(20−c)\frac{dc}{dt} = 10(20 - c)
  4. dcdt=20−c50\frac{dc}{dt} = 20 - \frac{c}{50}
Explanation: It is easier to first model the amount of salt, A(t)A(t). dAdt=(rate in)−(rate out)\frac{dA}{dt} = (\text{rate in}) - (\text{rate out}). Rate in = (20 g/L) × (10 L/min) = 200 g/min. Rate out = (concentration out) × (rate out) = c(t)c(t) g/L × 10 L/min = 10c10c g/min. So, dAdt=200−10c\frac{dA}{dt} = 200 - 10c. The concentration c(t)c(t) is related to the amount A(t)A(t) by A(t)=V⋅c(t)=500c(t)A(t) = V \cdot c(t) = 500c(t). Since the volume V=500V=500 is constant, dAdt=500dcdt\frac{dA}{dt} = 500\frac{dc}{dt}. Equating the two expressions for dAdt\frac{dA}{dt} gives 500dcdt=200−10c500\frac{dc}{dt} = 200 - 10c. Dividing by 500 gives dcdt=200−10c500=20−c50=150(20−c)\frac{dc}{dt} = \frac{200 - 10c}{500} = \frac{20 - c}{50} = \frac{1}{50}(20 - c).

Question 17

The price P(t)P(t) of a certain commodity changes at a rate proportional to the excess demand (Demand - Supply). The demand for the commodity is given by D(P)=1000−2PD(P) = 1000 - 2P, and the supply is given by S(P)=100+3PS(P) = 100 + 3P. Let k>0k > 0 be the proportionality constant. Which differential equation models the price of the commodity?

  1. dPdt=k(900−5P)\frac{dP}{dt} = k(900 - 5P) (correct answer)
  2. dPdt=k(1100+P)\frac{dP}{dt} = k(1100 + P)
  3. dPdt=k(5P−900)\frac{dP}{dt} = k(5P - 900)
  4. dPdt=k(1000−2P)\frac{dP}{dt} = k(1000 - 2P)
Explanation: The rate of change of price is proportional to the excess demand, so dPdt=k(D−S)\frac{dP}{dt} = k(D - S). First, calculate the excess demand: D−S=(1000−2P)−(100+3P)=1000−2P−100−3P=900−5PD - S = (1000 - 2P) - (100 + 3P) = 1000 - 2P - 100 - 3P = 900 - 5P. Substituting this into the proportionality relationship gives dPdt=k(900−5P)\frac{dP}{dt} = k(900 - 5P).

Question 18

In a radioactive decay chain, substance A decays into substance B at a rate proportional to the amount of A, with constant k1>0k_1 > 0. Substance B then decays into a stable substance C at a rate proportional to the amount of B, with constant k2>0k_2 > 0. Let A(t)A(t) and B(t)B(t) be the amounts of substances A and B at time tt. Which equation models the amount of substance B?

  1. dBdt=−k2B\frac{dB}{dt} = -k_2 B
  2. dBdt=k2B−k1A\frac{dB}{dt} = k_2 B - k_1 A
  3. dBdt=k1A−k2B\frac{dB}{dt} = k_1 A - k_2 B (correct answer)
  4. dBdt=k1A+k2B\frac{dB}{dt} = k_1 A + k_2 B
Explanation: The amount of substance B changes due to two processes. It is created from the decay of A, and it is consumed by its own decay into C. The rate of creation of B is the rate at which A decays, which is k1Ak_1 A. This is a positive term for dBdt\frac{dB}{dt}. The rate of decay of B is proportional to its own amount, which is k2Bk_2 B. Since this process consumes B, it is a negative term for dBdt\frac{dB}{dt}. Combining these gives dBdt=k1A−k2B\frac{dB}{dt} = k_1 A - k_2 B.

Question 19

A student is memorizing a list of MM new terms. The rate of memorization is proportional to the number of terms not yet memorized. Concurrently, the student forgets terms at a rate proportional to the number of terms they have already memorized. Let N(t)N(t) be the number of terms memorized at time tt, and let k1k_1 and k2k_2 be positive constants for memorization and forgetting, respectively. Which equation models N(t)N(t)?

  1. dNdt=k1(M−N)−k2N\frac{dN}{dt} = k_1(M-N) - k_2N (correct answer)
  2. dNdt=k1(M−N)+k2N\frac{dN}{dt} = k_1(M-N) + k_2N
  3. dNdt=k1N−k2(M−N)\frac{dN}{dt} = k_1N - k_2(M-N)
  4. dNdt=k1M−k2N\frac{dN}{dt} = k_1M - k_2N
Explanation: The rate of change dNdt\frac{dN}{dt} has two parts. The rate of memorization adds to NN. It is proportional to the number of terms not yet memorized, which is M−NM-N. So, this term is k1(M−N)k_1(M-N). The rate of forgetting subtracts from NN. It is proportional to the number of terms already memorized, NN. So, this term is −k2N-k_2N. Combining these two effects gives the overall rate: dNdt=k1(M−N)−k2N\frac{dN}{dt} = k_1(M-N) - k_2N.

Question 20

The balance B(t)B(t) in a savings account grows from two sources: continuous interest and regular deposits. The interest rate is not constant but decreases over time, given by r(t)=r01+tr(t) = \frac{r_0}{1+t}, where r0r_0 is the initial rate. Additionally, a constant amount DD is deposited into the account per unit of time. Which differential equation models the account balance?

  1. dBdt=r0B+D\frac{dB}{dt} = r_0 B + D
  2. dBdt=r01+tB+D\frac{dB}{dt} = \frac{r_0}{1+t} B + D (correct answer)
  3. dBdt=r0B1+t−D\frac{dB}{dt} = \frac{r_0 B}{1+t} - D
  4. dBdt=r0Bln⁡(1+t)+D\frac{dB}{dt} = r_0 B \ln(1+t) + D
Explanation: The rate of change of the balance, dBdt\frac{dB}{dt}, is the sum of the rate of growth from interest and the rate of deposits. The growth from interest is the product of the current balance BB and the interest rate r(t)r(t), which is r(t)B=r01+tBr(t)B = \frac{r_0}{1+t}B. The deposits add to the balance at a constant rate DD. Summing these two contributions gives dBdt=r01+tB+D\frac{dB}{dt} = \frac{r_0}{1+t}B + D.