Calculus 1 Quiz: Modeling With Differential Equations

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Modeling With Differential EquationsQuestion 1 of 20

The rate of change of a quantity $Y$ is proportional to the product of $Y$ and $(10 - \sqrt{Y})$ . Which differential equation describes this relationship?

\frac{dY}{dt} = k(10Y - Y^{3/2})

\frac{dY}{dt} = k(Y + (10 - \sqrt{Y}))

\frac{dY}{dt} = kY / (10 - \sqrt{Y})

\frac{dY}{dt} = k(10 - \sqrt{Y})

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Calculus 1 Quiz

Calculus 1 Quiz: Modeling With Differential Equations

Practice Modeling With Differential Equations in Calculus 1 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 1.

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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

The rate of change of a quantity $Y$ is proportional to the product of $Y$ and $(10 - \sqrt{Y})$ . Which differential equation describes this relationship?

$\frac{dY}{dt} = k(10Y - Y^{3/2})$ (correct answer)
$\frac{dY}{dt} = k(Y + (10 - \sqrt{Y}))$
$\frac{dY}{dt} = kY / (10 - \sqrt{Y})$
$\frac{dY}{dt} = k(10 - \sqrt{Y})$

Explanation: The problem states that the rate of change,

\frac{dY}{dt}

, is proportional to the product of two terms:

Y

and

(10 - \sqrt{Y})

. The product is

Y(10 - \sqrt{Y})

. Therefore,

\frac{dY}{dt} = k \cdot Y(10 - \sqrt{Y})

for some proportionality constant

k

. Distributing the

Y

gives

\frac{dY}{dt} = k(10Y - Y \cdot Y^{1/2}) = k(10Y - Y^{3/2})

Question 2

The population of a fish species, $P(t)$ , in a lake grows at a rate proportional to its current size. However, to maintain the ecosystem, fish are harvested at a rate proportional to the square root of the current population. Let $k_1$ and $k_2$ be positive constants of proportionality for growth and harvesting, respectively. Which differential equation models the population of the fish?

$\frac{dP}{dt} = k_1 P - k_2 P$
$\frac{dP}{dt} = k_1 P^2 - k_2 \sqrt{P}$
$\frac{dP}{dt} = k_1 P - k_2 \sqrt{P}$ (correct answer)
$\frac{dP}{dt} = k_1 \sqrt{P} - k_2 P$

Explanation: The problem describes two competing effects on the population. The growth rate is proportional to the current size, which is modeled by the term

k_1 P

. The harvesting rate (a decrease) is proportional to the square root of the population, modeled by the term

-k_2 \sqrt{P}

. The net rate of change is the sum of these two rates. Therefore, the differential equation is

\frac{dP}{dt} = k_1 P - k_2 \sqrt{P}

Question 3

In a community of $M$ people, the rate at which the number of people $P(t)$ who have heard a rumor increases is proportional to the product of the number of people who have heard the rumor and the number of people who have not yet heard it. Which differential equation describes this situation for some constant $k > 0$ ?

$\frac{dP}{dt} = kP$
$\frac{dP}{dt} = k(M-P)$
$\frac{dP}{dt} = kP(M-P)$ (correct answer)
$\frac{dP}{dt} = kP + k(M-P)$

Explanation: Let

P(t)

be the number of people who have heard the rumor. The total number of people is

M

. The number of people who have not yet heard the rumor is

M - P(t)

. The problem states that the rate of increase,

\frac{dP}{dt}

, is proportional to the product of these two quantities. Therefore,

\frac{dP}{dt} = k \cdot P \cdot (M-P)

for some positive constant of proportionality

k

Question 4

In a fixed population of $N$ individuals, the rate of change of the number of infected people, $I(t)$ , is proportional to the product of the number of infected people and the number of susceptible people ( $N-I$ ). Additionally, infected people are removed from the infected group at a rate proportional to the number of infected people. Let $k_1$ be the infection constant and $k_2$ be the removal constant. Which equation models $I(t)$ ?

$\frac{dI}{dt} = k_1 I(N-I) - k_2 I$ (correct answer)
$\frac{dI}{dt} = k_1 I(N-I) + k_2 I$
$\frac{dI}{dt} = k_1 N - k_2 I$
$\frac{dI}{dt} = k_1 I(N-I) - k_2$

Explanation: The rate of change

\frac{dI}{dt}

is composed of two processes. The rate of new infections increases

I

and is proportional to the product of infected and susceptible individuals, giving the term

k_1 I(N-I)

. The rate of removal decreases

I

and is proportional to the number of infected individuals, giving the term

-k_2 I

. Combining these results in the differential equation

\frac{dI}{dt} = k_1 I(N-I) - k_2 I

Question 5

A cylindrical water tank with a constant radius of 5 meters leaks water from a hole at its base. The rate at which the volume of water decreases is directly proportional to the square root of the height $h$ of the water in the tank. Let $h(t)$ be the height of the water at time $t$ . Which differential equation models this situation for some positive constant $k$ ?

$\frac{dh}{dt} = -k \sqrt{h}$ (correct answer)
$\frac{dh}{dt} = k \sqrt{h}$
$\frac{dh}{dt} = -k h$
$\frac{dh}{dt} = -k h^2$

Explanation: Let

V(t)

be the volume of water and

h(t)

be the height at time

t

. The volume of a cylinder is

V = \pi r^2 h

. Since the radius

r=5

is constant,

V = 25\pi h

. The rate of change of volume is

\frac{dV}{dt} = 25\pi \frac{dh}{dt}

. The problem states that the rate of decrease of volume is proportional to

\sqrt{h}

. Let this rate of decrease be

R_d

. Then

R_d = -\frac{dV}{dt}

. So,

-\frac{dV}{dt} = c \sqrt{h}

for some positive proportionality constant

c

. This implies

\frac{dV}{dt} = -c \sqrt{h}

. Substituting the expression for

\frac{dV}{dt}

, we get

25\pi \frac{dh}{dt} = -c \sqrt{h}

. Solving for

\frac{dh}{dt}

gives

\frac{dh}{dt} = -\frac{c}{25\pi} \sqrt{h}

. Since

c

and

25\pi

are positive constants, we can define a new positive constant

k = \frac{c}{25\pi}

. Thus, the model is

\frac{dh}{dt} = -k \sqrt{h}

Question 6

An economic model suggests that the rate of growth of a company's profit, $P(t)$ , is proportional to its current profit. However, due to market saturation, this growth is hindered by a factor that is proportional to the square of the current profit. Which differential equation models the profit $P(t)$ , where $k_1$ and $k_2$ are positive constants?

$\frac{dP}{dt} = k_1 P + k_2 P^2$
$\frac{dP}{dt} = k_1 P^2 - k_2 P$
$\frac{dP}{dt} = k_1 P - k_2 P^2$ (correct answer)
$\frac{dP}{dt} = \frac{k_1 P}{k_2 P^2}$

Explanation: The total rate of change of profit,

\frac{dP}{dt}

, is the sum of two effects. The growth proportional to the current profit is represented by a term

k_1 P

, where

k_1 > 0

. The hindering factor proportional to the square of the profit is represented by a term

-k_2 P^2

, where

k_2 > 0

. The negative sign indicates that this factor hinders, or reduces, the rate of growth. Combining these terms gives the differential equation

\frac{dP}{dt} = k_1 P - k_2 P^2

Question 7

An object of mass $m$ is dropped from a great height. The object is subject to the force of gravity, $mg$ , and an opposing air resistance force that is proportional to its velocity, $v(t)$ . According to Newton's Second Law ( $F_{net} = ma$ ), which differential equation models the velocity $v(t)$ of the object, assuming the downward direction is positive? Let $k$ be a positive constant of proportionality.

$\frac{dv}{dt} = g - \frac{k}{m}v$ (correct answer)
$\frac{dv}{dt} = mg - kv$
$\frac{dv}{dt} = g + \frac{k}{m}v$
$\frac{dv}{dt} = \frac{k}{m}v - g$

Explanation: The net force on the object is the sum of the gravitational force and the air resistance force. Taking the downward direction as positive, the gravitational force is

F_g = mg

. The air resistance opposes motion, so it acts in the upward (negative) direction. Its magnitude is proportional to velocity, so

F_{air} = -kv

. The net force is

F_{net} = mg - kv

. By Newton's Second Law,

F_{net} = ma

, where

a = \frac{dv}{dt}

. So,

m\frac{dv}{dt} = mg - kv

. To find the differential equation for

v(t)

, we divide by

m

\frac{dv}{dt} = g - \frac{k}{m}v

Question 8

The population of rabbits, $R(t)$ , in an ecosystem grows at a rate proportional to its current size. The rabbits are hunted by foxes, and the rate of predation is proportional to the product of the rabbit and fox populations. Assume the fox population is constant at $F$ individuals. Which differential equation models the rabbit population $R(t)$ , using positive constants of proportionality $k_1$ and $k_2$ ?

$\frac{dR}{dt} = k_1 R - k_2 F$
$\frac{dR}{dt} = k_1 R + k_2 RF$
$\frac{dR}{dt} = k_1 R - k_2 (R+F)$
$\frac{dR}{dt} = k_1 R - k_2 RF$ (correct answer)

Explanation: The rate of change of the rabbit population,

\frac{dR}{dt}

, is the natural growth rate minus the predation rate. The natural growth rate is proportional to the rabbit population, so it can be modeled as

k_1 R

for some

k_1 > 0

. The predation rate decreases the rabbit population and is proportional to the product of the rabbit and fox populations, so it can be modeled as

k_2 RF

for some

k_2 > 0

. Combining these gives

\frac{dR}{dt} = k_1 R - k_2 RF

Question 9

A spherical balloon is being inflated. The radius $r$ of the balloon increases at a rate that is inversely proportional to its surface area. Which of the following differential equations models the radius $r(t)$ of the balloon as a function of time $t$ , for some positive constant $k$ ?

$\frac{dr}{dt} = \frac{k}{r^2}$ (correct answer)
$\frac{dr}{dt} = kr^2$
$\frac{dr}{dt} = \frac{k}{r}$
$\frac{dr}{dt} = \frac{k}{r^3}$

Explanation: The rate of increase of the radius is

\frac{dr}{dt}

. The surface area of a sphere is

A = 4\pi r^2

. The problem states that

\frac{dr}{dt}

is inversely proportional to the surface area

A

. This can be written as

\frac{dr}{dt} = \frac{c}{A}

for some positive constant

c

. Substituting the formula for

A

, we get

\frac{dr}{dt} = \frac{c}{4\pi r^2}

. Since

c

and

4\pi

are constants, we can combine them into a single positive constant

k = \frac{c}{4\pi}

. Thus, the differential equation is

\frac{dr}{dt} = \frac{k}{r^2}

Question 10

An initial investment in a fund grows at a rate proportional to the amount of money, $M(t)$ , in the account. At the same time, the investor withdraws money at a rate proportional to the square root of the amount in the account. Which differential equation models the amount of money $M(t)$ , where $k_1$ and $k_2$ are positive constants?

$\frac{dM}{dt} = k_1 M - k_2 \sqrt{M}$ (correct answer)
$\frac{dM}{dt} = k_1 M + k_2 \sqrt{M}$
$\frac{dM}{dt} = k_1 \sqrt{M} - k_2 M$
$\frac{dM}{dt} = k_1 M - k_2 M^2$

Explanation: The rate of change of money in the account,

\frac{dM}{dt}

, is the growth rate from investment minus the withdrawal rate. The growth rate is proportional to the amount

M

, which is represented by the term

k_1 M

for some

k_1 > 0

. The withdrawal rate is proportional to the square root of the amount,

\sqrt{M}

, which is represented by the term

k_2 \sqrt{M}

for some

k_2 > 0

. Since this is a withdrawal, this rate is subtracted from the growth rate. The resulting model is

\frac{dM}{dt} = k_1 M - k_2 \sqrt{M}

Question 11

In a chemical reaction, one molecule of substance $A$ combines with one molecule of substance $B$ to form one molecule of substance $C$ . Let $c(t)$ be the concentration of $C$ . The rate of formation of $C$ is proportional to the product of the remaining concentrations of $A$ and $B$ . Initially, the concentrations are $[A] = a_0$ , $[B] = b_0$ , and $[C] = 0$ . Which differential equation models the concentration $c(t)$ for some $k > 0$ ?

$\frac{dc}{dt} = k(a_0 - c)(b_0 - c)$ (correct answer)
$\frac{dc}{dt} = k(a_0 - c) + k(b_0 - c)$
$\frac{dc}{dt} = k(c - a_0)(c - b_0)$
$\frac{dc}{dt} = k(a_0 + c)(b_0 + c)$

Explanation: The rate of formation of

C

\frac{dc}{dt}

. This rate is proportional to the product of the concentrations of

A

and

B

, so

\frac{dc}{dt} = k[A][B]

. To form a concentration

c

of substance

C

, a concentration

c

A

and a concentration

c

B

must be consumed. Therefore, the remaining concentration of

A

[A] = a_0 - c

, and the remaining concentration of

B

[B] = b_0 - c

. Substituting these into the rate equation gives

\frac{dc}{dt} = k(a_0 - c)(b_0 - c)

Question 12

A tank initially contains 1000 liters of brine with 50 kg of dissolved salt. Pure water enters the tank at a rate of 10 liters per minute. The solution is kept thoroughly mixed and drains from the tank at the same rate. Let $S(t)$ be the amount of salt in the tank (in kg) at time $t$ (in minutes). Which differential equation models this process?

$\frac{dS}{dt} = 10 - \frac{S}{100}$
$\frac{dS}{dt} = -\frac{S}{100}$ (correct answer)
$\frac{dS}{dt} = -\frac{S}{10}$
$\frac{dS}{dt} = \frac{S}{100}$

Explanation: The rate of change of salt,

\frac{dS}{dt}

, is the rate of salt in minus the rate of salt out. Since pure water enters, the rate of salt in is 0. The rate of salt out is the product of the salt concentration and the outflow rate. The volume of the tank is constant at 1000 L. The concentration of salt at time

t

C(t) = \frac{S(t)}{1000}

kg/L. The outflow rate is 10 L/min. So, the rate of salt out is

(\frac{S}{1000} \text{ kg/L}) \times (10 \text{ L/min}) = \frac{10S}{1000} = \frac{S}{100}

kg/min. Therefore,

\frac{dS}{dt} = 0 - \frac{S}{100} = -\frac{S}{100}

Question 13

The population $P$ of a certain species in a protected area is growing at a rate proportional to the current population. Additionally, due to a conservation program, 100 individuals of the species are introduced into the area per year. Which differential equation models the population $P(t)$ at time $t$ for some positive constant of proportionality $k$ ?

$\frac{dP}{dt} = kP - 100$
$\frac{dP}{dt} = k(P + 100)$
$\frac{dP}{dt} = kP + 100t$
$\frac{dP}{dt} = kP + 100$ (correct answer)

Explanation: The rate of change of the population,

\frac{dP}{dt}

, is the sum of the natural growth rate and the introduction rate. The natural growth rate is proportional to the population, which can be written as

kP

for some positive constant

k

. The introduction of 100 individuals per year is a constant positive rate of 100. Adding these two rates gives the total rate of change:

\frac{dP}{dt} = kP + 100

Question 14

According to Newton's Law of Cooling, an object's temperature changes at a rate proportional to the difference between its temperature and the ambient temperature. An object with temperature $T(t)$ is in an environment where the ambient temperature $T_a$ increases linearly with time, such that $T_a(t) = T_0 + at$ for positive constants $T_0$ and $a$ . Which equation models the object's temperature $T(t)$ , assuming the object cools when its temperature is higher than ambient? Use $k$ as a positive proportionality constant.

$\frac{dT}{dt} = -k(T - T_0)$
$\frac{dT}{dt} = -k(T - (T_0 + at))$ (correct answer)
$\frac{dT}{dt} = k(T - (T_0 + at))$
$\frac{dT}{dt} = -k(T - T_a) + a$

Explanation: Newton's Law of Cooling states that

\frac{dT}{dt}

is proportional to the difference between the object's temperature

T

and the ambient temperature

T_a

. This is written as

\frac{dT}{dt} = c(T - T_a)

. If the object is hotter than the environment (

T > T_a

), it should cool, meaning

\frac{dT}{dt}

must be negative. This requires the constant of proportionality

c

to be negative. Let

c = -k

where

k

is a positive constant. So,

\frac{dT}{dt} = -k(T - T_a)

. In this problem, the ambient temperature is not constant but is given by

T_a(t) = T_0 + at

. Substituting this expression for

T_a

gives the correct model:

\frac{dT}{dt} = -k(T - (T_0 + at))

Question 15

The amount of a certain drug, $A(t)$ , in a patient's bloodstream decreases at a rate proportional to the amount present. Simultaneously, the drug is administered intravenously at a constant rate of $R$ milligrams per hour. Which differential equation models the amount $A(t)$ of the drug in the bloodstream for some positive constant $k$ ?

$\frac{dA}{dt} = R + kA$
$\frac{dA}{dt} = kA - R$
$\frac{dA}{dt} = R - kA$ (correct answer)
$\frac{dA}{dt} = k(R - A)$

Explanation: The rate of change of the amount of the drug,

\frac{dA}{dt}

, is the rate at which the drug is administered minus the rate at which it is eliminated. The drug is administered at a constant rate

R

. The drug decreases (is eliminated) at a rate proportional to the amount present, which can be written as

kA

for some positive constant

k

. Since this term represents a decrease, it is subtracted. Therefore, the overall rate of change is

\frac{dA}{dt} = R - kA

Question 16

The rate at which a person learns a new skill, $S(t)$ , is proportional to the difference between a maximum potential skill level, $M$ , and their current skill level. Concurrently, they forget the skill at a rate proportional to their current skill level. Let $k_1$ be the learning constant and $k_2$ be the forgetting constant (both positive). Which equation models the rate of change of the skill level $S$ ?

$\frac{dS}{dt} = k_1(M-S) + k_2 S$
$\frac{dS}{dt} = k_1 M - k_2 S$
$\frac{dS}{dt} = k_1(M-S) - k_2 S$ (correct answer)
$\frac{dS}{dt} = k_1 S - k_2(M-S)$

Explanation: The net rate of change

\frac{dS}{dt}

is the sum of the rate of learning and the rate of forgetting. The rate of learning is proportional to

M-S

, which can be written as

k_1(M-S)

. The rate of forgetting represents a decrease in skill, so it's a negative term. It's proportional to the current skill level

S

, so it can be written as

-k_2 S

. Combining these gives the total rate of change:

\frac{dS}{dt} = k_1(M-S) - k_2 S

Question 17

Water is poured into a leaky cylindrical bucket of radius 5 cm at a rate of $10 \text{ cm}^3/\text{s}$ . The water leaks out at a rate proportional to the square root of the water's height $h$ , with proportionality constant $c > 0$ . Which differential equation models the height $h$ of the water in the bucket?

$\frac{dh}{dt} = 10 - c\sqrt{h}$
$\frac{dh}{dt} = \frac{10 - ch}{25\pi}$
$\frac{dh}{dt} = 25\pi(10 - c\sqrt{h})$
$\frac{dh}{dt} = \frac{10 - c\sqrt{h}}{25\pi}$ (correct answer)

Explanation: Let

V

be the volume of water. The net rate of change of volume is

\frac{dV}{dt} = (\text{rate in}) - (\text{rate out}) = 10 - c\sqrt{h}

. The volume of water in a cylinder is

V = \pi r^2 h = \pi(5^2)h = 25\pi h

. Differentiating with respect to time gives

\frac{dV}{dt} = 25\pi \frac{dh}{dt}

. Equating the two expressions for

\frac{dV}{dt}

gives

25\pi \frac{dh}{dt} = 10 - c\sqrt{h}

. Solving for

\frac{dh}{dt}

yields

\frac{dh}{dt} = \frac{10 - c\sqrt{h}}{25\pi}

Question 18

In an ecosystem, the population of prey, $x(t)$ , grows at a rate proportional to its size. The presence of a constant predator population, $P_0$ , causes the prey population to decrease at a rate proportional to the product of the prey and predator populations. Which differential equation models the prey population $x$ ?

$\frac{dx}{dt} = k_1 x - k_2 P_0$
$\frac{dx}{dt} = k_1 x + k_2 x P_0$
$\frac{dx}{dt} = k_1 x - k_2 x P_0$ (correct answer)
$\frac{dx}{dt} = k_1 - k_2(x+P_0)$

Explanation: The prey population's rate of change,

\frac{dx}{dt}

, has two parts. The natural growth is proportional to its size, which gives a term

k_1 x

. The decrease due to predation is proportional to the product of prey and predator populations,

x P_0

, leading to a term

-k_2 x P_0

. The net rate of change is the sum of these two effects:

\frac{dx}{dt} = k_1 x - k_2 x P_0

Question 19

A cup of coffee, with temperature $T(t)$ at time $t$ , is placed in a room where the ambient temperature is not constant. The room's temperature, $T_a$ , starts at 20°C and increases linearly at a rate of 0.1°C per minute. According to Newton's Law of Cooling, the rate of change of the coffee's temperature is proportional to the difference between its temperature and the ambient temperature. Which differential equation, with cooling constant $k > 0$ , models this scenario?

$\frac{dT}{dt} = -k(T - (20 + 0.1t))$ (correct answer)
$\frac{dT}{dt} = -k(T - 20) + 0.1$
$\frac{dT}{dt} = -k(T - 20)$
$\frac{dT}{dt} = k(T - (20 + 0.1t))$

Explanation: Newton's Law of Cooling states that

\frac{dT}{dt} = -k(T - T_a)

, where

T_a

is the ambient temperature and

k

is a positive constant. In this problem, the ambient temperature is a function of time. It starts at 20°C and increases at 0.1°C per minute, so

T_a(t) = 20 + 0.1t

. Substituting this expression for

T_a

into Newton's law gives

\frac{dT}{dt} = -k(T - (20 + 0.1t))

Question 20

In a community of $N$ people, the rate at which a rumor spreads is proportional to the product of the number of people who have heard the rumor, $y(t)$ , and the number of people who have not yet heard it. Which differential equation models this situation for some positive constant $k$ ?

$\frac{dy}{dt} = ky$
$\frac{dy}{dt} = k(N-y)$
$\frac{dy}{dt} = k(y + (N-y))$
$\frac{dy}{dt} = ky(N-y)$ (correct answer)

Explanation: The rate of spread is

\frac{dy}{dt}

. The number of people who have heard the rumor is

y

. The total number of people is

N

, so the number of people who have not heard the rumor is

N-y

. The problem states that the rate is proportional to the product of these two quantities. Therefore,

\frac{dy}{dt} = k \cdot y \cdot (N-y)

, which is the logistic growth model.