Modeling with Differential Equations - Calculus 1

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 $k$ ?
Answer: $\frac{dh}{dt} = -k \sqrt{h}$ 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 $\frac{dh}{dt} = -k \sqrt{h}$ .
A body of mass $m$ is launched vertically upward from the ground with an initial velocity $v_0$ . The air resistance is proportional to the velocity $v(t)$ . Let the upward direction be positive. Which differential equation models the velocity of the body for $t > 0$ while it is in the air?
Answer: $\frac{dv}{dt} = -g - \frac{k}{m}v$ Using Newton's Second Law, $m\frac{dv}{dt} = F_{net}$ . The upward direction is positive. The force of gravity is $-mg$ as it acts downward. Air resistance opposes motion. When the body is moving upward ( $v > 0$ ), the resistance force is downward, so it is $-kv$ . When the body is falling ( $v < 0$ ), the resistance force is upward, so it is also $-kv$ (since $v$ is negative, $-kv$ is positive). Thus, the force is always $-kv$ . The net force is $F_{net} = -mg - kv$ . Therefore, $m\frac{dv}{dt} = -mg - kv$ , which gives $\frac{dv}{dt} = -g - \frac{k}{m}v$ .
A population of bacteria, $P(t)$ , doubles every hour. However, the bacteria are being removed at a constant rate of 100 cells per hour. Which differential equation models this situation?
Answer: $\frac{dP}{dt} = (\ln 2)P - 100$ If a population doubles every hour, its growth is modeled by $P(t) = P_0 2^t$ . To find the corresponding differential equation, we write $2^t = e^{t \ln 2}$ , so $P(t) = P_0 e^{(\ln 2)t}$ . The differential equation for this exponential growth is $\frac{dP}{dt} = (\ln 2)P$ . The removal of bacteria occurs at a constant rate of 100, so we subtract this from the growth rate. The complete model is $\frac{dP}{dt} = (\ln 2)P - 100$ .
A 500-gallon tank contains brine with 100 kg of salt. Pure water enters at 10 gal/min, and the well-mixed solution exits at the same rate. Let $C(t)$ be the concentration of salt in kg/gal. Which differential equation models the concentration $C(t)$ ?
Answer: $\frac{dC}{dt} = -\frac{C}{50}$ Let $A(t)$ be the amount of salt. Since pure water enters, the rate of salt in is 0. The volume is constant at 500 gal. The concentration is $C = A/500$ . The rate of salt out is (concentration) × (outflow rate) = $(A/500) \times 10 = A/50$ . So, $\frac{dA}{dt} = -A/50$ . To find the DE for $C$ , substitute $A = 500C$ . Then $\frac{d(500C)}{dt} = -\frac{500C}{50}$ , which simplifies to $500 \frac{dC}{dt} = -10C$ . Dividing by 500 gives $\frac{dC}{dt} = -\frac{10}{500}C = -\frac{C}{50}$ .
The value of an investment, $V(t)$ , is growing at a rate proportional to its current value. However, the investor withdraws money continuously at a rate proportional to the square root of the investment's value. Which differential equation models the value of this investment?
Answer: $\frac{dV}{dt} = k_1 V - k_2 \sqrt{V}$ The rate of change of the investment's value, $\frac{dV}{dt}$ , is determined by two processes. The growth is proportional to the current value, which contributes a term $k_1 V$ . The withdrawals decrease the value at a rate proportional to the square root of the value, which contributes a term $-k_2 \sqrt{V}$ . The net rate of change is the sum of these two contributions: $\frac{dV}{dt} = k_1 V - k_2 \sqrt{V}$ .
A cylindrical tank with constant cross-sectional area $A$ and a small hole of area $a$ at the bottom is draining. By Torricelli's Law, the speed $v$ of the outflowing water is $v = c\sqrt{h}$ , where $h$ is the water height and $c$ is a constant. The rate of change of volume is $\frac{dV}{dt} = -av$ . Which differential equation models the height $h(t)$ of the water?
Answer: $\frac{dh}{dt} = -\frac{ac}{A}\sqrt{h}$ The volume of water in the cylinder is $V = Ah$ . Differentiating with respect to time gives $\frac{dV}{dt} = A \frac{dh}{dt}$ . We are also given $\frac{dV}{dt} = -av$ . Substituting $v = c\sqrt{h}$ gives $\frac{dV}{dt} = -ac\sqrt{h}$ . Equating the two expressions for $\frac{dV}{dt}$ yields $A \frac{dh}{dt} = -ac\sqrt{h}$ . Solving for $\frac{dh}{dt}$ gives $\frac{dh}{dt} = -\frac{ac}{A}\sqrt{h}$ .
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 $k > 0$ ?
Answer: $\frac{dc}{dt} = k(a_0 - c)(b_0 - c)$ The rate of formation of $C$ is $\frac{dc}{dt}$ . 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 $c$ of $A$ and a concentration $c$ of $B$ must be consumed. Therefore, the remaining concentration of $A$ is $[A] = a_0 - c$ , and the remaining concentration of $B$ is $[B] = b_0 - c$ . Substituting these into the rate equation gives $\frac{dc}{dt} = k(a_0 - c)(b_0 - c)$ .
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 $k$ ?
Answer: $\frac{dA}{dt} = R - kA$ 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 $\frac{dA}{dt} = R - kA$ .
The population $P(t)$ of a country grows due to a constant number of births per year, $B$ . People also immigrate to the country at a constant rate $I$ . The number of people who emigrate per year is proportional to the current population. Which differential equation models this population dynamic, where $k$ is a positive proportionality constant?
Answer: $\frac{dP}{dt} = B + I - kP$ The rate of change of the population, $\frac{dP}{dt}$ , is the sum of rates of increase and decrease. The birth rate is a constant increase $+B$ . The immigration rate is a constant increase $+I$ . The emigration rate is a decrease proportional to the population, $-kP$ . Summing these components gives the differential equation: $\frac{dP}{dt} = B + I - kP$ . Distractor A incorrectly models the constant birth rate $B$ as a proportional rate $BP$ .
A spherical snowball melts such that its surface area decreases at a rate proportional to its volume. Which differential equation models the radius $r$ of the snowball as a function of time $t$ for some positive constant $C$ ?
Answer: $\frac{dr}{dt} = -Cr^2$ Let $A$ be the surface area and $V$ be the volume. The geometric formulas are $A = 4\pi r^2$ and $V = \frac{4}{3}\pi r^3$ . The problem states $\frac{dA}{dt} = -kV$ for some positive constant $k$ . Using the chain rule, $\frac{dA}{dt} = \frac{dA}{dr} \frac{dr}{dt} = (8\pi r) \frac{dr}{dt}$ . Setting the two expressions for $\frac{dA}{dt}$ equal: $(8\pi r) \frac{dr}{dt} = -k(\frac{4}{3}\pi r^3)$ . Solving for $\frac{dr}{dt}$ gives $\frac{dr}{dt} = -\frac{4k\pi r^3}{3 \cdot 8\pi r} = -\frac{k}{6}r^2$ . Since $k/6$ is a positive constant, we can write this as $\frac{dr}{dt} = -Cr^2$ .
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?
Answer: $\frac{dP}{dt} = k_1 P - k_2 \sqrt{P}$ 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 $\frac{dP}{dt} = k_1 P - k_2 \sqrt{P}$ .
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 $k$ ?
Answer: $\frac{dr}{dt} = \frac{k}{r^2}$ 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 $\frac{dr}{dt} = \frac{k}{r^2}$ .
An object of mass $m$ is propelled across a surface with a constant force $F_0$ . The surface exerts a resistive force proportional to the cube root of the object's velocity, $v(t)$ . Which differential equation models the velocity of the object?
Answer: $\frac{dv}{dt} = \frac{F_0}{m} - \frac{k}{m} \sqrt[3]{v}$ Using Newton's Second Law, $m \frac{dv}{dt} = F_{net}$ . The net force is the sum of the propulsive force and the resistive force. The propulsive force is constant, $F_0$ . The resistive force opposes motion and is proportional to the cube root of the velocity, so it is $-k v^{1/3}$ for some constant $k>0$ . The net force is $F_0 - k v^{1/3}$ . Thus, $m \frac{dv}{dt} = F_0 - k v^{1/3}$ . Dividing by $m$ gives $\frac{dv}{dt} = \frac{F_0}{m} - \frac{k}{m} v^{1/3}$ .
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 $I(t)$ ?
Answer: $\frac{dI}{dt} = k_1 I(N-I) - k_2 I$ 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 $\frac{dI}{dt} = k_1 I(N-I) - k_2 I$ .
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?
Answer: $\frac{dY}{dt} = k(10Y - Y^{3/2})$ 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 $\frac{dY}{dt} = k(10Y - Y \cdot Y^{1/2}) = k(10Y - Y^{3/2})$ .
A tank initially contains 100 L of pure water. A salt solution with a concentration of 0.5 kg/L is pumped into the tank at a rate of 4 L/min. The well-mixed solution is pumped out at a rate of 3 L/min. Let A(t) be the amount of salt in kilograms in the tank at time t in minutes. Which of the following differential equations models this situation?
Answer: $\frac{dA}{dt} = 2 - \frac{3A}{100+t}$ The rate of change of the amount of salt, $\frac{dA}{dt}$ , is the rate at which salt enters minus the rate at which it leaves. Rate in = (concentration of inflow) × (rate of inflow) = (0.5 kg/L) × (4 L/min) = 2 kg/min. The volume of the solution in the tank changes over time. The initial volume is 100 L. Water flows in at 4 L/min and out at 3 L/min, so the net change is +1 L/min. The volume at time t is $V(t) = 100 + t$ . The concentration of salt in the tank at time t is $\frac{A(t)}{V(t)} = \frac{A(t)}{100+t}$ kg/L. Rate out = (concentration in tank) × (rate of outflow) = $\left(\frac{A}{100+t}\right) \text{kg/L} \times (3 \text{ L/min}) = \frac{3A}{100+t}$ kg/min. Therefore, the differential equation is $\frac{dA}{dt} = 2 - \frac{3A}{100+t}$ .
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 $S$ ?
Answer: $\frac{dS}{dt} = k_1(M-S) - k_2 S$ 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: $\frac{dS}{dt} = k_1(M-S) - k_2 S$ .
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.
Answer: $\frac{dT}{dt} = -k(T - (T_0 + at))$ 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: $\frac{dT}{dt} = -k(T - (T_0 + at))$ .
A drug is administered intravenously at a constant rate $R$ . The body eliminates the drug at a rate proportional to the amount of the drug, $Q(t)$ , currently in the body. Which differential equation models the amount of the drug in the body?
Answer: $\frac{dQ}{dt} = R - kQ$ The rate of change of the drug amount, $\frac{dQ}{dt}$ , is the rate in minus the rate out. The drug is administered at a constant rate $R$ , so the rate in is $R$ . The drug is eliminated at a rate proportional to the amount present, $Q$ , so the rate out is $kQ$ for some positive constant $k$ . Therefore, the net rate of change is $\frac{dQ}{dt} = R - kQ$ .
Money is deposited continuously into a savings account at a rate of $S$ dollars per year. The account earns interest at an annual rate of $r$ , compounded continuously. Let $A(t)$ be the amount of money in the account at time $t$ . Which differential equation models the value of the account?
Answer: $\frac{dA}{dt} = rA + S$ The rate of change of the amount, $\frac{dA}{dt}$ , has two components. First, the amount increases due to interest, which, when compounded continuously, is at a rate proportional to the current amount: $rA$ . Second, the amount increases due to continuous deposits at a constant rate of $S$ . The total rate of change is the sum of these two rates: $\frac{dA}{dt} = rA + S$ .

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