Differential Equations Quiz: Interpreting Solutions In Context

What this quiz covers

This quiz focuses on Interpreting Solutions In Context, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

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.

Question 1

1. The body's temperature is approximately $31.4\,\degree$C.
2. The body is cooling at a rate of approximately $0.91\,\degree$C per hour. (correct answer)
3. The body's temperature has dropped by a total of $5.6\,\degree$C from its initial value.
4. The rate of cooling is equal to the cooling constant, which is $0.08\,\degree$C per hour.

Explanation: The question asks for the interpretation of the rate of change at $t=5$. First, we need to find the derivative of $T(t)$: $T'(t) = \frac{d}{dt}(20 + 17e^{-0.08t}) = 17(-0.08)e^{-0.08t} = -1.36e^{-0.08t}$. Now, evaluate at $t=5$: $T'(5) = -1.36e^{-0.08(5)} = -1.36e^{-0.4} \approx -1.36(0.6703) \approx -0.9116$. The units of $T'(t)$ are $\degree$C/hour. The negative sign indicates the temperature is decreasing (cooling). Therefore, at $t=5$ hours, the body is cooling at a rate of approximately $0.91\,\degree$C per hour.

Question 2

1. The circuit is critically damped, satisfying the condition $R^2 = 4L/C$.
2. The circuit is overdamped, satisfying the condition $R^2 > 4L/C$.
3. The circuit is underdamped, satisfying the condition $R^2 < 4L/C$. (correct answer)
4. The circuit has no resistance ($R=0$), leading to pure sinusoidal oscillation.

Explanation: The solution is of the form $q(t) = e^{-\alpha t}(\text{sinusoidal term})$. The presence of the exponential decay term $e^{-3t}$ indicates that there is damping, and the presence of the sinusoidal term $(\cos(4t) + \sin(4t))$ indicates that the charge oscillates. This combination of oscillation with exponential decay is characteristic of an underdamped system. For an RLC circuit, the condition for underdamped behavior is $R^2 - 4L/C < 0$, which is equivalent to $R^2 < 4L/C$.

Question 3

1. It indicates that the velocity of the mass decreases by a factor of $e^{-0.2}$ every second.
2. It represents the frequency of the damped oscillations of the mass.
3. It describes the exponential decay of the amplitude of the oscillations. (correct answer)
4. It specifies that the position of the mass decreases by $0.2$ meters each second.

Explanation: In the solution for a damped harmonic oscillator, $x(t) = A_0 e^{-\alpha t} \cos(\omega t - \phi)$, the term $A(t) = A_0 e^{-\alpha t}$ is the time-varying amplitude. Here, $A(t) = 0.4e^{-0.2t}$. The term $e^{-0.2t}$ specifically governs the rate at which this amplitude decays over time due to the damping force (e.g., friction or air resistance). It does not represent velocity, frequency, or a linear decrease in position.

Question 4

1. 20,000 atoms (correct answer)
2. 1000 atoms
3. 50 atoms
4. An infinite number of atoms.

Explanation: When you encounter radioactive decay problems asking for the "total number" of decays over an infinite time period, you're looking at an improper integral. The activity function $A(t) = 1000e^{-0.05t}$ gives you the rate of decay at any time $t$, so to find the total decays, you need to integrate from $t = 0$ to $t = \infty$. Setting up the integral: $\int_0^{\infty} 1000e^{-0.05t} \, dt$ To evaluate this, find the antiderivative of $1000e^{-0.05t}$, which is $\frac{1000}{-0.05}e^{-0.05t} = -20000e^{-0.05t}$. Now apply the limits: $\lim_{b \to \infty} \left[-20000e^{-0.05t}\right]_0^b = \lim_{b \to \infty} \left(-20000e^{-0.05b} + 20000e^0\right)$ As $b \to \infty$, $e^{-0.05b} \to 0$, so this equals $0 + 20000 = 20000$ atoms. Choice A (20,000 atoms) is correct. Choice B (1000 atoms) represents a common error of confusing the initial activity rate with the total decay count. Choice C (50 atoms) might come from incorrectly using just the decay constant $0.05$ in calculations. Choice D (infinite atoms) reflects the misconception that since we're integrating to infinity, the result must be infinite—but exponential decay functions converge to finite values. Study tip: Remember that activity rates must be integrated over time to get total quantities. Practice recognizing when exponential decay integrals converge to finite values despite infinite time intervals.

Question 5

1. 2 mol/(L·s)
2. 5 mol/(L·s)
3. 10 mol/(L·s)
4. 50 mol/(L·s) (correct answer)

Explanation: The rate of a second-order reaction is given by Rate $= -\frac{d[A]}{dt} = k[A]^2$. First, let's find the initial concentration $[A]_0$ by setting $t=0$: $[A]_0 = \frac{5}{1+10(0)} = 5$ mol/L. To find the rate constant $k$, we can compare the given solution to the standard form for a second-order reaction, $\frac{1}{[A]} = kt + \frac{1}{[A]_0}$. Rearranging the given solution: $\frac{1}{[A](t)} = \frac{1+10t}{5} = \frac{1}{5} + \frac{10}{5}t = 0.2 + 2t$. Comparing forms, we see that $k=2$ L/(mol·s) and $\frac{1}{[A]_0} = 0.2$, which confirms $[A]_0=5$. The initial rate is at $t=0$: Rate$_0 = k[A]_0^2 = (2 \text{ L/(mol·s)}) \times (5 \text{ mol/L})^2 = 2 \times 25 = 50$ mol/(L·s).

Question 6

1. The initial amount of salt in the tank is 80 kg.
2. The concentration of the incoming brine solution is 80 kg/L.
3. The long-term amount of salt in the tank approaches 80 kg. (correct answer)
4. The volume of the tank is 80 liters.

Explanation: The solution is $A(t) = 80 - 80e^{-0.05t}$. To find the long-term behavior, we evaluate the limit as $t \to \infty$: $\lim_{t\to\infty} A(t) = \lim_{t\to\infty} (80 - 80e^{-0.05t}) = 80 - 80(0) = 80$. This means that as time goes on, the amount of salt in the tank approaches a steady-state value of 80 kg. This occurs when the concentration in the tank becomes equal to the concentration of the incoming brine.

Question 7

1. 0 mg/L, as the patient had no drug in their system initially.
2. 4 mg/L, from a previous dose that was still being eliminated. (correct answer)
3. 11 mg/L, representing the difference between the steady-state and initial values.
4. 15 mg/L, which is the steady-state concentration the infusion approaches.

Explanation: The question asks for the initial concentration, which is the value of $C(t)$ at $t=0$. We can find this by substituting $t=0$ into the given solution: $C(0) = 15(1 - e^{-0.25(0)}) + 4e^{-0.25(0)}$ $C(0) = 15(1 - e^0) + 4e^0$ $C(0) = 15(1 - 1) + 4(1)$ $C(0) = 15(0) + 4 = 4$. Thus, the initial concentration was 4 mg/L. This implies the patient had a pre-existing amount of the drug in their system when the new infusion started.

Question 8

1. 10 minutes (correct answer)
2. 7.5 minutes
3. 5 minutes
4. 15 minutes

Explanation: When you encounter Torricelli's law problems, you're dealing with proportional relationships and rates of change. The key insight here is that if the outflow rate is proportional to $\sqrt{h}$, then you need to find when $\sqrt{h}$ becomes half its initial value. First, let's find the initial outflow rate. At $t = 0$: $h(0) = (4 - 0.2(0))^2 = 16$ meters, so $\sqrt{h(0)} = 4$. Since the outflow rate is proportional to $\sqrt{h}$, the initial rate is proportional to 4. For the rate to be 50% of its initial value, we need $\sqrt{h(t)} = 0.5 \times 4 = 2$. This means $h(t) = 4$. Now solve: $(4 - 0.2t)^2 = 4$. Taking the square root of both sides: $4 - 0.2t = 2$ (we take the positive root since $4 - 0.2t > 0$ in our domain). Solving: $0.2t = 2$, so $t = 10$ minutes. Looking at the wrong answers: B) 7.5 minutes gives $h = (4-1.5)^2 = 6.25$, making $\sqrt{h} = 2.5$ (62.5% of initial rate). C) 5 minutes gives $h = (4-1)^2 = 9$, making $\sqrt{h} = 3$ (75% of initial rate). D) 15 minutes gives $h = (4-3)^2 = 1$, making $\sqrt{h} = 1$ (25% of initial rate). The answer is A) 10 minutes. Study tip: In proportional rate problems, always identify what quantity the rate depends on, then set up the proportion correctly. Don't confuse the rate itself with the variable it depends on.

Question 9

1. At $t = 0$ years, when the population is first introduced.
2. At approximately $t = 2.75$ years. (correct answer)
3. At approximately $t = 3.47$ years.
4. As $t \to \infty$, when the population approaches its maximum size.

Explanation: The logistic model is $P(t) = \frac{M}{1+Ae^{-kt}}$. The population grows fastest when it reaches half of the carrying capacity, $M/2$. From the given solution, the carrying capacity is $M=20$ hundred fish. The population is increasing most rapidly when $P(t) = M/2 = 10$. We need to solve for $t$: $10 = \frac{20}{1 + 3e^{-0.4t}}$ $1 + 3e^{-0.4t} = 2$ $3e^{-0.4t} = 1$ $e^{-0.4t} = 1/3$ $-0.4t = \ln(1/3) = -\ln(3)$ $t = \frac{\ln(3)}{0.4} \approx \frac{1.0986}{0.4} \approx 2.7465$ years.

Question 10

1. $v_2(t) = 100(1 - e^{-0.098t})$ (correct answer)
2. $v_2(t) = 100(1 - e^{-0.196t})$
3. $v_2(t) = 50(1 - e^{-0.098t})$
4. $v_2(t) = 50(1 - e^{-0.392t})$

Explanation: When you encounter a falling object with air resistance proportional to velocity, you're dealing with a classic differential equation that models terminal velocity behavior. The key insight is understanding how mass affects both the forces involved and the resulting motion. The differential equation for this scenario is $m\frac{dv}{dt} = mg - kv$, where $m$ is mass, $g$ is gravitational acceleration, and $k$ is the air resistance coefficient. For objects with the same shape and size, $k$ remains constant regardless of mass. The general solution has the form $v(t) = \frac{mg}{k}(1 - e^{-\frac{k}{m}t})$. From the given solution $v(t) = 50(1 - e^{-0.196t})$, we can identify that the terminal velocity $\frac{mg}{k} = 50$ and $\frac{k}{m} = 0.196$. When mass doubles, the terminal velocity doubles to $\frac{2mg}{k} = 100$, but the time constant changes to $\frac{k}{2m} = 0.098$. This gives us $v_2(t) = 100(1 - e^{-0.098t})$. Choice A is correct because it properly accounts for both effects of doubling mass. Choice B incorrectly keeps the same time constant. Choice C doubles the time constant instead of halving it while keeping the wrong terminal velocity. Choice D uses the wrong terminal velocity and incorrectly doubles the time constant. Remember: when mass increases in air resistance problems, terminal velocity increases proportionally, but the approach to terminal velocity becomes slower (smaller exponential coefficient).

Question 11

1. It is the initial amount invested in the account at $t=0$.
2. It is the total amount of money that will be withdrawn from the account.
3. It is the annual rate of deposits into the account, in dollars per year.
4. It is the negative of the equilibrium principal for the given interest and deposit rates. (correct answer)

Explanation: The solution is for the differential equation $dA/dt = rA + d$, where $r$ is the interest rate and $d$ is the deposit rate. The general solution is $A(t) = (A_0 + d/r)e^{rt} - d/r$. Comparing this to the given solution, we can identify $r=0.05$ and $-d/r = -100000$. The term $-d/r$ represents the equilibrium solution of the DE, which is the principal amount for which the interest earned ($rA$) would be exactly cancelled by withdrawals (a negative $d$). In this model with deposits, it's a theoretical negative principal where growth from interest would be perfectly offset by the deposits. Thus, it's the negative of the equilibrium principal.

Question 12

1. The object reaches 40°C after approximately 4.9 minutes; it started at 75°C in a 25°C environment with cooling constant 0.1 per minute
2. The object reaches 40°C after approximately 8.1 minutes; it started at 75°C in a 25°C environment with cooling constant 0.1 per hour
3. The object reaches 40°C after approximately 4.9 minutes; it started at 100°C in a 25°C environment with cooling constant 0.1 per minute
4. The object reaches 40°C after approximately 8.1 minutes; it started at 100°C in a 25°C environment with cooling constant 0.1 per minute (correct answer)

Explanation: Newton's law of cooling describes how objects cool toward ambient temperature, following the form $T(t) = T_{\text{ambient}} + (T_{\text{initial}} - T_{\text{ambient}})e^{-kt}$. When you see this type of equation, identify each component: the ambient temperature (what the object approaches), the initial temperature difference, and the cooling rate constant. From $T(t) = 25 + 75e^{-0.1t}$, you can read directly that the ambient temperature is 25°C (the constant term), and the initial temperature is $25 + 75 = 100°C$ (when $t = 0$). The cooling constant is 0.1 per minute since $t$ is given in minutes. To find when the object reaches 40°C, solve: $40 = 25 + 75e^{-0.1t}$. Subtracting 25: $15 = 75e^{-0.1t}$. Dividing by 75: $0.2 = e^{-0.1t}$. Taking the natural logarithm: $\ln(0.2) = -0.1t$, so $t = \frac{\ln(0.2)}{-0.1} \approx 8.1$ minutes. Option A incorrectly calculates the time and misidentifies the initial temperature as 75°C instead of 100°C. Option B has the wrong time units for the cooling constant and also mistakes the initial temperature. Option C gets the time wrong while correctly identifying some parameters. Only option D correctly identifies all values: 8.1 minutes, initial temperature of 100°C, ambient temperature of 25°C, and cooling constant of 0.1 per minute. Remember: in exponential decay problems, always check your parameter interpretations by substituting $t = 0$ to verify the initial condition, and identify the horizontal asymptote as your ambient value.

Question 13

1. The amplitude drops below 5% after approximately 15 seconds; the oscillation period is 6π seconds with initial displacement of about -0.14 meters
2. The amplitude drops below 5% after approximately 30 seconds; the oscillation period is 2π/3 seconds with initial displacement of about -0.14 meters
3. The amplitude drops below 5% after approximately 15 seconds; the oscillation period is 2π/3 seconds with initial displacement of about 0.14 meters
4. The amplitude drops below 5% after approximately 30 seconds; the oscillation period is 2π/3 seconds with initial displacement of about 0.14 meters (correct answer)

Explanation: When analyzing damped harmonic oscillators, you need to identify three key components: the amplitude envelope, the oscillation frequency, and initial conditions. The general form is $x(t) = A_0 e^{-\gamma t}\cos(\omega t + \phi)$ where $A_0 e^{-\gamma t}$ represents the decaying amplitude. For this problem, the amplitude at time $t$ is $0.2e^{-0.1t}$ meters. The initial amplitude is $0.2$ meters, so 5% of this is $0.01$ meters. To find when the amplitude drops to this level: $0.2e^{-0.1t} = 0.01$, which gives $e^{-0.1t} = 0.05$. Taking the natural logarithm: $-0.1t = \ln(0.05) \approx -2.996$, so $t \approx 30$ seconds. The oscillation period comes from the angular frequency $\omega = 3$ rad/s, giving $T = \frac{2\pi}{\omega} = \frac{2\pi}{3}$ seconds. For initial displacement, substitute $t = 0$: $x(0) = 0.2\cos(\frac{\pi}{4}) = 0.2 \times \frac{\sqrt{2}}{2} \approx 0.14$ meters. Answer A incorrectly calculates the time as 15 seconds and gets the period wrong as $6\pi$ seconds. Answer B has the wrong time (15 seconds) and wrong sign for initial displacement. Answer C has both the wrong time (15 seconds) and wrong sign for displacement. Answer D correctly identifies all three values: 30 seconds for amplitude decay, $\frac{2\pi}{3}$ seconds for the period, and positive 0.14 meters for initial displacement. Study tip: Always separate the exponential decay from the oscillatory motion—handle the amplitude envelope and periodic components independently, then combine your results.

Question 14

1. The envelope reaches 10% after approximately 0.46 seconds; this represents an underdamped circuit with damping coefficient 5 s⁻¹ and natural frequency 10 rad/s (correct answer)
2. The envelope reaches 10% after approximately 0.46 seconds; this represents an underdamped circuit with damping coefficient 10 s⁻¹ and natural frequency 5 rad/s
3. The envelope reaches 10% after approximately 0.23 seconds; this represents an underdamped circuit with damping coefficient 5 s⁻¹ and natural frequency 10 rad/s
4. The envelope reaches 10% after approximately 0.23 seconds; this represents a critically damped circuit with damping coefficient 5 s⁻¹ and oscillation frequency 10 rad/s

Explanation: The current envelope is 2e^(-5t). Initial maximum is 2 A. For 10%: 2e^(-5t) = 0.2, so e^(-5t) = 0.1, giving t = ln(10)/5 ≈ 0.46 seconds. Since the solution has the form Ae^(-αt)sin(ωt), this is underdamped with damping coefficient α = 5 s⁻¹ and damped frequency ω = 10 rad/s. Choice B confuses the coefficients, choice C uses ln(100) instead of ln(10), choice D misidentifies the damping type.

Question 15

1. The amplitude drops below 2 cm after approximately 1.35 seconds; the system has damping coefficient 2 s⁻¹ and effective frequency 4 rad/s with initial displacement of 0.3 m (correct answer)
2. The amplitude drops below 2 cm after approximately 1.35 seconds; the system has damping coefficient 4 s⁻¹ and effective frequency 2 rad/s with initial displacement of 0.3 m
3. The amplitude drops below 2 cm after approximately 2.08 seconds; the system has damping coefficient 2 s⁻¹ and effective frequency 4 rad/s with initial displacement of 0.3 m
4. The amplitude drops below 2 cm after approximately 2.08 seconds; the system has damping coefficient 2 s⁻¹ and natural frequency √20 rad/s with initial displacement of √0.34 m

Explanation: The envelope amplitude is 0.3e^(-2t)√(1² + 0.5²) = 0.3e^(-2t)√1.25 ≈ 0.335e^(-2t). For amplitude < 0.02 m: 0.335e^(-2t) = 0.02, so e^(-2t) = 0.02/0.335 ≈ 0.0597, giving t = ln(16.75)/2 ≈ 1.35 seconds. The damping coefficient is 2 s⁻¹, damped frequency is 4 rad/s, and x(0) = 0.3(1) = 0.3 m. Choice B confuses damping and frequency coefficients, choices C and D use incorrect amplitude calculations or time calculations.

Question 16

1. The object reaches 95% terminal velocity after approximately 23 seconds; terminal velocity is 50 m/s and the time constant indicates moderate air resistance with characteristic time 5 seconds
2. The object reaches 95% terminal velocity after approximately 15 seconds; terminal velocity is 50 m/s and the time constant indicates strong air resistance with characteristic time 0.2 seconds
3. The object reaches 95% terminal velocity after approximately 15 seconds; terminal velocity is 50 m/s and the time constant indicates moderate air resistance with characteristic time 5 seconds (correct answer)
4. The object reaches 95% terminal velocity after approximately 23 seconds; terminal velocity is 50 m/s and the air resistance coefficient is 0.05, indicating weak resistance

Explanation: When analyzing exponential models like this velocity function, you need to extract three key pieces of information: the terminal velocity, the time constant, and the time to reach a specific percentage of terminal velocity. The given model $v(t) = 50(1 - e^{-0.2t})$ represents exponential approach to terminal velocity. As $t \to \infty$, $e^{-0.2t} \to 0$, so the terminal velocity is 50 m/s. The time constant is $\tau = \frac{1}{0.2} = 5$ seconds, which represents the time needed to reach about 63% of terminal velocity and indicates the "speed" of approach. To find when the object reaches 95% of terminal velocity, set $0.95 \times 50 = 50(1 - e^{-0.2t})$. This gives $0.95 = 1 - e^{-0.2t}$, so $e^{-0.2t} = 0.05$. Taking the natural logarithm: $-0.2t = \ln(0.05) \approx -2.996$, yielding $t \approx 15$ seconds. Answer A incorrectly calculates the time as 23 seconds, likely confusing the relationship between the exponential decay and percentage calculations. Answer B correctly finds 15 seconds but misinterprets the time constant as 0.2 seconds (which is actually the decay rate $k$, not $\tau = 1/k$) and incorrectly describes this as "strong" resistance. Answer D makes the same timing error as A and introduces an irrelevant "air resistance coefficient." Remember: in exponential models $A(1 - e^{-kt})$, the time constant is $\tau = 1/k$, and larger time constants indicate slower approach to equilibrium, corresponding to weaker resistance effects.

Question 17

1. The process takes approximately 92 minutes, and the tank will contain about 0.01kg of salt per liter at the end
2. The process takes approximately 92 minutes, and the final salt concentration depends on the initial amount of salt present (correct answer)
3. The process takes approximately 46 minutes, and the tank will contain about 0.01kg of salt per liter at the end
4. The process takes approximately 46 minutes, and the final salt concentration depends on the initial amount of salt present

Explanation: The solution is S(t) = S₀e^(-0.05t). For S(t) = 0.01S₀, we need e^(-0.05t) = 0.01, so t = -ln(0.01)/0.05 = ln(100)/0.05 ≈ 92 minutes. The final concentration is S(t)/100 L = 0.01S₀/100 kg/L, which depends on the initial salt amount S₀. Choice A incorrectly assumes a specific initial concentration, choices C and D use wrong time calculation (confusing ln(100) with ln(10)).

Question 18

1. The reaction must be stopped after 30 minutes; the rate constant is approximately 0.0462 per minute, indicating a slow reaction rate
2. The reaction must be stopped after 45 minutes; the rate constant is approximately 0.0693 per minute, indicating a moderately fast reaction
3. The reaction must be stopped after 45 minutes; the rate constant is approximately 0.0462 per minute, indicating a moderately fast reaction (correct answer)
4. The reaction must be stopped after 30 minutes; the rate constant is approximately 0.0693 per minute, indicating a slow reaction rate

Explanation: When you encounter first-order kinetics problems, you're dealing with exponential decay where the rate of change is proportional to the current concentration. The key insight is connecting half-life to the rate constant, then using that to find when a specific concentration is reached. First, find the rate constant from the half-life. For first-order reactions, $k = \frac{\ln(2)}{t_{1/2}} = \frac{0.693}{15} = 0.0462$ per minute. This calculation is fundamental—memorize that $\ln(2) ≈ 0.693$. Next, solve for the time when concentration drops from 2.4 mol/L to 0.3 mol/L. The integrated first-order equation is $C(t) = C_0 e^{-kt}$. Substituting: $0.3 = 2.4 e^{-0.0462t}$. Solving: $\frac{0.3}{2.4} = e^{-0.0462t}$, so $\ln(0.125) = -0.0462t$, giving $t = 45$ minutes. Option C correctly identifies both values: 45 minutes and k = 0.0462 per minute. Option A uses the wrong time (30 minutes, which would be exactly two half-lives but doesn't account for the specific concentration ratio) and mischaracterizes the reaction rate. Option B has the correct time but wrong rate constant (0.0693 is $\ln(2)$, not the actual k-value). Option D combines both errors from A and B. Remember: always calculate the rate constant from half-life first, then use the integrated rate law. Don't confuse $\ln(2)$ with the actual rate constant—you must divide by the half-life period.