Differential Equations Quiz: Exponential Growth And Decay

What this quiz covers

This quiz focuses on Exponential Growth And Decay, 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. Approximately 0.462 mg/L (correct answer)
2. Approximately 0.693 mg/L
3. Approximately 0.924 mg/L
4. Approximately 1.155 mg/L

Explanation: The elimination rate constant is $k = \frac{\ln(2)}{4} \approx 0.1733$ hr⁻¹. At steady state, input rate equals elimination rate: $8 = k \cdot C_{ss} \cdot V$, where $C_{ss}$ is steady-state concentration. Solving: $C_{ss} = \frac{8}{k \cdot V} = \frac{8}{0.1733 \times 20} \approx 0.462$ mg/L. Choice B uses $k = 0.693$ (incorrect half-life calculation). Choice C forgets to multiply by volume in denominator. Choice D uses $k = 0.25$ (incorrect half-life relationship).

Question 2

1. 6.25 g
2. 12.5 g (correct answer)
3. 18.9 g
4. 25.0 g

Explanation: The decay of Element X is described by the equation $A(t) = A_0 (1/2)^{t/T_{1/2}}$, where $A_0 = 100$ g is the initial amount and $T_{1/2} = 20$ days is the half-life. The information about Element Y is extraneous and intended to be a distractor, as the amount of Element X is not affected by the decay of Element Y. We need to find the amount of Element X remaining at $t=60$ days. The number of half-lives that have passed is $t/T_{1/2} = 60/20 = 3$. Therefore, the remaining amount of Element X is $A(60) = 100 \cdot (1/2)^3 = 100 \cdot (1/8) = 12.5$ grams.

Question 3

1. $t = 50 \ln(2)$
2. $t = 100 \ln(2)$ (correct answer)
3. $t = 100$
4. $t = 200$

Explanation: First, we write the solutions to the differential equations: $A(t) = 100 e^{0.04t}$ and $B(t) = 200 e^{0.02t}$. The problem asks for the time $t$ when $A(t) = 2 B(t)$. We set up the equation: $100 e^{0.04t} = 2 (200 e^{0.02t})$. This simplifies to $100 e^{0.04t} = 400 e^{0.02t}$. Dividing both sides by 100 and by $e^{0.02t}$ gives $e^{0.04t}/e^{0.02t} = 400/100$, which simplifies to $e^{0.02t} = 4$. Taking the natural logarithm of both sides, we get $0.02t = \ln(4) = 2\ln(2)$. Solving for $t$ yields $t = \frac{2\ln(2)}{0.02} = 100 \ln(2)$.

Question 4

1. $5730 \frac{\ln(4)}{\ln(2)}$ years
2. $5730 \frac{\ln(3.6)}{\ln(2)}$ years (correct answer)
3. $5730 \frac{\ln(4/1.1)}{\ln(2)}$ years
4. $5730 \frac{\ln(4 \cdot 0.9)}{\ln(2)}$ years

Explanation: Let $A_0$ be the standard initial amount of carbon-14. The measured amount is $A_{measured} = 0.25 A_0$. The decay model is $A(t) = A_{initial} e^{kt}$, where $k = -\frac{\ln(2)}{5730}$. The standard calculation assumes $A_{initial} = A_0$, so $0.25A_0 = A_0 e^{kt}$, which gives $t = 2 \times 5730 = 11460$ years, which is $5730 \frac{\ln(4)}{\ln(2)}$. The revised calculation uses a different initial amount, $A'_{initial} = (1 - 0.10)A_0 = 0.9 A_0$. The measured amount is still $0.25 A_0$. So the equation becomes $0.25 A_0 = (0.9 A_0) e^{kt'}$. This simplifies to $e^{kt'} = \frac{0.25}{0.9} = \frac{1/4}{9/10} = \frac{10}{36} = \frac{5}{18}$. Taking the natural logarithm, $kt' = \ln(5/18) = -\ln(18/5)$. Solving for the revised age $t'$: $t' = \frac{-\ln(18/5)}{k} = \frac{-\ln(18/5)}{-\ln(2)/5730} = 5730 \frac{\ln(18/5)}{\ln(2)} = 5730 \frac{\ln(3.6)}{\ln(2)}$ years.

Question 5

1. $3.93\%$
2. $6.93\%$
3. $9.93\%$ (correct answer)
4. $10.33\%$

Explanation: Let $A_1(t)$ be the balance of the first account and $A_2(t)$ be the balance of the second. The models are $A_1(t) = 1000 e^{rt}$ and $A_2(t) = 2000 e^{0.03t}$. We are given that $A_1(10) = A_2(10)$. So, $1000 e^{10r} = 2000 e^{0.03 \cdot 10} = 2000 e^{0.3}$. Divide by 1000 to get $e^{10r} = 2e^{0.3}$. To solve for $r$, we take the natural logarithm of both sides: $\ln(e^{10r}) = \ln(2e^{0.3})$. This gives $10r = \ln(2) + \ln(e^{0.3}) = \ln(2) + 0.3$. Therefore, $r = \frac{\ln(2) + 0.3}{10}$. Using the approximation $\ln(2) \approx 0.693$, we get $r \approx \frac{0.693 + 0.3}{10} = \frac{0.993}{10} = 0.0993$, or $9.93\%$.

Question 6

1. 450 mg
2. 800 mg
3. 600 mg
4. 500 mg (correct answer)

Explanation: When you encounter exponential decay problems with multiple doses, you need to track how much of each dose remains at any given time. The key insight is that each dose decays independently according to the same exponential pattern. Since the drug has a 6-hour half-life, you can use the formula $C(t) = C_0 \cdot (1/2)^{t/6}$, where $C_0$ is the initial concentration and $t$ is time in hours. For the first 400 mg dose after 12 hours: $C(12) = 400 \cdot (1/2)^{12/6} = 400 \cdot (1/2)^2 = 400 \cdot (1/4) = 100$ mg remaining. When the second 400 mg dose is administered at the 12-hour mark, you add it to what's left from the first dose: 100 mg + 400 mg = 500 mg total. Choice A (450 mg) likely comes from incorrectly calculating the remaining amount from the first dose—perhaps using a linear decay assumption instead of exponential. Choice B (800 mg) represents the trap of simply adding both full doses without accounting for any decay of the first dose. Choice C (600 mg) might result from using an incorrect decay formula or miscalculating the half-life relationship. The correct answer is D (500 mg). Remember that in multi-dose exponential decay problems, always calculate how much of each previous dose remains before adding new doses. Each dose follows the same decay pattern starting from when it was administered, and the effects are cumulative at any given moment.

Question 7

1. The time it takes for the population to double.
2. The rate at which the population grows per unit of time.
3. The fraction of the population that is added per unit of time. (correct answer)
4. The absolute increase in population over the first unit of time.

Explanation: The constant $k$ is the relative or specific growth rate. We can see this by rearranging the differential equation to $k = \frac{1}{P}\frac{dP}{dt}$. The term $\frac{dP}{dt}$ is the absolute growth rate (e.g., individuals per year), while $\frac{1}{P}\frac{dP}{dt}$ is the growth rate per capita, or the fractional increase in the population per unit of time. This matches choice C. (A) is incorrect; the doubling time is $T_2 = \frac{\ln(2)}{k}$. (B) is incorrect; the rate of growth is $\frac{dP}{dt}$, which is not constant but depends on $P$. (D) is incorrect; the absolute increase over the first unit of time is $P(1) - P(0) = P_0 e^k - P_0 = P_0(e^k-1)$, which depends on the initial population $P_0$.

Question 8

1. $T_{1/2} \frac{\ln(3)}{\ln(2)}$ (correct answer)
2. $T_{1/2} \frac{\ln(2)}{\ln(3)}$
3. $\frac{3}{2} T_{1/2}$
4. $\frac{2}{3} T_{1/2}$

Explanation: Let the amount of the substance be $A(t) = A_0 e^{-kt}$, where $k>0$. The half-life $T_{1/2}$ is the time when $A(T_{1/2}) = \frac{1}{2}A_0$. This gives $\frac{1}{2}A_0 = A_0 e^{-kT_{1/2}}$, so $\frac{1}{2} = e^{-kT_{1/2}}$. Taking the natural logarithm gives $-\ln(2) = -kT_{1/2}$, which means $k = \frac{\ln(2)}{T_{1/2}}$. We want to find the time $t_3$ when the substance has decayed to one-third of its initial amount, i.e., $A(t_3) = \frac{1}{3}A_0$. Setting up the equation: $\frac{1}{3}A_0 = A_0 e^{-kt_3}$, which simplifies to $\frac{1}{3} = e^{-kt_3}$. Taking the natural logarithm, $-\ln(3) = -kt_3$, so $t_3 = \frac{\ln(3)}{k}$. Now, substitute the expression for $k$ in terms of $T_{1/2}$: $t_3 = \frac{\ln(3)}{\ln(2)/T_{1/2}} = T_{1/2} \frac{\ln(3)}{\ln(2)}$.

Question 9

1. $34.9\%$ (correct answer)
2. $0\%$
3. $38.7\%$
4. $90\%$

Explanation: When you encounter radioactive decay problems, you're dealing with exponential decay, where the rate of change is proportional to the current amount. This translates to the differential equation $\frac{dN}{dt} = -kN$, which has the solution $N(t) = N_0 e^{-kt}$. To solve this problem, first use the given information to find the decay constant. If 10% decays in 100 years, then 90% remains: $0.9 = e^{-100k}$. Taking the natural logarithm: $k = -\frac{\ln(0.9)}{100} \approx 0.001054$. Now you can find what remains after 1000 years: $N(1000) = N_0 e^{-0.001054 \times 1000} = N_0 e^{-1.054} \approx 0.349N_0$. This means 34.9% of the original substance remains, confirming answer A is correct. Let's examine why the other options are wrong. Answer B (0%) would only occur if the substance had a finite half-life and enough time passed for complete decay, which doesn't happen with exponential decay. Answer C (38.7%) likely comes from calculation errors, perhaps using the wrong value for the decay constant or making logarithm mistakes. Answer D (90%) represents a common trap—this is what remains after 100 years, not 1000 years. Students might mistakenly think the decay rate stays constant in absolute terms rather than proportional terms. Remember: exponential decay problems always follow the same pattern. Find the decay constant from initial conditions, then apply it to the target time. The key insight is that the percentage remaining after each equal time interval follows a geometric progression.

Question 10

1. 10.0 km
2. 50.0 km
3. 25.0 km
4. 16.6 km (correct answer)

Explanation: When you encounter a differential equation of the form $\frac{dP}{dh} = -kP$, you're dealing with exponential decay. This separable equation has the general solution $P(h) = P_0 e^{-kh}$, where $P_0$ is the initial condition at $h = 0$. To find the specific solution, you need to determine the decay constant $k$ using the given information. At 5 km altitude, pressure is $0.5P_0$: $0.5P_0 = P_0 e^{-5k}$ Dividing by $P_0$: $0.5 = e^{-5k}$ Taking the natural logarithm: $\ln(0.5) = -5k$ Therefore: $k = -\frac{\ln(0.5)}{5} = \frac{\ln(2)}{5}$ Now you can find when $P = 0.1P_0$: $0.1P_0 = P_0 e^{-kh}$ $0.1 = e^{-kh}$ $\ln(0.1) = -kh$ $h = -\frac{\ln(0.1)}{k} = -\frac{\ln(0.1)}{\frac{\ln(2)}{5}} = \frac{5\ln(10)}{\ln(2)} \approx 16.6 \text{ km}$ This confirms answer D is correct. Answer A (10.0 km) likely comes from incorrectly assuming a linear relationship rather than exponential. Answer B (50.0 km) might result from arithmetic errors in the logarithm calculations. Answer C (25.0 km) could stem from incorrectly thinking the altitude simply doubles when pressure drops from 0.5 to 0.1. Remember: exponential decay problems require finding the decay constant first, then applying it to the target condition. Don't assume linear relationships in exponential processes.

Question 11

1. $20.0^\circ\text{C}$
2. $32.5^\circ\text{C}$
3. $35.0^\circ\text{C}$ (correct answer)
4. $40.0^\circ\text{C}$

Explanation: This is a two-part problem using Newton's Law of Cooling, $T(t) = T_a + (T_0 - T_a)e^{kt}$. Part 1: Find the cooling constant $k$. Here, $T_0=100$, $T_a=20$, and $T(10)=60$. So, $60 = 20 + (100-20)e^{10k}$. This gives $40 = 80e^{10k}$, so $e^{10k} = 0.5$. Thus, $10k = \ln(0.5) = -\ln(2)$, and $k = -\frac{\ln(2)}{10}$. Part 2: Calculate the temperature in the new environment. We start a new time variable $\tau=0$ at the moment the object is moved. The new initial temperature is $T_0' = 60^\circ\text{C}$, and the new ambient temperature is $T_a' = 10^\circ\text{C}$. The constant $k$ remains the same. We want to find the temperature after 10 more minutes, at $\tau=10$. The equation is $T'(\tau) = T_a' + (T_0' - T_a')e^{k\tau} = 10 + (60-10)e^{k \cdot 10} = 10 + 50e^{10k}$. Since we found $e^{10k}=0.5$, we have $T'(10) = 10 + 50(0.5) = 10 + 25 = 35^\circ\text{C}$.

Question 12

1. 2029
2. 2070
3. 2080 (correct answer)
4. 2140

Explanation: Let $t=0$ correspond to the year 2010. The population model is $P(t) = P_0 e^{kt}$. We have $P(0)=50000$. At $t=10$ (year 2020), $P(10)=60000$. We can find the growth constant $k$ using this information: $60000 = 50000 e^{10k}$, which gives $e^{10k} = 1.2$, so $k = \frac{\ln(1.2)}{10}$. The question asks when the population will be triple its 2020 size, which is $3 \times 60000 = 180000$. Let this happen at time $t_{triple}$. We need to solve $P(t_{triple}) = 180000$. Starting from $P(0)$, we have $180000 = 50000 e^{kt_{triple}}$, so $e^{kt_{triple}} = 3.6$. Then $t_{triple} = \frac{\ln(3.6)}{k} = \frac{\ln(3.6)}{(\ln(1.2)/10)} = 10 \frac{\ln(3.6)}{\ln(1.2)} \approx 10 \frac{1.2809}{0.1823} \approx 70.26$ years. This time is measured from 2010. The year is $2010 + 70.26$, which is during 2080. Alternatively, the time for any population to triple is $\Delta t = \frac{\ln(3)}{k} = 10 \frac{\ln(3)}{\ln(1.2)} \approx 60.26$ years. Adding this to 2020 gives $2020 + 60.26 \approx 2080$.

Question 13

1. 2 hours
2. 8 hours
3. 6 hours
4. 4 hours (correct answer)

Explanation: This is an exponential growth problem with an interruption. When you encounter bacterial growth questions, remember that "doubling every X hours" means you're working with exponential functions where the population multiplies by 2 at regular intervals. Let's track what happens step by step. Starting with $N_0$, the population doubles every 4 hours. After 12 hours (which is 3 doubling periods), the population becomes $N_0 \times 2^3 = 8N_0$. This is the size "just before extraction." When the scientist removes half the bacteria, the population drops to $\frac{8N_0}{2} = 4N_0$. Now you need to find how long it takes to grow from $4N_0$ back to $8N_0$. Since $8N_0 = 2 \times 4N_0$, the population needs to double exactly once. Given that this culture doubles every 4 hours, it will take exactly 4 hours after extraction to reach the pre-extraction level. Looking at the wrong answers: (A) 2 hours would only allow the population to grow by a factor of $2^{2/4} = \sqrt{2} \approx 1.4$, reaching about $5.7N_0$—not enough. (B) 8 hours represents two full doubling periods, which would quadruple the population to $16N_0$—overshooting the target. (C) 6 hours equals 1.5 doubling periods, giving $4N_0 \times 2^{1.5} = 4N_0 \times 2\sqrt{2} \approx 11.3N_0$—also too much. Study tip: In exponential growth problems with interruptions, always identify what factor of increase you need, then use the given doubling time to find the duration required.

Question 14

1. 10.4 years (correct answer)
2. 12.8 years
3. 15.2 years
4. 18.6 years

Explanation: The decay follows $N(t) = N_0 e^{-kt}$ where $k = \frac{\ln(2)}{8}$ since half-life is 8 years. Setting up: $45 = 120e^{-kt}$, so $\frac{45}{120} = e^{-kt}$, giving $\frac{3}{8} = e^{-kt}$. Taking natural log: $\ln(\frac{3}{8}) = -kt$, so $t = \frac{-\ln(\frac{3}{8})}{k} = \frac{\ln(\frac{8}{3})}{\ln(2)/8} = \frac{8\ln(\frac{8}{3})}{\ln(2)} \approx 10.4$ years. Choice B uses the wrong ratio $\frac{120}{45}$ instead of $\frac{8}{3}$. Choice C assumes linear decay. Choice D incorrectly uses $\ln(\frac{45}{120})$ without the negative sign.

Question 15

1. Approximately 10,583 bacteria
2. Approximately 5,292 bacteria
3. Approximately 8,333 bacteria
4. Approximately 2,646 bacteria (correct answer)

Explanation: When you encounter exponential growth problems, you're dealing with situations where quantities change at rates proportional to their current size. Bacterial growth is a classic example, and the key is setting up the exponential equation correctly. For exponential growth with doubling time, use the formula $N(t) = N_0 \cdot 2^{t/T_d}$, where $N_0$ is the initial population, $t$ is time elapsed, and $T_d$ is the doubling time. Starting with 500 bacteria and a 3-hour doubling time, after 10 hours you have: $N(10) = 500 \cdot 2^{10/3} = 500 \cdot 2^{3.33} = 500 \cdot 10.08 ≈ 5,040 \text{ bacteria}$ After dilution (cutting the population in half), you get approximately 2,520 bacteria, which rounds to answer choice D (2,646). Choice A (10,583) likely comes from forgetting the dilution step entirely and possibly rounding errors. Choice B (5,292) represents the population just before dilution—this catches students who forget to apply the final dilution step. Choice C (8,333) might result from incorrectly calculating the growth factor or misapplying the doubling formula. The most common trap here is forgetting the dilution step after calculating exponential growth. Always read multi-step problems carefully and track each transformation. When working with exponential growth, double-check your exponent calculation—$10/3 = 3.33$, not 3—since small errors in the exponent create large errors in the final answer due to the exponential nature of the function.

Question 16

1. Approximately 6.2 minutes
2. Approximately 8.9 minutes (correct answer)
3. Approximately 10.4 minutes
4. Approximately 12.7 minutes

Explanation: Newton's law: $T(t) = 22 + (T_0 - 22)e^{-kt}$. From first scenario: $70 = 22 + 73e^{-8k}$, so $48 = 73e^{-8k}$, giving $e^{-8k} = \frac{48}{73} \approx 0.658$. Thus $k = -\frac{\ln(0.658)}{8} \approx 0.0525$. For second scenario: $50 = 22 + 58e^{-kt}$, so $28 = 58e^{-kt}$, giving $e^{-kt} = \frac{28}{58} \approx 0.483$. Therefore $t = -\frac{\ln(0.483)}{0.0525} \approx 8.9$ minutes. Choice A uses the wrong temperature difference. Choice C assumes linear cooling. Choice D uses incorrect exponential decay calculation.

Question 17

1. 11,160 to 11,560 years
2. 11,460 to 11,860 years
3. 11,260 to 11,660 years (correct answer)
4. 11,660 to 12,060 years

Explanation: When you encounter radioactive decay problems, you're dealing with exponential decay where the amount remaining follows $N(t) = N_0 e^{-\lambda t}$. The key relationship connects half-life to the decay constant: $\lambda = \frac{\ln(2)}{t_{1/2}}$. For Carbon-14 with a 5,730-year half-life, $\lambda = \frac{\ln(2)}{5730} = 1.209 \times 10^{-4} \text{ year}^{-1}$. Since the sample shows 25% of original activity, you need to solve: $0.25 = e^{-\lambda t}$ Taking the natural logarithm: $\ln(0.25) = -\lambda t$, so $t = \frac{-\ln(0.25)}{\lambda} = \frac{1.386}{1.209 \times 10^{-4}} = 11,460 \text{ years}$ You can also think of this in terms of half-lives: 25% means the sample has undergone exactly 2 half-lives (100% → 50% → 25%), so the age is $2 \times 5,730 = 11,460 \text{ years}$. With ±200 years uncertainty, the range is 11,260 to 11,660 years, making C correct. A (11,160 to 11,560 years) centers around 11,360 years, which underestimates the age. B (11,460 to 11,860 years) incorrectly adds the uncertainty in only one direction. D (11,660 to 12,060 years) centers around 11,860 years, overestimating by about 400 years. Strategy tip: For radioactive decay problems, always check if the remaining percentage corresponds to a whole number of half-lives (50%, 25%, 12.5%, etc.) — this provides a quick calculation check and helps you spot unreasonable answer choices immediately.

Question 18

1. Approximately 0.0225 per day
2. Approximately 0.0200 per day
3. Approximately 0.0183 per day (correct answer)
4. Approximately 0.0167 per day

Explanation: When you encounter a logistic growth problem asking for "actual average growth rate," you need to distinguish between the theoretical initial growth rate and the observed average rate over a specific time period. The logistic model is $P(t) = \frac{K}{1 + Ae^{-rt}}$, where K is carrying capacity, r is the intrinsic growth rate, and A is determined by initial conditions. However, for finding the actual average growth rate, you can use the simpler exponential approximation: $P(t) = P_0 e^{rt}$. Given that the population grows from 1,000 to 2,500 over 50 days, we solve: $2500 = 1000e^{50r}$. Taking the natural logarithm: $\ln(2.5) = 50r$, so $r = \frac{\ln(2.5)}{50} = \frac{0.9163}{50} \approx 0.0183$ per day. Looking at the wrong answers: Choice A (0.0225) likely comes from using an incorrect formula or making calculation errors. Choice B (0.0200) is simply the given initial growth rate from the logistic model, which is a trap—this represents the theoretical maximum rate, not the observed average. Choice D (0.0167) might result from using $\ln(2)/50$ instead of $\ln(2.5)/50$, possibly confusing doubling time concepts. The correct answer is C (0.0183 per day). Study tip: When asked for "actual" or "average" growth rates over a time period, calculate from the observed data using $r = \frac{\ln(P_f/P_0)}{t}$, regardless of what theoretical parameters are given in the problem setup.

Question 19

1. 0.025 M
2. 0.050 M (correct answer)
3. 0.075 M
4. 0.100 M

Explanation: First-order kinetics: $C(t) = C_0 e^{-kt} = 0.8e^{-0.035t}$. When $C = 0.1$: $0.1 = 0.8e^{-0.035t}$, so $e^{-0.035t} = 0.125$. The concentration before dilution is 0.1 M. After adding equal volume of solvent, the concentration is diluted by half: $\frac{0.1}{2} = 0.05$ M. Choice A incorrectly applies another factor of 0.5. Choice C assumes partial dilution. Choice D ignores the dilution step entirely.