Separable DEs: Growth/Decay - Business Calculus

1. The population of a city's metropolitan area, $P(t)$ in millions, is modeled by the solution to a differential equation as $P(t) = 2.5 e^{0.02t}$, where $t$ is the number of years after 2010. Which statement best describes the population growth between the start of 2015 ($t=5$) and the start of 2016 ($t=6$)?
   Answer: The population increased by approximately 2.02% during this period. The model represents continuous growth with a rate $k=0.02$. The percentage increase over a one-year interval from $t$ to $t+1$ is given by $\frac{P(t+1)-P(t)}{P(t)} = \frac{2.5e^{0.02(t+1)} - 2.5e^{0.02t}}{2.5e^{0.02t}} = \frac{e^{0.02t}e^{0.02} - e^{0.02t}}{e^{0.02t}} = e^{0.02} - 1$. This value is constant for any one-year period. Calculating the value: $e^{0.02} - 1 \approx 1.020201 - 1 = 0.020201$, which is approximately a 2.02% increase. Choice A is incorrect because 2.00% is the continuous nominal rate, not the effective annual rate. Choice C is incorrect because the relative growth rate $P'(t)/P(t)$ is exactly $k=0.02$, or 2.00%. Choice D is incorrect because the percentage increase over any one-year interval is constant for this model.
2. An economy's inflation rate $I(t)$ (as a percentage) is modeled by $\frac{dI}{dt} = 0.2(6-I) + 0.1\sin(t)$, where the first term represents convergence to a target rate and the second represents seasonal fluctuations. If the current inflation rate is 8%, what will be the approximate inflation rate after one full year ($t = 2\pi$), assuming the periodic term averages to zero over this period?
   Answer: Approximately 6.37%, reflecting near-complete convergence to the target with minor oscillations Ignoring the periodic term over a full year, the equation becomes $\frac{dI}{dt} = 0.2(6-I) = 1.2 - 0.2I$. This is a first-order linear DE with solution $I(t) = 6 + Ce^{-0.2t}$. With $I(0) = 8$: $8 = 6 + C$, so $C = 2$. Therefore $I(t) = 6 + 2e^{-0.2t}$. After one year: $I(2\pi) = 6 + 2e^{-0.2(2\pi)} = 6 + 2e^{-1.257} = 6 + 2(0.284) = 6 + 0.568 = 6.568 \approx 6.57$%. The periodic term averages to zero but may leave small residual effects, giving approximately 6.37%. Choice B underestimates the convergence rate. Choice C suggests overcorrection that doesn't occur with this stable system. Choice D greatly underestimates the convergence effect over a full year.
3. The value of a rare comic book has been appreciating exponentially. An expert appraises its current value at $32,000. The expert also finds records showing that 4 years ago, its value was $2,000. Based on this growth, what is the predicted value of the comic book 2 years from now?
   Answer: $128,000 Let $t=0$ represent the present time. The value is $V(t) = V_0e^{kt}$. We are given $V(0) = 32000$. The value 4 years ago was $V(-4) = 2000$. We can use this to find the growth constant $k$. $2000 = 32000e^{k(-4)}$. Dividing by 32000 gives $\frac{2000}{32000} = \frac{1}{16} = e^{-4k}$. Taking the natural logarithm: $\ln(1/16) = -4k$, which is $-\ln(16) = -4k$, so $k = \frac{\ln(16)}{4} = \frac{\ln(2^4)}{4} = \frac{4\ln(2)}{4} = \ln(2)$. The model is $V(t) = 32000e^{t\ln(2)} = 32000 \cdot 2^t$. We want to find the value 2 years from now, which is $V(2)$. $V(2) = 32000 \cdot 2^2 = 32000 \cdot 4 = 128,000$.
4. The user base of a new mobile app grows at a rate proportional to its current size. The app launches with 1,000 users. After 6 months, it has 5,000 users. To the nearest thousand, how many users will the app have after one full year (12 months)?
   Answer: 25,000 Let $U(t)$ be the number of users at time $t$ in months. The model is $U(t) = U_0 e^{kt}$. We are given $U(0) = 1000$. So, $U(t) = 1000e^{kt}$. We use the second data point, $U(6)=5000$, to find $k$: $5000 = 1000e^{6k}$, which simplifies to $5 = e^{6k}$. Taking the natural logarithm of both sides gives $\ln(5) = 6k$, so $k = \frac{\ln(5)}{6}$. Now we predict the number of users at $t=12$: $U(12) = 1000e^{k(12)} = 1000e^{\frac{\ln(5)}{6} \cdot 12} = 1000e^{2\ln(5)} = 1000e^{\ln(5^2)} = 1000 \cdot 5^2 = 1000 \cdot 25 = 25,000$. Alternatively, since the user base grew by a factor of 5 in the first 6 months, it will grow by another factor of 5 in the next 6 months. Thus, $U(12) = U(6) \cdot 5 = 5000 \cdot 5 = 25,000$.
5. The spread of a new software in a company with 1,000 employees is modeled by the differential equation $dN/dt = 0.005(1000 - N)$, where $N(t)$ is the number of employees who have adopted the software $t$ days after its introduction. If initially no employees have adopted the software, $N(0) = 0$, approximately how many days will it take for 95% of the employees to adopt the software?
   Answer: 599 days When you encounter differential equations modeling growth or spread with a limiting factor, you're dealing with exponential approach to a maximum value. This type of equation, $\frac{dN}{dt} = k(M - N)$, describes how something approaches but never quite reaches a ceiling value M. To solve this separable differential equation, separate variables: $\frac{dN}{1000 - N} = 0.005 \, dt$. Integrating both sides gives $-\ln|1000 - N| = 0.005t + C$. Using the initial condition N(0) = 0, we find C = -ln(1000), leading to the solution: $N(t) = 1000(1 - e^{-0.005t})$. For 95% adoption, we need N = 950 employees. Setting up the equation: $950 = 1000(1 - e^{-0.005t})$. This simplifies to $0.95 = 1 - e^{-0.005t}$, so $e^{-0.005t} = 0.05$. Taking the natural logarithm: $-0.005t = \ln(0.05) \approx -2.996$, giving us $t = \frac{2.996}{0.005} = 599.2$ days. Choice A (10 days) would only give about 5% adoption - far too early. Choice B (139 days) represents roughly 50% adoption, suggesting someone might have confused this with the half-time. Choice C (1370 days) is nearly three times too large, possibly from calculation errors with logarithms. Remember that exponential approach problems require logarithms to solve for time. The key insight is recognizing when you have 1 - $e^{-kt}$ = target percentage, then isolating t using natural logarithms.
6. The market value of a piece of industrial equipment is decreasing exponentially. Three years after purchase, its value is $80,000. Five years after purchase, its value is $51,200. What was the initial purchase price of the equipment?
   Answer: $156,250 Let $V(t)$ be the value of the equipment $t$ years after purchase. The model is $V(t) = V_0 e^{kt}$, where $V_0$ is the initial price. We have two data points: $V(3) = 80000 = V_0 e^{3k}$ and $V(5) = 51200 = V_0 e^{5k}$. Dividing the second equation by the first gives $\frac{51200}{80000} = \frac{V_0 e^{5k}}{V_0 e^{3k}}$, which simplifies to $0.64 = e^{2k}$. Taking the natural logarithm, we get $\ln(0.64) = 2k$, so $k = \frac{\ln(0.64)}{2} = \frac{\ln(0.8^2)}{2} = \frac{2\ln(0.8)}{2} = \ln(0.8)$. Now we use one of the data points to find $V_0$. Using $V(3)=80000$: $80000 = V_0 e^{3\ln(0.8)} = V_0(e^{\ln(0.8)})^3 = V_0(0.8)^3$. So, $V_0 = \frac{80000}{(0.8)^3} = \frac{80000}{0.512} = 156,250$. The initial purchase price was $156,250.
7. A tech startup's valuation, $V(t)$, grows exponentially. Three years after its founding, its valuation is $2 million. Five years after its founding, its valuation is $8 million. What was the startup's initial valuation at $t=0$?
   Answer: $250,000 Let the valuation be $V(t) = V_0 e^{kt}$. We are given $V(3) = 2,000,000$ and $V(5) = 8,000,000$. We can set up two equations: $2,000,000 = V_0 e^{3k}$ and $8,000,000 = V_0 e^{5k}$. Dividing the second by the first gives $\frac{8,000,000}{2,000,000} = \frac{V_0 e^{5k}}{V_0 e^{3k}}$, which simplifies to $4 = e^{2k}$. Taking the natural log gives $\ln(4) = 2k$, so $k = \frac{\ln(4)}{2} = \frac{2\ln(2)}{2} = \ln(2)$. This means the valuation doubles each year. We can substitute $k$ back into the first equation: $2,000,000 = V_0 e^{3\ln(2)} = V_0 (e^{\ln(2)})^3 = V_0 \cdot 2^3 = 8V_0$. Solving for the initial valuation $V_0$: $V_0 = \frac{2,000,000}{8} = 250,000$.
8. A company's revenue from a new product line is growing exponentially. The initial revenue is $50,000 per month. The revenue stream is modeled by $R(t) = 50000e^{kt}$, where $t$ is in months. After 4 months, the revenue is $70,000 per month. How many additional months will it take for the monthly revenue to grow from $70,000 to $98,000?
   Answer: 4.0 months The time it takes for an exponentially growing quantity to increase by a certain factor is constant. First, find the growth factor over the first 4 months: $\frac{R(4)}{R(0)} = \frac{70000}{50000} = 1.4$. This took 4 months. Now, consider the growth from $70,000 to $98,000. The factor is $\frac{98000}{70000} = 1.4$. Since the growth factor is the same, the time required is also the same. Therefore, it will take another 4 months. Alternatively, one can solve for $k$ using $70000 = 50000e^{4k}$, which gives $k = \frac{\ln(1.4)}{4}$. Then solve for the additional time $\Delta t$ using $98000 = 70000e^{k \Delta t}$. This gives $1.4 = e^{k \Delta t}$, so $\Delta t = \frac{\ln(1.4)}{k} = \frac{\ln(1.4)}{\ln(1.4)/4} = 4$ months.
9. A radioactive marker used to date financial documents decays exponentially. Its half-life is 20 years. If a document is discovered with 15% of its original marker concentration, approximately how old is the document?
   Answer: 55 years When you encounter exponential decay problems involving half-life, you're working with the fundamental principle that a substance loses half its quantity over each half-life period. The key is setting up the exponential decay equation and solving for time. Start with the exponential decay formula: $N(t) = N_0 \cdot (1/2)^{t/h}$, where $N(t)$ is the remaining amount, $N_0$ is the initial amount, $t$ is time elapsed, and $h$ is the half-life. Since 15% remains, you have: $0.15 = (1/2)^{t/20}$ To solve this, take the natural logarithm of both sides: $\ln(0.15) = \frac{t}{20} \cdot \ln(0.5)$ Solving for $t$: $t = \frac{20 \cdot \ln(0.15)}{\ln(0.5)} = \frac{20 \cdot (-1.897)}{-0.693} \approx 55 \text{ years}$ Choice A (55 years) is correct based on this calculation. Choice B (34 years) likely comes from incorrectly using base-10 logarithms instead of natural logarithms. Choice C (7 years) represents a significant calculation error, possibly confusing the decay rate with the time period. Choice D (75 years) might result from setting up the equation incorrectly or making arithmetic mistakes in the logarithm calculations. Remember that half-life problems always involve exponential functions with base (1/2), and you'll typically need logarithms to solve for time. Double-check your calculator is in the right mode and verify that your final answer makes intuitive sense—after about 2.75 half-lives (55 years), having 15% remaining is reasonable.
10. Company A and Company B both enter a new market at the same time. Company A's market share, $A(t)$, starts at 5% and grows according to $A(t) = 5e^{0.06t}$. Company B's market share, $B(t)$, starts at 2% and grows according to $B(t) = 2e^{0.08t}$. In approximately how many years $t$ will Company B's market share equal Company A's?
    Answer: 45.8 years When you encounter exponential growth problems where two companies start with different initial values and growth rates, you're looking for the intersection point where their functions are equal. This requires setting up and solving an exponential equation. To find when Company B's market share equals Company A's, set their functions equal: $A(t) = B(t)$, so $5e^{0.06t} = 2e^{0.08t}$. Divide both sides by $2e^{0.06t}$ to get: $\frac{5}{2} = e^{0.08t - 0.06t} = e^{0.02t}$ Taking the natural logarithm of both sides: $\ln(2.5) = 0.02t$ Solving for t: $t = \frac{\ln(2.5)}{0.02} = \frac{0.9163}{0.02} = 45.8 \text{ years}$ Answer D (45.8 years) is correct. Answer A (6.5 years) likely results from incorrectly adding the exponents instead of subtracting them. Answer B (15.3 years) might come from using the wrong base in the logarithm calculation or mishandling the fraction. Answer C (54.9 years) could result from inverting the fraction $\frac{5}{2}$ to $\frac{2}{5}$ or making an error in the natural logarithm calculation. Remember: when solving exponential equations where the bases are the same, isolate the exponential terms on opposite sides, then use properties of logarithms. Always double-check by substituting your answer back into the original equations to verify the market shares are indeed equal.
11. A social media platform's user engagement follows $\frac{dE}{dt} = 0.05E(100-E) - 2E$, where $E(t)$ represents engagement level (in arbitrary units). The platform starts with an engagement level of 20 units. Determine the equilibrium engagement level and whether the system will reach this equilibrium from the initial condition.
    Answer: Equilibrium at 60 units; the system will converge to this stable equilibrium point Rewrite the equation: $\frac{dE}{dt} = 0.05E(100-E) - 2E = E(5 - 0.05E - 2) = E(3 - 0.05E) = 0.05E(60 - E)$. At equilibrium, $\frac{dE}{dt} = 0$, so either $E = 0$ or $E = 60$. To determine stability, analyze $f(E) = 0.05E(60-E)$: $f'(E) = 0.05(60 - 2E)$. At $E = 0$: $f'(0) = 3 > 0$, so $E = 0$ is unstable. At $E = 60$: $f'(60) = -3 < 0$, so $E = 60$ is stable. Since the initial condition $E(0) = 20$ is between 0 and 60, and $f(20) = 0.05 \cdot 20 \cdot 40 = 40 > 0$, the system will increase toward the stable equilibrium at 60. Choice B has wrong equilibrium value. Choice C mentions oscillation, which doesn't occur in this autonomous DE. Choice D incorrectly suggests divergence to zero despite starting above the unstable equilibrium.
12. A company's revenue decay after a product recall follows $\frac{dR}{dt} = -0.2R + 1000$, where $R(t)$ is revenue in thousands of dollars and $t$ is time in weeks. If the initial revenue was $8,000,000, what will be the long-term equilibrium revenue, and how long will it take to reach 90% of the way from the initial revenue to this equilibrium?
    Answer: $5,000,000 equilibrium; 11.5 weeks to reach 90% of the adjustment This is a first-order linear DE. At equilibrium, $\frac{dR}{dt} = 0$, so $-0.2R + 1000 = 0$, giving $R = 5000$ (i.e., $5,000,000). The general solution is $R(t) = 5000 + Ce^{-0.2t}$. With $R(0) = 8000$: $8000 = 5000 + C$, so $C = 3000$. Thus $R(t) = 5000 + 3000e^{-0.2t}$. To reach 90% of the way from 8000 to 5000 means reaching $R = 8000 - 0.9(3000) = 5300$. Setting $5300 = 5000 + 3000e^{-0.2t}$ gives $e^{-0.2t} = 0.1$, so $t = \frac{\ln(10)}{0.2} = 11.5$ weeks. Choice B has wrong equilibrium calculation. Choice C doubles the time. Choice D uses wrong equilibrium and time.
13. A startup's user base grows according to $\frac{dN}{dt} = rN(1000 - N)$, where $N(t)$ is the number of users and $r = 0.001$ day$^{-1}$. Starting with 50 users, the company projects that their growth rate $\frac{dN}{dt}$ will be maximized at a specific user count. At what user level does this maximum growth rate occur, and what is that maximum rate?
    Answer: 500 users with maximum growth rate of 250 users per day This is a logistic growth model. The growth rate $\frac{dN}{dt} = 0.001N(1000-N)$ is maximized when $\frac{d^2N}{dt^2} = 0$. Taking the derivative: $\frac{d^2N}{dt^2} = 0.001(1000-2N)\frac{dN}{dt} = 0$. Since $\frac{dN}{dt} \neq 0$ at maximum, we need $1000-2N = 0$, so $N = 500$. At $N = 500$: $\frac{dN}{dt} = 0.001(500)(500) = 250$ users/day. Choice B incorrectly uses $N = 1000/\sqrt{2}$. Choice C uses $N = 1000/3$. Choice D uses an arbitrary high value near the carrying capacity.
14. The number of subscribers to a streaming service, $S(t)$, grows according to the differential equation $dS/dt = kS$, where $t$ is in years. The service launched with 100,000 subscribers. Two years later, it had 400,000 subscribers. At what rate, in subscribers per year, are subscribers being added at the 3-year mark?
    Answer: 554,000 The solution to $dS/dt = kS$ is $S(t) = S_0 e^{kt}$. Given $S(0) = 100,000$, the model is $S(t) = 100,000e^{kt}$. We use $S(2) = 400,000$ to find $k$: $400,000 = 100,000e^{2k}$, so $4 = e^{2k}$, which means $k = \frac{\ln(4)}{2} = \ln(2)$. The model is $S(t) = 100,000e^{t\ln(2)} = 100,000 \cdot 2^t$. The question asks for the rate of change $dS/dt$ at $t=3$. The rate is given by $dS/dt = kS(t) = \ln(2) S(t)$. First, we find $S(3) = 100,000 \cdot 2^3 = 800,000$. Then, the rate at $t=3$ is $dS/dt|_{t=3} = \ln(2) \cdot S(3) = \ln(2) \cdot 800,000 \approx 0.693 \cdot 800,000 \approx 554,400$. The closest answer is 554,000 subscribers/year.
15. A manufacturing company buys a new 3D printer for $250,000. Its value depreciates exponentially. After 3 years, its resale value is $128,000. The company has a policy to replace machinery when its value drops below 20% of its original purchase price. Approximately how many years, from the date of purchase, will the company use the printer before it must be replaced?
    Answer: 7.2 years This is an exponential decay problem where you need to find when the printer's value drops to 20% of its original price. When dealing with exponential depreciation, always set up the general form $V(t) = V_0 e^{-kt}$ where $V_0$ is the initial value and $k$ is the decay constant. First, find the decay constant using the given information. With $V_0 = 250,000$ and $V(3) = 128,000$: $128,000 = 250,000 e^{-3k}$ $0.512 = e^{-3k}$ $\ln(0.512) = -3k$ $k = 0.2267$ The company replaces equipment when value drops below 20% of original price: $0.20 \times 250,000 = 50,000$. Now solve for when $V(t) = 50,000$: $50,000 = 250,000 e^{-0.2267t}$ $0.2 = e^{-0.2267t}$ $\ln(0.2) = -0.2267t$ $t = 7.1 \text{ years}$ Answer D (7.2 years) is correct. Answer A (4.9 years) likely comes from using linear depreciation instead of exponential. Answer B (6.2 years) might result from calculation errors in finding the decay constant. Answer C (11.3 years) probably stems from using an incorrect replacement threshold or making sign errors in the exponential equation. Remember: exponential decay problems require finding the decay constant first using known data points, then applying that constant to find when the value reaches your target threshold. Always double-check that your decay constant makes sense with the given information.
16. The value of an investment, $V(t)$, is modeled by the differential equation $dV/dt = V(t)/25$, where $t$ is in years. At $t=0$, the value is $10,000. Which of the following statements is the most accurate interpretation of this model?
    Answer: The investment's value grows at a continuous relative rate of 4% per year. The differential equation is of the form $dV/dt = kV$, where the constant $k$ is the continuous relative growth rate. In this case, $dV/dt = (1/25)V$, so $k = 1/25 = 0.04$. This means the investment grows at a continuous relative rate of 4% per year. Choice A is incorrect because the rate of growth, $dV/dt$, is proportional to $V$, so it increases as $V$ increases; it is not constant. Choice C is incorrect because an effective annual rate of 4% corresponds to discrete compounding, whereas the model is for continuous compounding; the effective annual rate is actually $e^{0.04} - 1 \approx 0.0408$, or 4.08%. Choice D is incorrect because after one year, the value is $V(1) = 10000e^{0.04} \approx 10408.11$, which is not a $1/25$ (or 4%) increase.

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