Business Calculus Quiz: Separable Des Growth Decay

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Separable Des Growth DecayQuestion 1 of 16

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?

500 users with maximum growth rate of 250 users per day

707 users with maximum growth rate of 207 users per day

333 users with maximum growth rate of 222 users per day

866 users with maximum growth rate of 116 users per day

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Business Calculus Quiz

Business Calculus Quiz: Separable Des Growth Decay

Practice Separable Des Growth Decay in Business Calculus 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 Separable Des Growth Decay, giving you a quick way to practice the rules, question types, and explanations that matter most for Business Calculus.

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.

All questions

Question 1

500 users with maximum growth rate of 250 users per day (correct answer)
707 users with maximum growth rate of 207 users per day
333 users with maximum growth rate of 222 users per day
866 users with maximum growth rate of 116 users per day

Explanation: 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.

Question 2

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)?

9,000
10,000
25,000 (correct answer)
50,000

Explanation: 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

Question 3

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?

2.8 months
4.0 months (correct answer)
5.6 months
8.0 months

Explanation: 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

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.

Question 4

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?

55 years (correct answer)
34 years
7 years
75 years

Explanation: 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.

Question 5

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?

6.5 years
15.3 years
54.9 years
45.8 years (correct answer)

Explanation: 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}

\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.

Question 6

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?

The investment's value grows by a constant amount of $400 per year.
The investment's value grows at a continuous relative rate of 4% per year. (correct answer)
The investment's value increases by an effective annual rate of exactly 4%.
The investment's total value will be exactly $1/25$ larger after one year.

Explanation: 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.

Question 7

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?

$47,000
$64,000
$128,000 (correct answer)
$512,000

Explanation: 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

Question 8

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?

10 days
139 days
1370 days
599 days (correct answer)

Explanation: 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.

Question 9

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 ( $t=6$ )?

The population increased by exactly 2.00% during this period.
The population increased by approximately 2.02% during this period. (correct answer)
The population's relative rate of growth was exactly 2.02% during this year.
The population increased by a larger percentage than between $t=4$ and $t=5$ .

Explanation: The model represents continuous growth with a rate

k=0.02

. The percentage increase over a one-year interval from

t

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.

Question 10

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 $t=0$?

$31,250
$250,000 (correct answer)
$500,000
$1,000,000

Explanation: 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

Question 11

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?

4.9 years
6.2 years
11.3 years
7.2 years (correct answer)

Explanation: 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.

Question 12

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?

$156,250 (correct answer)
$125,000
$123,200
$40,960

Explanation: 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.

Question 13

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?

277,000
400,000
554,000 (correct answer)
800,000

Explanation: The solution to

dS/dt = kS

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

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

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.

Question 14

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?

Approximately 6.37%, reflecting near-complete convergence to the target with minor oscillations (correct answer)
Approximately 7.15%, representing partial adjustment toward target with residual deviation
Approximately 5.82%, indicating overcorrection below target due to strong convergence forces
Approximately 8.43%, showing minimal change due to competing seasonal and target effects

Explanation: 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.

Question 15

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.

Equilibrium at 60 units; the system will converge to this stable equilibrium point (correct answer)
Equilibrium at 40 units; the system will diverge away from this unstable equilibrium
Equilibrium at 80 units; the system will oscillate around this equilibrium indefinitely
Equilibrium at 60 units; the system will diverge to zero due to the linear decay term

Explanation: 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

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.

Question 16

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?

$5,000,000 equilibrium; 11.5 weeks to reach 90% of the adjustment (correct answer)
$3,200,000 equilibrium; 8.7 weeks to reach 90% of the adjustment
$5,000,000 equilibrium; 23.0 weeks to reach 90% of the adjustment
$6,400,000 equilibrium; 15.3 weeks to reach 90% of the adjustment

Explanation: 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.