Calculus 2 Quiz: Exponential Growth And Decay Models

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Exponential Growth And Decay ModelsQuestion 1 of 20

A chemical reaction follows first-order kinetics: $\frac{d[A]}{dt} = -k[A]$ , where $[A]$ is the concentration of reactant A. If the reaction is 80% complete after 45 minutes, what is the half-life of the reaction?

t_{1/2} = 45\ln(2) \approx 31.2

minutes

t_{1/2} = \frac{45}{2} = 22.5

minutes

t_{1/2} = \frac{45\ln(2)}{\ln(0.2)} \approx 19.4

minutes

t_{1/2} = \frac{45\ln(2)}{\ln(5)} \approx 19.4

minutes

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Calculus 2 Quiz

Calculus 2 Quiz: Exponential Growth And Decay Models

Practice Exponential Growth And Decay Models in Calculus 2 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 Exponential Growth And Decay Models, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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

$t_{1/2} = 45\ln(2) \approx 31.2$ minutes
$t_{1/2} = \frac{45}{2} = 22.5$ minutes
$t_{1/2} = \frac{45\ln(2)}{\ln(0.2)} \approx 19.4$ minutes
$t_{1/2} = \frac{45\ln(2)}{\ln(5)} \approx 19.4$ minutes (correct answer)

Explanation: When you encounter first-order kinetics problems, you're dealing with exponential decay where the rate of change is proportional to the current amount. The key is solving the differential equation and connecting different time points. Starting with

\frac{d[A]}{dt} = -k[A]

, separation of variables gives us

\frac{d[A]}{[A]} = -k \, dt

. Integrating both sides yields

\ln[A] = -kt + C

, or

[A] = [A]_0 e^{-kt}

where

[A]_0

is the initial concentration. If the reaction is 80% complete after 45 minutes, then 20% of the original reactant remains:

[A] = 0.2[A]_0

. Substituting into our equation:

0.2[A]_0 = [A]_0 e^{-45k}

, which simplifies to

0.2 = e^{-45k}

. Taking the natural logarithm:

\ln(0.2) = -45k

, so

k = -\frac{\ln(0.2)}{45} = \frac{\ln(5)}{45}

. For half-life, we need the time when

[A] = 0.5[A]_0

0.5 = e^{-kt_{1/2}}

. This gives us

\ln(0.5) = -kt_{1/2}

, so

t_{1/2} = \frac{\ln(2)}{k} = \frac{45\ln(2)}{\ln(5)}

, which is answer D. A incorrectly multiplies by

\ln(2)

instead of dividing by the rate constant. B naively assumes linear decay rather than exponential. C uses

\ln(0.2)

in the denominator but forgets that

\ln(0.2) = -\ln(5)

, missing the sign change. Remember: for first-order kinetics, always solve for the rate constant first using given information, then apply it to find other time-related quantities.

Question 2

Carbon-14 dating is used to determine the age of an ancient artifact. The decay model is $\frac{dC}{dt} = -\lambda C$ with half-life 5730 years. If the artifact currently contains 25% of the original carbon-14, approximately how old is the artifact?

11,460 years (exactly 2 half-lives) (correct answer)
8,267 years (using $t = \frac{\ln(4)}{\lambda}$ )
14,325 years (exactly 2.5 half-lives)
22,920 years (exactly 4 half-lives)

Explanation: If 25% remains, then 75% has decayed. Since

C(t) = C_0 e^{-\lambda t}

, we have

0.25C_0 = C_0 e^{-\lambda t}

, so

0.25 = e^{-\lambda t}

. Taking ln:

\ln(0.25) = -\lambda t

, giving

t = \frac{\ln(4)}{\lambda}

. Since

\lambda = \frac{\ln(2)}{5730}

, we get

t = \frac{\ln(4) \cdot 5730}{\ln(2)} = \frac{2\ln(2) \cdot 5730}{\ln(2)} = 11,460

years. This is exactly 2 half-lives since

(1/2)^2 = 1/4 = 0.25

. Choice B uses the correct formula but miscalculates. Choice C assumes 2.5 half-lives. Choice D assumes 4 half-lives.

Question 3

An exponentially growing population $P(t)$ is modeled by the differential equation $\frac{dP}{dt} = k(t)P$ , where the relative growth rate is given by $k(t) = \frac{a}{1+t}$ for $t \ge 0$ and constant $a > 0$ . If $P(0) = P_0$ , which of the following is an expression for $P(t)$ ?

$P_0 e^{at}$
$P_0 e^{a \ln(1+t)}$
$P_0(1+t)^a$ (correct answer)
$P_0(1+at)$

Explanation: The differential equation is separable:

\frac{1}{P} dP = k(t) dt = \frac{a}{1+t} dt

. Integrating both sides gives

\int \frac{1}{P} dP = \int \frac{a}{1+t} dt

, which yields

\ln|P| = a \ln(1+t) + C

. Exponentiating both sides, we get

P(t) = e^{a \ln(1+t) + C} = e^C e^{\ln((1+t)^a)} = A(1+t)^a

, where

A = e^C

. Using the initial condition

P(0) = P_0

, we have

P_0 = A(1+0)^a = A

. Thus, the solution is

P(t) = P_0(1+t)^a

Question 4

A cup of coffee at 90°C is placed in a room. After 10 minutes, its temperature is 70°C. After another 10 minutes, its temperature is 55°C. Assuming the coffee cools according to Newton's Law of Cooling, what is the ambient temperature of the room?

5°C
10°C (correct answer)
15°C
20°C

Explanation: Let

T(t)

be the temperature and

T_s

be the room temperature. Newton's Law of Cooling is

\frac{dT}{dt} = k(T - T_s)

, with solution

T(t) = T_s + (T_0 - T_s)e^{kt}

. We have

T(0)=90

T(10)=70

, and

T(20)=55

. From these, we get the equations:

70 = T_s + (90-T_s)e^{10k}

and

55 = T_s + (90-T_s)e^{20k}

. Let

x = e^{10k}

. The equations become

70 - T_s = (90-T_s)x

and

55 - T_s = (90-T_s)x^2

. From the first equation,

x = \frac{70-T_s}{90-T_s}

. Substituting this into the second equation gives

55 - T_s = (90 - T_s) \left(\frac{70-T_s}{90-T_s}\right)^2

, which simplifies to

(55-T_s)(90-T_s) = (70-T_s)^2

. Expanding both sides yields

4950 - 145T_s + T_s^2 = 4900 - 140T_s + T_s^2

. Solving for

T_s

gives

50 = 5T_s

, so

T_s = 10

Question 5

The population $P(t)$ of a certain species grows in such a way that the rate of change of its relative growth rate is proportional to the current population. The relative growth rate is defined as $\frac{1}{P}\frac{dP}{dt}$ . Which differential equation correctly models this scenario, where $c$ is a constant of proportionality?

$\frac{d^2 P}{dt^2} = cP$
$\frac{dP}{dt} = cP^2$
$\frac{d}{dt}\left( \frac{1}{P}\frac{dP}{dt} \right) = cP$ (correct answer)
$\frac{d}{dt}\left( \frac{dP}{dt} \right) = cP$

Explanation: Let's translate the statement piece by piece. The 'relative growth rate' is given as

R = \frac{1}{P}\frac{dP}{dt}

. The 'rate of change of its relative growth rate' is the derivative of

R

with respect to time,

\frac{dR}{dt} = \frac{d}{dt}\left( \frac{1}{P}\frac{dP}{dt} \right)

. The statement says this quantity is 'proportional to the current population', which means it equals

cP

for some constant

c

. Therefore, the correct differential equation is

\frac{d}{dt}\left( \frac{1}{P}\frac{dP}{dt} \right) = cP

Question 6

A 1000-liter tank initially contains 500 liters of brine with 50 kg of dissolved salt. Brine containing 0.2 kg of salt per liter is pumped into the tank at a rate of 10 L/min. The well-mixed solution is pumped out at a rate of 5 L/min. Which differential equation models the amount of salt $A(t)$ in kilograms in the tank at time $t$ in minutes?

$\frac{dA}{dt} = 2 - \frac{A}{100}$
$\frac{dA}{dt} = 2 - \frac{A}{100+t}$ (correct answer)
$\frac{dA}{dt} = 10 - \frac{A}{100+t}$
$\frac{dA}{dt} = 2 - \frac{A}{500+5t}$

Explanation: The rate of change of salt is

\frac{dA}{dt} = (\text{rate in}) - (\text{rate out})

. The rate of salt flowing in is

(0.2 \text{ kg/L}) \times (10 \text{ L/min}) = 2 \text{ kg/min}

. The volume of the solution in the tank changes over time. Initially

V(0)=500

L. Since liquid flows in at 10 L/min and out at 5 L/min, the net change is +5 L/min. So,

V(t) = 500 + 5t

. The concentration of salt at time

t

C(t) = \frac{A(t)}{V(t)} = \frac{A(t)}{500+5t}

kg/L. The rate of salt flowing out is

C(t) \times (\text{outflow rate}) = \frac{A(t)}{500+5t} \times 5 = \frac{5A(t)}{500+5t} = \frac{A(t)}{100+t}

. Combining these gives

\frac{dA}{dt} = 2 - \frac{A}{100+t}

Question 7

Substance A has a half-life of $h_A$ years and Substance B has a half-life of $h_B$ years. An initial sample contains equal masses of A and B. After a time $t$ such that $t = 2h_A = 3h_B$ , what is the ratio of the remaining mass of Substance A to the remaining mass of Substance B?

$\frac{2}{3}$
$\frac{3}{2}$
2 (correct answer)
$\frac{1}{2}$

Explanation: The mass of a substance remaining after time

t

with half-life

h

is given by

m(t) = m_0 (1/2)^{t/h}

. Let the initial equal masses be

m_0

. The mass of substance A at time

t

m_A(t) = m_0 (1/2)^{t/h_A}

. The mass of substance B is

m_B(t) = m_0 (1/2)^{t/h_B}

. We are given

t = 2h_A

and

t = 3h_B

. So,

m_A(t) = m_0 (1/2)^{2h_A/h_A} = m_0 (1/2)^2 = m_0/4

. And

m_B(t) = m_0 (1/2)^{3h_B/h_B} = m_0 (1/2)^3 = m_0/8

. The ratio of the mass of A to the mass of B is

\frac{m_A(t)}{m_B(t)} = \frac{m_0/4}{m_0/8} = \frac{8}{4} = 2

Question 8

An account balance $B(t)$ earns interest at a rate proportional to the current balance, with an annual proportionality constant $r$ . Simultaneously, continuous withdrawals are made at a rate proportional to the square root of the balance, with a proportionality constant $w$ . Which differential equation correctly models the balance $B(t)$ over time?

$\frac{dB}{dt} = rB - wB^2$
$\frac{dB}{dt} = rB^2 - w\sqrt{B}$
$\frac{dB}{dt} = rB - w\sqrt{B}$ (correct answer)
$\frac{dB}{dt} = r\sqrt{B} - wB$

Explanation: The differential equation is built by considering the rates of increase and decrease of the balance. The phrase 'earns interest at a rate proportional to the current balance' translates to an interest rate term of

+rB

. The phrase 'continuous withdrawals are made at a rate proportional to the square root of the balance' translates to a withdrawal term of

-w\sqrt{B}

. Combining these two terms gives the net rate of change of the balance:

\frac{dB}{dt} = rB - w\sqrt{B}

Question 9

A hot object is placed in a room with a constant ambient temperature. According to Newton's Law of Cooling, the temperature difference between the object and the room decreases by 50% in the first 10 minutes. By what percentage does this temperature difference decrease over the first 20 minutes?

25%
50%
75% (correct answer)
100%

Explanation: Let

D(t) = T(t) - T_s

be the temperature difference. It follows the model

\frac{dD}{dt} = kD

, with solution

D(t) = D_0 e^{kt}

. A 50% decrease in 10 minutes means

D(10) = 0.5 D_0

. This is the definition of half-life, so the half-life of the temperature difference is 10 minutes. After another 10 minutes (for a total of 20 minutes), another half-life has passed. The difference will be halved again:

D(20) = 0.5 D(10) = 0.5(0.5 D_0) = 0.25 D_0

. If 25% of the difference remains, the total decrease is

100% - 25% = 75%

Question 10

A quantity $y$ grows exponentially according to $y(t) = y_0 e^{kt}$ . It is observed that the quantity doubles from its initial value at time $t=T$ , and it triples from its initial value at time $t=T+5$ . What is the value of the doubling time $T$ ?

5 years
$5 \frac{\ln(2)}{\ln(3) - \ln(2)}$ (correct answer)
$5 \frac{\ln(1.5)}{\ln(2)}$
$5 (\ln(3) - \ln(2))$

Explanation: We are given two conditions:

y(T) = 2y_0

and

y(T+5) = 3y_0

. Using the solution form, these become

y_0 e^{kT} = 2y_0

and

y_0 e^{k(T+5)} = 3y_0

. The first equation simplifies to

e^{kT} = 2

. The second becomes

e^{kT}e^{5k} = 3

. Substituting

e^{kT}=2

into the second simplified equation gives

2e^{5k} = 3

, so

e^{5k} = 1.5

. From

e^{kT}=2

, we have

kT = \ln(2)

, so

k = \frac{\ln(2)}{T}

. From

e^{5k} = 1.5

, we have

5k = \ln(1.5)

, so

k = \frac{\ln(1.5)}{5}

. Equating the two expressions for

k

gives

\frac{\ln(2)}{T} = \frac{\ln(1.5)}{5}

. Solving for

T

yields

T = 5 \frac{\ln(2)}{\ln(1.5)} = 5 \frac{\ln(2)}{\ln(3/2)} = 5 \frac{\ln(2)}{\ln(3) - \ln(2)}

Question 11

Two populations, $P_1(t)$ and $P_2(t)$ , grow exponentially with respective positive growth constants $k_1$ and $k_2$ . Initially, at $t=0$ , the size of the first population is twice the size of the second, i.e., $P_1(0) = 2P_2(0)$ . After 5 years, the populations are equal. Which equation correctly relates the growth constants?

$k_2 - k_1 = \frac{\ln(2)}{5}$ (correct answer)
$k_1 - k_2 = \frac{\ln(2)}{5}$
$k_2 = 2k_1$
$k_1 = 2k_2$

Explanation: The population models are

P_1(t) = P_1(0)e^{k_1 t}

and

P_2(t) = P_2(0)e^{k_2 t}

. Let

P_2(0) = A

, so

P_1(0) = 2A

. At

t=5

, we have

P_1(5) = P_2(5)

. Substituting the models gives

(2A)e^{5k_1} = (A)e^{5k_2}

. Dividing by

A

(since

A>0

) gives

2e^{5k_1} = e^{5k_2}

. Taking the natural logarithm of both sides gives

\ln(2e^{5k_1}) = \ln(e^{5k_2})

, which simplifies to

\ln(2) + 5k_1 = 5k_2

. Rearranging the terms to relate

k_1

and

k_2

gives

5k_2 - 5k_1 = \ln(2)

, or

k_2 - k_1 = \frac{\ln(2)}{5}

Question 12

The differential equation $\frac{dy}{dt} = -0.02(y - 10)$ with the initial condition $y(0) = 50$ models the temperature of an object. Which of the following statements best describes the long-term behavior of $y(t)$ as $t \to \infty$ ?

The value of $y(t)$ approaches 10 from above. (correct answer)
The value of $y(t)$ approaches 10 from below.
The value of $y(t)$ approaches 0.
The value of $y(t)$ decreases without bound.

Explanation: This is a form of Newton's Law of Cooling,

\frac{dy}{dt} = k(y - T_s)

, where

k=-0.02

and the ambient temperature is

T_s=10

. The general solution is

y(t) = T_s + C e^{kt} = 10 + C e^{-0.02t}

. Using the initial condition

y(0)=50

, we find

50 = 10 + C e^0 \Rightarrow C=40

. So,

y(t) = 10 + 40e^{-0.02t}

. As

t \to \infty

, the term

e^{-0.02t}

approaches 0. Therefore,

\lim_{t \to \infty} y(t) = 10

. Since the term

40e^{-0.02t}

is always positive for finite

t

y(t)

is always greater than 10. Thus,

y(t)

approaches 10 from above.

Question 13

The concentration of a drug in the bloodstream, $C(t)$ , decays exponentially after an injection. The initial concentration is 1.2 mg/L. After 4 hours, the concentration is 0.3 mg/L. To be effective, the concentration must be at least 0.1 mg/L. For how many total hours after the injection is the drug effective?

6
$4 + 2\log_2(3)$ (correct answer)
$2\log_2(1.2)$
$4\log_3(12)$

Explanation: The model is

C(t) = C_0 e^{kt}

with

C_0 = 1.2

. We are given

C(4) = 0.3

. So,

0.3 = 1.2 e^{4k}

, which gives

e^{4k} = 0.3/1.2 = 1/4

. Taking the natural log,

4k = \ln(1/4) = -2\ln(2)

, so

k = -\frac{\ln(2)}{2}

. This implies a half-life of 2 hours. We need to find the time

t

for which

C(t) = 0.1

. So,

0.1 = 1.2 e^{kt}

, which gives

e^{kt} = \frac{0.1}{1.2} = \frac{1}{12}

. Taking the log,

kt = \ln(1/12) = -\ln(12)

. Substituting for

k

(-\frac{\ln(2)}{2})t = -\ln(12)

. Solving for

t

gives

t = \frac{2\ln(12)}{\ln(2)} = 2\log_2(12)

. We can simplify this:

t = 2\log_2(4 \cdot 3) = 2(\log_2(4) + \log_2(3)) = 2(2 + \log_2(3)) = 4 + 2\log_2(3)

Question 14

The value of a machine, $V(t)$ in dollars, depreciates according to the differential equation $\frac{dV}{dt} = -0.15V$ , where $t$ is in years. The initial value is $V(0) = \$ 20,000 $. What is the correct interpretation of the quantity$ \frac{1}{5}\int_0^5 V(t) dt$?

The total depreciation of the machine over the first 5 years.
The instantaneous rate of depreciation at $t=5$ years.
The average value of the machine over the first 5 years. (correct answer)
The value of the machine after 5 years.

Explanation: The expression

\frac{1}{b-a}\int_a^b f(x) dx

is the definition of the average value of a function

f(x)

on the interval

[a, b]

. In this context, with

V(t)

representing the value of the machine at time

t

, the expression

\frac{1}{5-0}\int_0^5 V(t) dt

represents the average value of the machine over the time interval from

t=0

t=5

years. The other choices represent different quantities: total depreciation is

V(0)-V(5)

, instantaneous rate is

V'(5)

, and value at 5 years is

V(5)

Question 15

A fossil is found to have a Carbon-14 to Carbon-12 ratio that is 30% of the ratio found in living organisms. The half-life of Carbon-14 is 5730 years. A second fossil from the same species is found in a deeper geological layer, and its C-14 to C-12 ratio is 15% of the living ratio. What is the approximate age difference between the two fossils?

2865 years
5730 years (correct answer)
8600 years
11460 years

Explanation: Let

R(t)

be the C-14 ratio at time

t

after death. The decay model is

R(t) = R_0 (1/2)^{t/T_{1/2}}

, where

T_{1/2} = 5730

years. Let

t_1

be the age of the first fossil and

t_2

be the age of the second. For the first fossil,

0.30 R_0 = R_0 (1/2)^{t_1/5730}

. For the second,

0.15 R_0 = R_0 (1/2)^{t_2/5730}

. Dividing the second equation by the first gives

\frac{0.15}{0.30} = \frac{(1/2)^{t_2/5730}}{(1/2)^{t_1/5730}}

, which simplifies to

1/2 = (1/2)^{(t_2 - t_1)/5730}

. For the equality to hold, the exponents must be equal:

1 = \frac{t_2 - t_1}{5730}

. Therefore, the age difference

t_2 - t_1

is 5730 years. This makes sense conceptually: the C-14 ratio was halved, so one half-life must have passed.

Question 16

Let $y(t)$ be a solution to the differential equation $\frac{dy}{dt} = ky$ with $k \neq 0$ and initial condition $y(0) = y_0 > 0$ . Consider the function $L(t) = \ln(y(t))$ . Which of the following statements is true about the function $L(t)$ ?

$L(t)$ is an exponential function of $t$ with base $e^k$ .
$L(t)$ is a linear function of $t$ with slope $k$ and intercept $\ln(y_0)$ . (correct answer)
$L(t)$ is a logarithmic function of $t$ of the form $k \ln(t) + C$ .
$L(t)$ is a constant function with value $\ln(y_0)$ .

Explanation: The solution to

\frac{dy}{dt} = ky

with

y(0) = y_0

y(t) = y_0 e^{kt}

. The function

L(t)

is defined as

L(t) = \ln(y(t))

. Substituting the expression for

y(t)

, we get

L(t) = \ln(y_0 e^{kt})

. Using the properties of logarithms, this becomes

L(t) = \ln(y_0) + \ln(e^{kt}) = \ln(y_0) + kt

. This is the equation of a line,

L(t) = kt + \ln(y_0)

, in the form

mx+b

, where the slope

m

k

and the y-intercept

b

\ln(y_0)

Question 17

A culture of bacteria grows at a rate proportional to the number of bacteria present. The number of bacteria triples in 6 hours. By what factor will the number of bacteria increase in 18 hours?

9
12
18
27 (correct answer)

Explanation: Let

P(t)

be the number of bacteria. The model is

P(t) = P_0 e^{kt}

. We are given that

P(6) = 3P_0

. So,

3P_0 = P_0 e^{6k}

, which implies

e^{6k} = 3

. We want to find the factor of increase in 18 hours, which is

\frac{P(18)}{P_0}

P(18) = P_0 e^{18k} = P_0 e^{6k \cdot 3} = P_0 (e^{6k})^3

. Since we know

e^{6k} = 3

, we have

P(18) = P_0 (3)^3 = 27P_0

. Therefore, the number of bacteria will increase by a factor of 27.

Question 18

The logistic growth model is given by $\frac{dP}{dt} = rP(1 - \frac{P}{K})$ , where $r > 0$ is the intrinsic growth rate and $K > 0$ is the carrying capacity. For which of the following conditions does the logistic model most closely approximate the simple exponential growth model, $\frac{dP}{dt} = rP$ ?

When the population $P$ is very small compared to the carrying capacity $K$ . (correct answer)
When the population $P$ is very close to the carrying capacity $K$ .
When the population $P$ is equal to half the carrying capacity, $K/2$ .
When the intrinsic growth rate $r$ is very close to zero.

Explanation: The logistic model equation is

\frac{dP}{dt} = rP - \frac{r}{K}P^2

. The simple exponential growth model is

\frac{dP}{dt} = rP

. The two models are approximately equal when the second term in the logistic equation,

-\frac{r}{K}P^2

, is negligible compared to the first term,

rP

. This occurs when the factor

(1 - \frac{P}{K})

is close to 1. This condition is met when

\frac{P}{K}

is very close to 0, which means that the population

P

is very small compared to the carrying capacity

K

Question 19

A bank account with continuous compounding at an annual rate $r$ is modeled by $\frac{dB}{dt} = rB$ . If an initial deposit of $P$ quadruples in 18 years, in how many years will the initial deposit triple?

9 years
12 years
$18 \frac{\ln(4)}{\ln(3)}$ years
$18 \frac{\ln(3)}{\ln(4)}$ years (correct answer)

Explanation: When you encounter differential equations modeling exponential growth like

\frac{dB}{dt} = rB

, you're dealing with continuous compounding problems that follow the pattern

B(t) = Pe^{rt}

. The key insight is using given information to find relationships between different time scenarios. Since the account quadruples in 18 years, we have

4P = Pe^{18r}

, which simplifies to

4 = e^{18r}

. Taking the natural logarithm:

\ln(4) = 18r

, so

r = \frac{\ln(4)}{18}

. Now we need to find when the account triples:

3P = Pe^{rt}

, giving us

3 = e^{rt}

. Taking the natural logarithm:

\ln(3) = rt = t \cdot \frac{\ln(4)}{18}

. Solving for t:

t = \frac{18\ln(3)}{\ln(4)}

. Looking at the wrong answers: Choice A (9 years) incorrectly assumes tripling takes half the time of quadrupling, which only works for linear growth, not exponential. Choice B (12 years) might come from incorrectly thinking tripling takes

\frac{3}{4}

the time of quadrupling (

18 \times \frac{3}{4} = 13.5

, rounded to 12). Choice C has the logarithms inverted—this would give the wrong relationship between the growth rates. The correct answer is D:

18 \frac{\ln(3)}{\ln(4)}

years. Study tip: In exponential growth problems, always use the given condition to find the growth rate first, then apply it to the desired scenario. Remember that time ratios in exponential growth involve logarithms, not simple proportions.

Question 20

The rate of change of a quantity $y$ is directly proportional to $y$ . It is known that at time $t_1$ , the value of the quantity is $a$ , and at time $t_2$ , its value is $b$ . What is the value of the quantity at time $2t_2 - t_1$ ?

$2b-a$
$\frac{b}{a}$
$\frac{a^2}{b}$
$\frac{b^2}{a}$ (correct answer)

Explanation: The governing equation is

\frac{dy}{dt} = ky

, so

y(t) = y_0 e^{kt}

. We are given

y(t_1) = a = y_0 e^{kt_1}

and

y(t_2) = b = y_0 e^{kt_2}

. Dividing these gives

\frac{b}{a} = \frac{y_0 e^{kt_2}}{y_0 e^{kt_1}} = e^{k(t_2 - t_1)}

. We want to find

y(2t_2 - t_1) = y_0 e^{k(2t_2 - t_1)}

. We can rewrite the exponent as

k(t_2 + (t_2 - t_1))

. So,

y(2t_2-t_1) = y_0 e^{kt_2} e^{k(t_2 - t_1)}

. We know

y_0 e^{kt_2} = y(t_2) = b

and

e^{k(t_2 - t_1)} = \frac{b}{a}

. Therefore,

y(2t_2 - t_1) = b \cdot \frac{b}{a} = \frac{b^2}{a}