Calculus 1 Quiz: Exponential De Models

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Exponential De ModelsQuestion 1 of 20

The rate of growth of a cell culture is proportional to its size. Initially, the culture contains 200 cells, and after 3 hours, it contains 500 cells. How many cells are in the culture after 6 hours?

1250

800

1000

1100

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Calculus 1 Quiz

Calculus 1 Quiz: Exponential De Models

Practice Exponential De Models in Calculus 1 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 De Models, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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.

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Question 1

The rate of growth of a cell culture is proportional to its size. Initially, the culture contains 200 cells, and after 3 hours, it contains 500 cells. How many cells are in the culture after 6 hours?

1250 (correct answer)
800
1000
1100

Explanation: Let

P(t)

be the number of cells. The model is

P(t) = P_0 e^{kt}

. We have

P_0 = 200

. So,

P(t) = 200 e^{kt}

. We are given

P(3) = 500

, so

500 = 200 e^{3k}

, which means

e^{3k} = 500/200 = 2.5

. We want to find

P(6)

P(6) = 200 e^{6k} = 200 (e^{3k})^2

. Substituting the value of

e^{3k}

, we get

P(6) = 200 (2.5)^2 = 200 (6.25) = 1250

Question 2

Let $f(t)$ be a solution to $y' = ky$ with $k>0$ . The line tangent to the graph of $y=f(t)$ at $t=a$ passes through the origin. Which of the following must be true?

$a = k$
$ka = 1$ (correct answer)
$f(a) = k$
$a = 0$

Explanation: This problem combines differential equations with tangent line geometry. When you see a differential equation like

y' = ky

paired with geometric conditions, you need to find the general solution first, then apply the constraint. The differential equation

y' = ky

has the general solution

f(t) = Ce^{kt}

for some constant

C

. At point

(a, f(a))

, the slope of the tangent line is

f'(a) = kf(a) = kCe^{ka}

. Since this tangent line passes through both the point

(a, f(a))

and the origin

(0,0)

, you can find its slope using the slope formula:

\text{slope} = \frac{f(a) - 0}{a - 0} = \frac{Ce^{ka}}{a}

. Setting the two expressions for slope equal:

kCe^{ka} = \frac{Ce^{ka}}{a}

. Assuming

Ce^{ka} \neq 0

(which must be true since the tangent line has a well-defined slope), you can divide both sides by

Ce^{ka}

to get

k = \frac{1}{a}

, which means

ka = 1

. Looking at the wrong answers: (A)

a = k

would only be true if

k = 1

, but

k

can be any positive value. (C)

f(a) = k

has no necessary relationship to our geometric constraint. (D)

a = 0

would make the tangent line undefined since we'd have a vertical line through the origin. Study tip: When tangent lines pass through special points like the origin, always set up two expressions for the slope—one using the derivative and one using the point-slope relationship—then solve the resulting equation.

Question 3

The population of a bacterial colony, $P$ , grows according to the differential equation $\frac{dP}{dt} = kP$ . The initial population doubles in 3 hours. At what time, in hours, will the population be 10 times its initial size?

$\frac{3 \ln(10)}{\ln(2)}$ (correct answer)
$\frac{3 \ln(2)}{\ln(10)}$
15
$\frac{10 \ln(3)}{\ln(2)}$

Explanation: The solution is

P(t) = P_0 e^{kt}

. The doubling time is 3 hours, so

2P_0 = P_0 e^{k \cdot 3}

, which simplifies to

2 = e^{3k}

. Taking the natural logarithm gives

\ln(2) = 3k

, so

k = \frac{\ln(2)}{3}

. We want to find the time

t

when

P(t) = 10P_0

. So,

10P_0 = P_0 e^{kt}

, which means

10 = e^{kt}

. Taking the natural logarithm gives

\ln(10) = kt

. Substituting the value of

k

, we get

\ln(10) = \left(\frac{\ln(2)}{3}\right)t

. Solving for

t

gives

t = \frac{3 \ln(10)}{\ln(2)}

Question 4

A quantity $Q(t)$ changes at a rate proportional to its current value. If $Q(2) = 20$ and $Q(5) = 160$ , what is the instantaneous rate of change of $Q$ at $t=2$ ?

$20 \ln(2)$ (correct answer)
$\ln(2)$
$5 \ln(2)$
80

Explanation: The model is

\frac{dQ}{dt} = kQ

, with solution

Q(t) = Q_0 e^{kt}

. We have

20 = Q_0 e^{2k}

and

160 = Q_0 e^{5k}

. Dividing the second equation by the first gives

\frac{160}{20} = \frac{e^{5k}}{e^{2k}}

, which simplifies to

8 = e^{3k}

. Taking the natural logarithm,

\ln(8) = 3k

, so

3\ln(2) = 3k

, which means

k = \ln(2)

. The question asks for the rate of change at

t=2

, which is

\frac{dQ}{dt}|_{t=2}

. Using the differential equation,

\frac{dQ}{dt}|_{t=2} = k \cdot Q(2) = \ln(2) \cdot 20 = 20 \ln(2)

Question 5

Two radioactive substances, A and B, decay according to the law $\frac{dy}{dt} = ky$ . Substance A has a half-life of 20 years. Substance B has a half-life of 30 years. If samples of A and B initially contain the same number of atoms, what is the ratio of the number of atoms of A to the number of atoms of B after 60 years?

$1/2$ (correct answer)
2
$2/3$
$4/9$

Explanation: After 60 years, substance A has undergone

60/20 = 3

half-lives. The fraction of substance A remaining is

(1/2)^3 = 1/8

. After 60 years, substance B has undergone

60/30 = 2

half-lives. The fraction of substance B remaining is

(1/2)^2 = 1/4

. Let the initial number of atoms be

N_0

for both. After 60 years,

N_A = N_0/8

and

N_B = N_0/4

. The ratio of A to B is

\frac{N_A}{N_B} = \frac{N_0/8}{N_0/4} = \frac{1/8}{1/4} = \frac{4}{8} = 1/2

Question 6

A cup of coffee at $180^{\circ}$ F is placed in a room with a constant temperature of $70^{\circ}$ F. The coffee's temperature, $T(t)$ , is modeled by the differential equation $\frac{dT}{dt} = k(T - 70)$ . If the coffee cools to $130^{\circ}$ F in 10 minutes, which expression represents the temperature $T(t)$ after $t$ minutes?

$T(t) = 70 + 110 (\frac{6}{11})^{t/10}$ (correct answer)
$T(t) = 70 + 180 (\frac{6}{11})^{t/10}$
$T(t) = 70 + 110 e^{t \ln(6/11)}$
$T(t) = 180 (\frac{13}{18})^{t/10}$

Explanation: The differential equation

\frac{dT}{dt} = k(T - 70)

has the solution

T(t) - 70 = (T_0 - 70)e^{kt}

, where

T_0

is the initial temperature. With

T(0)=180

, this becomes

T(t) = 70 + (180 - 70)e^{kt} = 70 + 110e^{kt}

. We use the point

(10, 130)

to find

k

130 = 70 + 110e^{10k}

, which simplifies to

60 = 110e^{10k}

, so

\frac{6}{11} = e^{10k}

. We can write

e^k = (\frac{6}{11})^{1/10}

. Substituting this back into the solution gives

T(t) = 70 + 110(e^k)^t = 70 + 110 ((\frac{6}{11})^{1/10})^t = 70 + 110 (\frac{6}{11})^{t/10}

Question 7

A bank account balance, $B$ , grows according to the differential equation $\frac{dB}{dt} = 0.04B$ , where $t$ is in years. An initial deposit of $100 is made at $t=0$ , and a second deposit of $100 is made at $t=5$ . What is the balance in the account at $t=10$ ?

$100e^{0.4} + 100e^{0.2}$ (correct answer)
$200e^{0.4}$
$100e^{0.4} + 100e^{0.6}$
$100e^{0.2} + 100e^{0.2}$

Explanation: We can model the growth of each deposit separately. The first $100 is invested for the full 10 years. Its value at

t=10

100e^{0.04 \cdot 10} = 100e^{0.4}

. The second $100 is deposited at

t=5

and is invested for

10-5=5

years. Its value at

t=10

100e^{0.04 \cdot 5} = 100e^{0.2}

. The total balance at

t=10

is the sum of the values of these two deposits, which is

100e^{0.4} + 100e^{0.2}

Question 8

Let $y(t)$ be a solution to $\frac{dy}{dt} = ky$ with $k<0$ and $y(0)>0$ . Let $t_h$ be the half-life. Which of the following statements is true about the time $t_q$ it takes for the quantity to reduce to one-quarter of its original size?

$t_q = 4t_h$
$t_q = (t_h)^2$
$t_q = t_h + \ln(2)/|k|$
$t_q = 2t_h$ (correct answer)

Explanation: When you encounter differential equations of the form

\frac{dy}{dt} = ky

with

k < 0

, you're dealing with exponential decay. The solution is

y(t) = y(0)e^{kt}

, where

y(0)

is the initial value. The key insight is understanding the relationship between half-life and quarter-life. The half-life

t_h

is when

y(t_h) = \frac{1}{2}y(0)

. Setting up the equation:

\frac{1}{2}y(0) = y(0)e^{kt_h}

, which simplifies to

\frac{1}{2} = e^{kt_h}

. Taking the natural log:

kt_h = \ln(1/2) = -\ln(2)

, so

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

. For the quarter-life

t_q

, we need

y(t_q) = \frac{1}{4}y(0)

. Notice that

\frac{1}{4} = \frac{1}{2} \times \frac{1}{2}

, meaning the quantity halves twice. Since each halving takes time

t_h

, the total time is

t_q = 2t_h

. This makes D correct. Let's examine why the other answers fail: A suggests

t_q = 4t_h

, which would correspond to reducing to

(\frac{1}{2})^4 = \frac{1}{16}

of the original size. B proposes

t_q = (t_h)^2

, which has no physical basis in exponential decay. C gives

t_q = t_h + \ln(2)/|k|

, but since

t_h = \ln(2)/|k|

, this would equal

2t_h

anyway—though the form suggests a misunderstanding of the additive nature of half-lives. Remember: in exponential decay, each additional half-life period halves the remaining quantity, so multiple reductions are additive in time.

Question 9

A quantity $P(t)$ grows according to the differential equation $\frac{dP}{dt} = kP$ . If $P(t)$ increases from $A$ to $B$ over a time interval of length $T_1$ , and from $B$ to $C$ over a time interval of length $T_2$ , where $A, B, C$ are positive constants, which of the following is equal to $\frac{T_1}{T_2}$ ?

$\frac{\ln(B-A)}{\ln(C-B)}$
$\frac{B - A}{C - B}$
$\frac{\ln(B/A)}{C/B}$
$\frac{\ln(B) - \ln(A)}{\ln(C) - \ln(B)}$ (correct answer)

Explanation: When you encounter a differential equation of the form

\frac{dP}{dt} = kP

, you're dealing with exponential growth. This equation has the solution

P(t) = P_0 e^{kt}

, where

P_0

is the initial value. To find the time intervals, you need to use the relationship between consecutive values. If

P

grows from

A

B

in time

T_1

, then

B = Ae^{kT_1}

. Taking the natural logarithm:

\ln(B) = \ln(A) + kT_1

, so

T_1 = \frac{\ln(B) - \ln(A)}{k}

. Similarly, if

P

grows from

B

C

in time

T_2

, then

T_2 = \frac{\ln(C) - \ln(B)}{k}

. Therefore,

\frac{T_1}{T_2} = \frac{\ln(B) - \ln(A)}{\ln(C) - \ln(B)}

, which is answer choice D. Let's examine why the other options are wrong: A)

\frac{\ln(B-A)}{\ln(C-B)}

incorrectly uses differences inside the logarithm rather than differences of logarithms. This would apply to additive growth, not exponential growth. B)

\frac{B-A}{C-B}

is the ratio of absolute changes, which would be correct for linear growth (

\frac{dP}{dt} = k

), but not exponential growth. C)

\frac{\ln(B/A)}{C/B}

mixes logarithmic and linear ratios inappropriately, combining the correct numerator form with an incorrect denominator. Remember: for exponential growth problems, time intervals are proportional to differences of natural logarithms, not differences of the original values. Always solve the differential equation first to establish the exponential relationship.

Question 10

The pressure $P$ of a gas in a container is decreasing at a rate proportional to the pressure itself. The initial pressure is $P_0$ . The pressure is measured at $t=1$ to be $P_1$ and at $t=3$ to be $P_3$ . Which of the following is an expression for $P_0$ ?

$\sqrt{P_1 P_3}$
$\frac{3P_1 - P_3}{2}$
$P_1 \sqrt{P_1/P_3}$ (correct answer)
$P_3 + (P_1 - P_3)/2$

Explanation: When you encounter a problem stating that a quantity decreases "at a rate proportional to itself," you're dealing with exponential decay. This translates to the differential equation

\frac{dP}{dt} = -kP

, where k is a positive constant. The solution to this equation is

P(t) = P_0 e^{-kt}

, where

P_0

is the initial pressure at t = 0. Using the given data points, we have:

At t = 1: $P_1 = P_0 e^{-k}$
At t = 3: $P_3 = P_0 e^{-3k}$

To find

P_0

, first solve for k using the ratio

\frac{P_3}{P_1} = \frac{P_0 e^{-3k}}{P_0 e^{-k}} = e^{-2k}

. This gives us

e^{-k} = \sqrt{\frac{P_3}{P_1}}

. Substituting back into

P_1 = P_0 e^{-k}

P_0 = \frac{P_1}{e^{-k}} = \frac{P_1}{\sqrt{P_3/P_1}} = P_1 \sqrt{\frac{P_1}{P_3}}

This confirms answer C is correct. Answer A,

\sqrt{P_1 P_3}

, would be the geometric mean but doesn't account for the time intervals properly. Answer B,

\frac{3P_1 - P_3}{2}

, suggests linear decay rather than exponential. Answer D,

P_3 + \frac{P_1 - P_3}{2}

, is simply the arithmetic mean of the two measured values. Study tip: Whenever you see "rate proportional to the quantity itself," immediately think exponential growth or decay. Set up

\frac{dy}{dt} = \pm ky

and use the given data points to work backwards to find unknown parameters.

Question 11

Carbon-14 decays according to the equation $\frac{dy}{dt} = -0.00012 y$ , where $t$ is in years. A sample of a fossil is found to contain 25% of its original carbon-14. Which expression represents the age of the fossil in years?

$\frac{\ln(0.25)}{0.00012}$
$\frac{\ln(4)}{0.00012}$ (correct answer)
$\frac{4}{0.00012}$
$4 \times 5730$

Explanation: When you encounter exponential decay problems, you're dealing with differential equations where the rate of change is proportional to the current amount. The equation

\frac{dy}{dt} = -0.00012y

tells us that carbon-14 decreases at a rate proportional to how much is present. To solve this separable differential equation, separate variables:

\frac{dy}{y} = -0.00012 \, dt

. Integrating both sides gives

\ln|y| = -0.00012t + C

. Using the initial condition that at

t = 0

y = y_0

(the original amount), we find

C = \ln(y_0)

. Therefore:

y = y_0 e^{-0.00012t}

. Since the fossil contains 25% of its original carbon-14, we have

0.25y_0 = y_0 e^{-0.00012t}

. Dividing by

y_0

gives

0.25 = e^{-0.00012t}

. Taking the natural logarithm:

\ln(0.25) = -0.00012t

, so

t = \frac{\ln(0.25)}{-0.00012}

. Since

\ln(0.25) = \ln(\frac{1}{4}) = -\ln(4)

, we get

t = \frac{-\ln(4)}{-0.00012} = \frac{\ln(4)}{0.00012}

. This matches choice B. Choice A has the wrong sign—it would give a negative age. Choice C omits the logarithm entirely, treating this like a linear rather than exponential relationship. Choice D uses the half-life formula incorrectly, multiplying 4 by carbon-14's half-life without proper logarithmic adjustment. Study tip: In exponential decay problems, always set up the ratio equation first, then use natural logarithms to solve for time. Watch for sign errors when rearranging.

Question 12

Let $y_1(t)$ be the solution to $\frac{dy}{dt} = -0.5y$ with $y(0)=100$ , and let $y_2(t)$ be the solution to $\frac{dy}{dt} = -0.2y$ with $y(0)=100$ . For $t>0$ , let $D(t) = y_2(t) - y_1(t)$ . At what time $t$ does $D(t)$ reach its maximum value?

$\frac{10}{7} \ln(\frac{5}{2})$
$0$
$\frac{10}{3} \ln(\frac{5}{2})$ (correct answer)
The function $D(t)$ has no maximum value for $t > 0$ .

Explanation: When you encounter differential equations of the form

\frac{dy}{dt} = -ky

, you're dealing with exponential decay. These have solutions of the form

y(t) = y_0 e^{-kt}

where

y_0

is the initial value. For this problem, you need to find the solutions first. Since both equations have

y(0) = 100

$y_1(t) = 100e^{-0.5t}$
$y_2(t) = 100e^{-0.2t}$

Therefore:

D(t) = y_2(t) - y_1(t) = 100e^{-0.2t} - 100e^{-0.5t} = 100(e^{-0.2t} - e^{-0.5t})

To find the maximum, set

D'(t) = 0

D'(t) = 100(-0.2e^{-0.2t} + 0.5e^{-0.5t}) = 0

This gives us:

0.5e^{-0.5t} = 0.2e^{-0.2t}

Dividing both sides by

0.2e^{-0.5t}

\frac{0.5}{0.2} = e^{-0.2t + 0.5t} = e^{0.3t}

2.5 = e^{0.3t}

, which means

\ln(2.5) = 0.3t

Therefore:

t = \frac{\ln(2.5)}{0.3} = \frac{\ln(5/2)}{3/10} = \frac{10}{3}\ln\left(\frac{5}{2}\right)

Answer choice A uses

\frac{10}{7}

instead of

\frac{10}{3}

in the denominator—a calculation error. Choice B suggests the maximum occurs at

t = 0

, but

D(0) = 0

, which is clearly not the maximum since

D(t) > 0

for small positive

t

. Choice D incorrectly assumes no maximum exists, but since both exponentials decay to zero,

D(t)

must have a maximum. Study tip: For difference functions involving exponential decay, always check the derivative and remember that different decay rates will create a maximum at some finite time.

Question 13

The concentration $C(t)$ of a drug in the bloodstream decays exponentially according to $dC/dt = kC$ . The initial concentration is $C_0$ . After 2 hours, the concentration is $\frac{C_0}{e}$ . At what time $t$ is the instantaneous rate of decay, $-dC/dt$ , equal to half of the initial rate of decay?

$2e$ hours
$4$ hours
$\ln(2)$ hours
$2\ln(2)$ hours (correct answer)

Explanation: When you encounter exponential decay problems, you're dealing with differential equations where the rate of change is proportional to the current amount. Here,

\frac{dC}{dt} = kC

tells us the concentration follows exponential decay. First, solve for the general solution:

C(t) = C_0 e^{kt}

. Using the given condition that

C(2) = \frac{C_0}{e}

, we get

\frac{C_0}{e} = C_0 e^{2k}

, which gives us

e^{2k} = \frac{1}{e}

, so

k = -\frac{1}{2}

. Therefore,

C(t) = C_0 e^{-t/2}

. The instantaneous rate of decay is

-\frac{dC}{dt} = -kC = \frac{1}{2}C_0 e^{-t/2}

. The initial rate of decay (at

t = 0

) is

\frac{1}{2}C_0

. We need to find when the rate equals half of this initial rate:

\frac{1}{2}C_0 e^{-t/2} = \frac{1}{2} \cdot \frac{1}{2}C_0 = \frac{1}{4}C_0

. Solving:

e^{-t/2} = \frac{1}{2}

, so

-\frac{t}{2} = \ln(\frac{1}{2}) = -\ln(2)

, giving us

t = 2\ln(2)

. Choice A (

2e

hours) confuses the exponential base with the natural logarithm. Choice B (4 hours) likely comes from doubling the given 2-hour timeframe without proper calculation. Choice C (

\ln(2)

hours) misses the factor of 2 that comes from our decay constant

k = -\frac{1}{2}

. Remember: in exponential decay problems, always establish your decay constant first using given conditions, then carefully set up equations for rates rather than concentrations when the question asks about rates of change.

Question 14

The value of an investment, $V(t)$ , grows according to $\frac{dV}{dt} = kV$ . The investment triples in 15 years. What is the instantaneous rate of growth of the investment, in dollars per year, at the exact moment its value is $9000?

$600 \ln(3)$ (correct answer)
$9000 \ln(3)$
$27000 \ln(3)$
$600$

Explanation: First, find the constant

k

. We are given that the investment triples in 15 years. So,

3V_0 = V_0 e^{k \cdot 15}

, which implies

3 = e^{15k}

. Taking the natural logarithm gives

\ln(3) = 15k

, so

k = \frac{\ln(3)}{15}

. The instantaneous rate of growth is given by

\frac{dV}{dt} = kV

. We want to find this rate when

V = 9000

. Substituting the values of

k

and

V

gives

\frac{dV}{dt} = \left(\frac{\ln(3)}{15}\right) (9000) = 600 \ln(3)

. The initial value of the investment is not needed to solve the problem.

Question 15

The relationship between the atmospheric pressure $P$ and the altitude $h$ is modeled by the differential equation $\frac{dP}{dh} = -kP$ , where $k$ is a positive constant. If the pressure at sea level ( $h=0$ ) is $P_0$ and the pressure at altitude $H$ is $0.5 P_0$ , what is the pressure at altitude $3H$ ?

$0.25 P_0$
$0.125 P_0$ (correct answer)
$P_0 e^{-3kH}$
The pressure cannot be determined without knowing $k$ .

Explanation: When you encounter a differential equation like

\frac{dP}{dh} = -kP

, you're dealing with exponential decay. This separable equation tells us that pressure decreases at a rate proportional to the current pressure as altitude increases. To solve this, separate variables:

\frac{dP}{P} = -k \, dh

. Integrating both sides gives

\ln P = -kh + C

, so

P = Ae^{-kh}

where

A

is a constant. Using the initial condition

P(0) = P_0

, we find

A = P_0

, giving us

P(h) = P_0 e^{-kh}

. Now here's the key insight: at altitude

H

, we know

P(H) = 0.5P_0

. This means

P_0 e^{-kH} = 0.5P_0

, so

e^{-kH} = 0.5

. At altitude

3H

P(3H) = P_0 e^{-k(3H)} = P_0 (e^{-kH})^3 = P_0 (0.5)^3 = 0.125P_0

. Answer A (

0.25P_0

) represents what you'd get if you mistakenly thought the relationship was linear rather than exponential. Answer C (

P_0 e^{-3kH}

) is technically correct but unnecessarily complicated since we can determine the numerical value. Answer D is wrong because we can find the pressure without explicitly knowing

k

- the given condition at altitude

H

provides all the information we need. Study tip: In exponential decay problems, look for ways to use given conditions to eliminate unknown constants. The key is recognizing that

(e^{-kH})^3 = (0.5)^3

when you know

e^{-kH} = 0.5

Question 16

A radioactive substance decays according to the differential equation $\frac{dM}{dt} = kM$ , where $M$ is the mass in grams and $t$ is the time in years. The mass of the substance is measured to be $M_1$ at time $t_1$ and $M_2$ at time $t_2 = t_1 + \Delta t$ . What is the half-life of the substance?

$\frac{\Delta t \ln(2)}{\ln(M_1) - \ln(M_2)}$ (correct answer)
$\frac{\Delta t \ln(2)}{\ln(M_2) - \ln(M_1)}$
$\frac{\ln(M_1/M_2)}{\Delta t \ln(2)}$
$\Delta t \frac{M_1}{M_2} \ln(2)$

Explanation: The solution to the differential equation is

M(t) = M_0 e^{kt}

. We have

M_1 = M(t_1)

and

M_2 = M(t_1 + \Delta t)

. Dividing the two equations gives

\frac{M_2}{M_1} = \frac{M_0 e^{k(t_1 + \Delta t)}}{M_0 e^{kt_1}} = e^{k \Delta t}

. Solving for

k

yields

k = \frac{1}{\Delta t}\ln(\frac{M_2}{M_1})

. The half-life

T_{1/2}

is the time it takes for the substance to decay to half its amount, which is given by the formula

T_{1/2} = \frac{\ln(1/2)}{k} = \frac{-\ln(2)}{k}

. Substituting our expression for

k

, we get

T_{1/2} = \frac{-\ln(2)}{\frac{1}{\Delta t}\ln(M_2/M_1)} = \frac{-\Delta t \ln(2)}{\ln(M_2) - \ln(M_1)} = \frac{\Delta t \ln(2)}{\ln(M_1) - \ln(M_2)}

Question 17

The rate of change of a quantity $y$ is directly proportional to $y$ . If $y$ decreases by 20% in 4 years, what is the constant of proportionality $k$ ?

$\frac{\ln(0.8)}{4}$ (correct answer)
$\frac{\ln(0.2)}{4}$
$-0.05$
$4\ln(0.8)$

Explanation: The model is

y(t) = y_0 e^{kt}

. A decrease of 20% means that the value at

t=4

is 80% of the initial value. So,

y(4) = 0.8 y_0

. Substituting this into the model equation gives

0.8 y_0 = y_0 e^{k \cdot 4}

. Dividing by

y_0

gives

0.8 = e^{4k}

. To solve for

k

, we take the natural logarithm of both sides:

\ln(0.8) = 4k

. Therefore,

k = \frac{\ln(0.8)}{4}

Question 18

The population of a species is governed by $dP/dt = kP$ . The time it takes for the population to grow from 100 to 200 is $T$ . How long does it take for the population to grow from 200 to 800?

$6T$
$3T$
$4T$
$2T$ (correct answer)

Explanation: When you encounter a differential equation like

\frac{dP}{dt} = kP

, you're dealing with exponential growth. This equation tells you that the rate of population change is proportional to the current population, which produces the classic exponential pattern where populations double, triple, etc. in consistent time intervals. To solve this, start with the general solution:

P(t) = P_0 e^{kt}

, where

P_0

is the initial population. Since the population grows from 100 to 200 in time

T

, you can write:

200 = 100e^{kT}

, which simplifies to

2 = e^{kT}

. Now for the key insight: you want to find how long it takes to go from 200 to 800. Notice that

800 = 200 \times 4 = 200 \times 2^2

. Since the population doubles every

T

time units (from

2 = e^{kT}

), going from 200 to 800 requires exactly 2 doublings: 200 → 400 → 800. Therefore, this takes

2T

. Choice A (

6T

) would represent 6 doublings, reaching

200 \times 2^6 = 12,800

. Choice B (

3T

) represents 3 doublings, reaching

200 \times 2^3 = 1,600

. Choice C (

4T

) represents 4 doublings, reaching

200 \times 2^4 = 3,200

. All of these overshoot the target of 800. Study tip: In exponential growth problems, always identify the "doubling time" first. Then count how many doublings you need to reach your target. This approach is much faster than working with the exponential formula directly.

Question 19

The rate of decay of a radioactive isotope is proportional to the amount present. If the half-life of the isotope is 100 years, what fraction of the original amount remains after 50 years?

$\frac{3}{4}$
$\frac{1}{2}$
$\frac{1}{\sqrt{2}}$ (correct answer)
$\frac{1}{4}$

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 creates the differential equation

\frac{dN}{dt} = -kN

, which has the solution

N(t) = N_0 e^{-kt}

, where

N_0

is the initial amount. The key insight is using the half-life to find the decay constant. Since half the material remains after 100 years:

\frac{1}{2} = e^{-100k}

. Taking the natural logarithm:

k = \frac{\ln(2)}{100}

. Now you can find the amount remaining after 50 years:

N(50) = N_0 e^{-k \cdot 50} = N_0 e^{-\frac{\ln(2)}{100} \cdot 50} = N_0 e^{-\frac{\ln(2)}{2}}

Since

e^{-\frac{\ln(2)}{2}} = (e^{\ln(2)})^{-1/2} = 2^{-1/2} = \frac{1}{\sqrt{2}}

, the fraction remaining is

\frac{1}{\sqrt{2}}

, making C correct. Looking at the wrong answers: A)

\frac{3}{4}

might tempt you if you incorrectly assume linear decay rather than exponential. B)

\frac{1}{2}

is the amount remaining at the half-life (100 years), not at 50 years. D)

\frac{1}{4}

would be correct at 200 years (two half-lives), not 50 years. Study tip: Remember that exponential decay means equal time periods correspond to equal fractional changes, not equal absolute changes. After one half-life you have

\frac{1}{2}

, after half a half-life you have

\frac{1}{\sqrt{2}}

, and the pattern continues exponentially.

Question 20

A solution $y=f(t)$ to $y'=ky$ satisfies $f(1)=3$ and $f(2)=12$ . What is the value of $f(0)$ ?

$3/2$
$3/4$ (correct answer)
$1/2$
$-6$

Explanation: When you encounter a differential equation of the form

y' = ky

, you're dealing with exponential growth or decay. The general solution is always

y = Ce^{kt}

, where

C

and

k

are constants determined by the given conditions. To find these constants, use your two data points:

f(1) = 3

and

f(2) = 12

. Substituting into

y = Ce^{kt}

3 = Ce^{k \cdot 1} = Ce^k

12 = Ce^{k \cdot 2} = Ce^{2k}

Divide the second equation by the first to eliminate

C

\frac{12}{3} = \frac{Ce^{2k}}{Ce^k} = e^k

e^k = 4

, which means

k = \ln(4)

. Now find

C

using the first equation:

3 = Ce^k = C \cdot 4

, so

C = \frac{3}{4}

. Therefore,

f(t) = \frac{3}{4}e^{t\ln(4)} = \frac{3}{4} \cdot 4^t

. At

t = 0

f(0) = \frac{3}{4} \cdot 4^0 = \frac{3}{4} \cdot 1 = \frac{3}{4}

. The answer is B. Choice A (

\frac{3}{2}

) likely comes from incorrectly assuming

C = \frac{3}{2}

without properly using both conditions. Choice C (

\frac{1}{2}

) might result from computational errors in finding

C

. Choice D (

-6

) could stem from sign errors or completely misapplying the exponential formula. Study tip: For exponential differential equations, always use the ratio method to find

k

first—it eliminates

C

and simplifies your work. Then substitute back to find

C