# Exponential Growth and Decay Study Guide Problems

## Problem 1

This question tests your understanding of the wording of questions.

Recall the two equations for exponential growth and decay.

$N = N_O \cdot e^{k\cdot t}$ or $ln \Bigl(\frac {N}{N_o} \Bigr) = k \cdot t$

(a) If the population increases by a factor of 1.6, then which of the following is true?

A. $N = 2.6 N_o$
B. $N = N_o/2.6$
C. $N = 1.16 N_o$
D. $N = 1.6 N_o$
E. $N = N_o /1.6$

(b) If the population decreases by a factor of 3 every 2 days, which of the following is true?

A. $3 = e^{-2k}$
B.  $ln (1/3) = -2 k$
C.  $2/3 = e^{-2k}$
D.  $ln (3/2) = -3 k$
E.  $ln (1/3) = -3 k$

(c) If the population decreases by 33.3% every 2 days, which of the following is true?

A. $3 = e^{-2k}$
B.  $ln (1/3) = -2 k$
C.  $2/3 = e^{-2k}$
D.  $ln (3/2) = -3 k$
E.  $ln (1/3) = -3 k$

(a)

A. $N = 2.6 N_o$ - No. If you are having trouble with this type of problem, review the section on exponential growth.
B. $N = N_o/2.6$ - No. If you are having trouble with this type of problem, review the section on exponential growth.
C. $N = 1.16 N_o$ - No. If you are having trouble with this type of problem, review the section on exponential growth.
D. $N = 1.6 N_o$ - Correct!
E. $N = N_o /1.6$ - No. If you are having trouble with this type of problem, review the section on exponential growth.

(b)

A. $3 = e^{-2k}$ - No. If there is a decrease by a factor of 3, then after 2 days (when t=2) there will only be 1/3 of the initial number remaining.
B.  $ln (1/3) = -2 k$ - Correct!
C.  $2/3 = e^{-2k}$ - No. If there is a decrease by a factor of 3, then after 2 days (when t=2) there will only be 1/3 of the initial number remaining.
D.  $ln (3/2) = -3 k$ - No. If there is a decrease by a factor of 3, then after 2 days (when t=2) there will only be 1/3 of the initial number remaining.
E.  $ln (1/3) = -3 k$ - No. If there is a decrease by a factor of 3, then after 2 days (when t=2) there will only be 1/3 of the initial number remaining.

(c)

A. $3 = e^{-2k}$ - No. If 33.3% have been lost, then 1/3 of the original have been lost. There must be 2/3 remaining when t = 2 (ie. after 2 days).
B.  $ln (1/3) = -2 k$ - No. If 33.3% have been lost, then 1/3 of the original have been lost. There must be 2/3 remaining when t = 2 (ie. after 2 days).
C.  $2/3 = e^{-2k}$ - Correct!
D.  $ln (3/2) = -3 k$ - No. If 33.3% have been lost, then 1/3 of the original have been lost. There must be 2/3 remaining when t = 2 (ie. after 2 days).
E.  $ln (1/3) = -3 k$ - No. If 33.3% have been lost, then 1/3 of the original have been lost. There must be 2/3 remaining when t = 2 (ie. after 2 days).

## Problem 2

(a) A wolf population increases with a growth constant of $0.278 \;\mathrm {year}^{-1}$. After $5.00$ years, by what factor has the population increased?

Don't be concerned because you do not know the initial population. It is only the ratio of final to initial populations (ie. $N/N_o$) that really matters. If this ratio were 7, then it would mean that be population had increased by a factor of 7.
From this information you should be able to calculate $N/N_o$.

#### Do the calculation before going on to the solution.

$\frac {N}{N_o} = e ^{(0.278)\cdot(5)}$

So the population has increased by a factor of $4.01$.

(b) By what percentage has it increased?

From part (a), we know that for every animal at the beginning, there will be $4.01$ animals after 5 years. This gives us an increase of $3.01$.

$\frac {N_\mathrm{final}-N_\mathrm {initial}} {N_\mathrm{initial}}\cdot (100) = \% \mathrm{change}$

This gives us $(3.01/1)(100) = 301\%$

So, the wolf population had a $301\%$ increase.

## Problem 3

A deer population is declining with a decay constant of magnitude $0.327 \mathrm {year}^{-1}$. After $6.00$ years,

(a) By what factor has the population declined?

Here, the ratio of final population to inital population will be less than one $(1)$. If, for instance, it were $0.2$, then the population would have decreased by a factor of $5$ (because $0.2 = 1/5$). That is, for every $5$ original animals, we end up with only one.

$\frac {N}{N_o} = 0.2$

And the reciprocal of this, $(0.2)^{-1}$, gives the factor $5$.

#### Now try the question, using this information before going on to the solution.

You should have:

$\frac {N}{No} = e^{(-0.327)\cdot(6)} = 0.141$

The factor then, will be the reciprocal of this, that is, $(0.141)^{-1} = 7.11$

(b) by what percentage has it declined?

If you start with $100$ animals, then after 6 years, there would be $100/7.11 = 14.1$ deer. Round this off to $14$, and calculate the percentage decline in the population.

#### Try it before going to the solution.

Remember, you want to use the equation:

$\frac {N_\mathrm{final}-N_\mathrm {initial}} {N_\mathrm{initial}}\cdot (100) = \% \mathrm{change}$

So, $(86/100)(100) = 86\%$

## Problem 4

A cell culture has a population of $1.00 \times 10^3$; culture B has a population of $1.50 \times 10^3$. Culture A has a growth constant of $0.0500 \; \mathrm {day}^{-1}$. Culture B doubles in size every $3.00$ days. How long will it be until culture B has $\pi$ times as many cells as culture A?

1. One of the most common mistakes in this problem lies in getting the information in the final sentence backwards. Write down the equation that is implied in the last sentence.

You should get: $NB = \pi$  $NA$, where '$NB$' means "the number in culture B", and '$NA$' is "the number in culture A".

2. The next step is to find the growth constant for culture B. It is

A. $0.0500 \; day ^{-1}$
B. $0.231 \; day ^{-1}$
C. $0.667 \; day ^{-1}$
D. $1.50 \; day ^{-1}$
E. $3.00 \; day ^{-1}$

The time when $NB = \pi$  $NA$, is:

A. 1 day
B. 2.3 days
C. 4.1 days
D. 5.7 days

$N_A = N_A e^{k(A)t}$

$N_A = 1000e^{0.05t}$

$1.500e^{0.231t} = \pi e^{0.05t}$

$t = 4.1 \; \mathrm {days}$

$N_B = N_B e ^{k(B)t}$

$N_B = 1500e^{0.231t} \mathrm{because} \; k(B) = 0.693/3 = 0.231 \; \mathrm {day}^{-1}$

## Problem 5

The number of walruses in a certain bay is increasing exponentially. On April 1, 1979, the population was 323. On April 16, 1978, the population was 298.

(a) How many months ago (before April 1, 1979) was the population 200? (Assume that 1 month = 30.0 days.)

(b) What would the population have been on August 16, 1974?

1. In this problem, you will make the easiest progress if you work in months. Thus the population increased from 298 to 323 in 11.5 months from April 16, 1978 to April 1, 1979. With this information, you should be able to calculate the growth constant, 'k'.

2. The growth constant is:
A. $-0.0070 \; \mathrm {month} ^{-1}$
B. $0.0070 \; \mathrm {month} ^{-1}$
C. $-0.010 \; \mathrm {month} ^{-1}$
D. $0.010 \; \mathrm {month} ^{-1}$

In part (a) the answer is
A. 8.42 mo
B. 9.59 mo
C. 10.36 mo
D. 12.01 mo

A. $-0.0070 \; \mathrm {month} ^{-1}$ - No. k is defined as a growth constant and is always positive.
B. $0.0070 \; \mathrm {month} ^{-1}$ - Correct!
C. $-0.010 \; \mathrm {month} ^{-1}$ - No. k is defined as a growth constant and is always positive.
D. $0.010 \; \mathrm {month} ^{-1}$ - No. The numerical value is incorrect.

In part (a) the answer is
A. 8.42 mo - No. The numerical value is incorrect.
B. 9.59 mo - Correct!
C. 10.36 mo - No. The numerical value is incorrect.
D. 12.01 mo - No. The numerical value is incorrect.

The time between the 2 dates is 11.5 mo

$323 = 298e^{k11.5}$

$1.08389 = e^{k11.5}$

$k = 7.0 \times 10^{-3} \;mo^{-1}$

$200 = 298e^{7.0 \times (10-3)t}$

$t = 9.59$ mo earlier

Aug. 16 1978 is 44 months earlier

$298 = N_0e^{7 \times(10-3) \times 44}$

$298 = N_0 (1.36)$

$N_0 = 219$

## Question 6

The physical half-life of an isotope is 15 days. What is the corresponding physical decay constant in $\mathrm {day}^{-1}$ and in $s^{-1}$

We will use $T_p$ to represent the physical half-life, and $\lambda _p$ to represent the physical decay constant.

#### Try the question before going to the solution.

The appropriate equation  is:

$\lambda_p \cdot T_p = 0.693$

Since we know the half life, $T_p$, then we can solve for the decay constant. You should find that $\lambda _p = 0.046 \; \mathrm {day}^{-1}$.

To get the answer in terms of $s^{-1}$, we need to convert 15 days to seconds, and repeat the above calculation. You should get $\lambda _p = 5.34 \times 10^{-7} s^{-1}$

## Problem 7

The values of $\lambda _p$ and $\lambda _b$ for a particular radioactive isotope in humans is $0.080 \; \mathrm {day}^{-1}$ and $0.050 \; \mathrm {day}^{-1}$ respectively. What is the effective half life of this isotope?

This problem can be done in different ways, but the following solution is the simplest to calculate. First we will find the effective decay constant, $\lambda _{eff}$, and then use this to caculate $T_{eff}$.

The effective decay constant, $\lambda _{eff}$, is:

A. $0.0040 \; \mathrm {day}^{-1}$
B. $0.031 \; \mathrm {day}^{-1}$
C. $0.067 \; \mathrm {day}^{-1}$
D. $0.13 \; \mathrm {day}^{-1}$
E. $3.25 \; \mathrm {day}^{-1}$

A. $0.0040 \; \mathrm {day}^{-1}$ - No. The relevant relation is $\lambda _e = \lambda b + \lambda p$
B. $0.031 \; \mathrm {day}^{-1}$ -  No. The relevant relation is $\lambda _e = \lambda b + \lambda p$
C. $0.067 \; \mathrm {day}^{-1}$ -  No. The relevant relation is $\lambda _e = \lambda b + \lambda p$
D. $0.13 \; \mathrm {day}^{-1}$ - Correct!
E. $3.25 \; \mathrm {day}^{-1}$ -  No. The relevant relation is $\lambda _e = \lambda b + \lambda p$

The effective half life is therefore:
A. 173 day
B. 22.day
C. 10.3 day
D. 5.3 day
E. 0.21 day

A. 173 day - No. The relevant relation is: $\lambda_e T_e = ln\; 2 = 0.693$
B. 22.day -  No. The relevant relation is: $\lambda_e T_e = ln\; 2 = 0.693$
C. 10.3 day -  No. The relevant relation is: $\lambda_e T_e = ln\; 2 = 0.693$
D. 5.3 day - Correct!
E. 0.21 day -  No. The relevant relation is: $\lambda_e T_e = ln\; 2 = 0.693$

## Problem 8

Two goats at Port Hopeless, Ontario accidentally get into a field containing radioactive grass. The goat 'Aristophanes' eats twice as much grass as the goat 'Demetrius'. However, Aristophanes metabolism is $1.5$ times faster than that of Demetrius, and the ingested isotope in this case has a biological decay constant which is directly proportional to metabolism. If the biological decay constant is $0.10 \; \mathrm {day}^{-1}$ in Demetrius, how long will it be until the two goats contain the same concentration of isotope? Assume that the physical half-life is very large compared to the biological half-lives.

This problem looks pretty bad at first, but if you procede carefully, you will find it less difficult than it appears.

Consider the last sentence. If $T_p >> T_b$, then we know

$\lambda_p << \lambda_b$

This implies that $\lambda_p$ is so small, we can ignore it. Consequently, the effective and biological decay constants will be equal for any one goat.

The time is:

A. 0.53 day
B. 1.46 day
C. 19 day
D. 27 day
E. 146 day

A. 0.53 day - No.
The relevant conditions are
$N_{0A} = 2N_{0D}$
$\lambda _A = 1.5\lambda _D$
Required is the time when
$N_A = N_D$

B. 1.46 day - No.
The relevant conditions are
$N_{0A} = 2N_{0D}$
$\lambda _A = 1.5\lambda _D$
Required is the time when
$N_A = N_D$

C. 19 day - Correct!

D. 27 day - No.
The relevant conditions are
$N_{0A} = 2N_{0D}$
$\lambda _A = 1.5\lambda _D$
Required is the time when
$N_A = N_D$

E. 146 day - No.
The relevant conditions are
$N_{0A} = 2N_{0D}$
$\lambda _A = 1.5\lambda _D$
Required is the time when
$N_A = N_D$

## Question 9

Three bats, 'Bewitched', 'Bothered', and 'Bemildred', each drink one cup of radioactive coffee, leaving exactly one cup of coffee in the pot. Forty (40.0) minutes later, two of the bats are eaten by 'Howland Owl'. Six (6.00) hours after this meal, the owl has a total body count of 200 disintegrations per minute, which, by coincidence, is exactly the same count rate as the cold coffee at that time. The effective half-life for the radioactive isotope is 4.00 hours in an owl, and 1 hour and 20 minutes in a bat.

(a) What is the initial amount ingested by the owl?
(b) What was the initial amount in the pot?
(c) What is the physical half life of the isotope?
(d) What is the biological half-life for this isotope in a bat?

This is a difficult problem and if you can do it without looking at the solution you are well prepared.

There are 3 timelines to be followed and described mathematically:

1. The one quarter portion of coffee in the pot decays physically for 400 min.
2. The bats contribute to the decay of 2 portions for 40 min.
3. What is left in the bats becomes the initial amount for the owl where it decreases for 360 min.

If you write down the correct equation for each of these timelines you will have the answers. Of the 3 equations only one will have one unknown-clearly you must start the numerical part of the solution with that one.

There are three different time periods. You will have to set up an equation to describe what happens in each of them. First, write down an equation to describe the 400 minute decay of the cup of coffee left in the pot.

You should have come up with:

[1] $200 = N_o \cdot e ^{(-\lambda_p)\cdot 400}$

Where $N_0$ is one quarter of the initial amount in the pot. It is $\lambda_p$ because there are no biological eliminations for the pot i.e.$\lambda_b = 0$

Next, you should be able to get an equation to describe the 360 minute time the isotope was in the owl. The effective half-life in the owl is 4.00 hours, or 240 minutes. Therefore the effective half-life in the owl will be

$\lambda_{eff} = 0.693/240\; \mathrm {min}^{-1}$

and

[2] $200 = N_0 ^1 \cdot e^{- \Bigl( \frac {0.693}{240} \Bigr) \cdot 360}$

Where $N'\;_0$ is the amount in both bats at the moment they are eaten. Therefore $0.5N'\; _0$ is the final amount in the bat.
Next, you need an equation for the 40 minutes after the bats drank the coffee and before they are eaten.

[3] $N_0 \; ^1 = 2 \cdot e^{-\Bigl( \frac {0.693}{80} \Bigr)\cdot 40}$

From [2] $N'\;_0 = 566$ {Answer to (a)}

Using this in [3]

$N_0 = 400$

So the initial amount in the pot was 4X400 = 1600 counts per minute. {Answer to (b)}

Now using [1]

$\lambda_p = (0.693)/400 = 1.73\times10^{-3} \; \mathrm {min}^{-1}$ {Answer to (c)}

The fourth and final equation you will need relates the three decay constants to each other.

[4] $\lambda _{eff} = \lambda_p + \lambda_b$

$\lambda_{bB} = \lambda_e -\lambda_{pB} = (8.66 - 1.73)\times 10^{-3} \; min^{-1}$  {Answer to (d)}