Integration: Solving Separable Differential Equations - Population Growth

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Needs to be tested

2 years, 9 months ago

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Christian Lawson-Perfect 2 years, 7 months ago

Published this.

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Evi Papadaki 2 years, 9 months ago

Gave some feedback: Needs to be tested

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Evi Papadaki 2 years, 9 months ago

Created this as a copy of Trigonometry: Introduction to Wave Functions - Drug Concentration.

Name	Status	Author	Last Modified
Trigonometry: Introduction to Wave Functions - Drug Concentration	Needs to be tested	Evi Papadaki	12/04/2024 19:28
Integration: Solving Separable Differential Equations - Population Growth	Needs to be tested	Evi Papadaki	20/12/2022 18:55
Trigonometry: Introduction to Wave Functions - Electromagnetic Waves	Needs to be tested	Evi Papadaki	20/12/2022 19:15

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Question statement

A population of $N$ individuals grows over time, $t$ years, according to the equation:
$\frac{\mathrm{d}N}{\mathrm{d}t}=\left(\var{a}+ e^{\simplify{-{b}t}}\right)N $

At time $t=0$, the population contains $\var{N}$ individuals.

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	1
b	integer	2
n	integer	380
t0	integer	6
ans	integer	252753
C	number	6.4401712527
step1	number	0.0000030721
step2	number	12.4401681806
step3	number	252753.0440714767

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How many individuals are in the population $\var{t0}$ years later?

The population is Unnamed gap individuals.

If needed, round your answer to the nearest integer.

Advice

To calculate the population $\var{N}$ years later, we need an expression for how the population, $N$, changes in terms of time, $t$. This is given by the differential equation,

\[ \frac{dN}{dt}=\left(\var{a}+ e^{\simplify{-{b}t}}\right)N \]

which we can solve using separation of variables.

First, we need to separate the variables between the two sides:

\[ \frac{1}{N}dN=\left( \var{a}+e^{\simplify{-{b}t}}\right)dt \]

Now, we integrate both sides:

\[ \begin{split} \int\frac{1}{N}dN &=\int\left( \var{a}+e^{\simplify{-{b}t}}\right)dt \\ \ln{N} &= \simplify{{a}*t}+\left(-\simplify{1/{b}*e}^{\simplify{-{b}t}}\right)+c \\ \ln{N} &= \simplify{{a}*t}-\simplify{1/{b}*e}^{\simplify{-{b}t}}+c \end{split} \]

To find the value of $c$, we can use the fact that when $t=0$, $N=\var{N}$. By substituting in the values in the equation we get:

\[ \begin{split} \ln(\var{N}) &= \var{a} \times 0-\simplify{1/{b}*e}^{-\simplify{{b}* 0}}+c \\ \ln(\var{N}) &= -\simplify{1/{b}*e}^{0}+c \\ \ln(\var{N}) &= -\simplify{1/{b}* 1}+c \\ \ln(\var{N}) &= -\simplify{1/{b}}+c \\ c&=\simplify{ ln{{N}}+1/{b}}\end{split} \]

Therefore $c=\simplify{ln{{N}}+1/{b}}=\var{c}$. So,

\[ \begin{split} \ln({N}) &= \simplify{{a}*t}-\simplify{1/{b}*e}^{-\simplify{{b}*t}}+\simplify{ln{{N}}+1/{b}} \end{split} \]

Taking the exponential of both sides:

\[ \begin{split} N &= e^{\simplify{{a}*t}-\simplify{1/{b}e}^{-\simplify{{b}*t}}+\simplify{ln{{N}}+1/{b}}} \end{split} \]

We can now substitute $t=\var{t0}$ to calculate the population:

\[ \begin{split} N &= e^{\var{a} \times \var{t0}-\simplify{1/{b}*e}^{-\var{b} \times \var{t0}}+\simplify{ln{{N}}+1/{b}}} \\ N&=\var{step3} \end{split} \]

Thus, the population $\var{t0}$ years later will be $\var{ans}$ individuals.

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This question is used in the following exams:

Further Integration by Evi Papadaki in MASH Bath: Moodle quizzes and TS.
Natural Sciences by Evi Papadaki in Evi's workspace.
Natural Sciences by Evi Papadaki in Evi's workspace.

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Integration: Solving Separable Differential Equations - Population Growth

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Christian Lawson-Perfect 2 years, 7 months ago

Evi Papadaki 2 years, 9 months ago

Evi Papadaki 2 years, 9 months ago

Error in variable testing condition

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Generated value: `integer`

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Integration: Solving Separable Differential Equations - Population Growth

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Christian Lawson-Perfect 2 years, 7 months ago

Evi Papadaki 2 years, 9 months ago

Evi Papadaki 2 years, 9 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`