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Exponential Growth and Decay

Use the exponential growth model in applications, including population growth and compound interest.

Exponential Growth Model

Many systems exhibit exponential growth. These systems follow a model of the form $y={y}_{0}{e}^{kt},$ where ${y}_{0}$ represents the initial state of the system and $k$ is a positive constant, called the growth constant. Notice that in an exponential growth model, we have

\[{y}^{'}=k{y}_{0}{e}^{kt}=ky.\]

That is, the rate of growth is proportional to the current function value. This is a key feature of exponential growth. involves derivatives and is called a differential equation. We learn more about differential equations in Introduction to Differential Equations.

Population growth is a common example of exponential growth. Consider a population of bacteria, for instance. It seems plausible that the rate of population growth would be proportional to the size of the population. After all, the more bacteria there are to reproduce, the faster the population grows. and represent the growth of a population of bacteria with an initial population of $200$ bacteria and a growth constant of $0.02.$ Notice that after only $2$ hours $(120$ minutes), the population is $10$ times its original size!

Time (min)	Population Size (no. of bacteria)
$10$	$244$
$20$	$298$
$30$	$364$
$40$	$445$
$50$	$544$
$60$	$664$
$70$	$811$
$80$	$991$
$90$	$1210$
$100$	$1478$
$110$	$1805$
$120$	$2205$

Note that we are using a continuous function to model what is inherently discrete behavior. At any given time, the real-world population contains a whole number of bacteria, although the model takes on noninteger values. When using exponential growth models, we must always be careful to interpret the function values in the context of the phenomenon we are modeling.

Example

Try it.

Consider the population of bacteria described earlier. This population grows according to the function $f(t)=200{e}^{0.02t},$ where t is measured in minutes. How many bacteria are present in the population after $5$ hours $(300$ minutes)? When does the population reach $100,000$ bacteria?

Solution

We have $f(t)=200{e}^{0.02t}.$ Then

\[f(300)=200{e}^{0.02(300)}\approx 80,686.\]

There are $80,686$ bacteria in the population after $5$ hours.

To find when the population reaches $100,000$ bacteria, we solve the equation

\[\begin{array}{lll}100,000 & = & 200{e}^{0.02t} \\ 500 & = & {e}^{0.02t} \\ \text{ln}\ 500 & = & 0.02t \\ t & = & \frac{\text{ln}\ 500}{0.02}\approx 310.73.\end{array}\]

The population reaches $100,000$ bacteria after $310.73$ minutes.

\[1000(1+0.02)=\text{\$}1020.\]\[\text{Balance}=P{e}^{rt}.\]

Condensed — the full section is in OpenStax Calculus Volume 2.

Exponential Decay Model

Exponential functions can also be used to model populations that shrink (from disease, for example), or chemical compounds that break down over time. We say that such systems exhibit exponential decay, rather than exponential growth. The model is nearly the same, except there is a negative sign in the exponent. Thus, for some positive constant $k,$ we have $y={y}_{0}{e}^{\text{-}kt}.$

As with exponential growth, there is a differential equation associated with exponential decay. We have

\[{y}^{'}=\text{-}k{y}_{0}{e}^{\text{-}kt}=\text{-}ky.\]

The following figure shows a graph of a representative exponential decay function.

Let’s look at a physical application of exponential decay. Newton’s law of cooling says that an object cools at a rate proportional to the difference between the temperature of the object and the temperature of the surroundings. In other words, if $T$ represents the temperature of the object and ${T}_{a}$ represents the ambient temperature in a room, then

\[{T}^{'}=\text{-}k(T-{T}_{a}).\]

Note that this is not quite the right model for exponential decay. We want the derivative to be proportional to the function, and this expression has the additional ${T}_{a}$ term. Fortunately, we can make a change of variables that resolves this issue. Let $y(t)=T(t)-{T}_{a}.$ Then ${y}^{'}(t)={T}^{'}(t)-0={T}^{'}(t),$ and our equation becomes

\[{y}^{'}=\text{-}ky.\]

From our previous work, we know this relationship between y and its derivative leads to exponential decay. Thus,

\[y={y}_{0}{e}^{\text{-}kt},\]

and we see that

\[\begin{array}{lll}T-{T}_{a} & = & ({T}_{0}-{T}_{a}){e}^{\text{-}kt} \\ T & = & ({T}_{0}-{T}_{a}){e}^{\text{-}kt}+{T}_{a}\end{array}\]\[\begin{array}{lll}\frac{{y}_{0}}{2} & = & {y}_{0}{e}^{\text{-}kt} \\ \frac{1}{2} & = & {e}^{\text{-}kt} \\ -\text{ln}\ 2 & = & \text{-}kt \\ t & = & \frac{\text{ln}\ 2}{k}.\end{array}\]

Condensed — the full section is in OpenStax Calculus Volume 2.

Key Concepts

Exponential growth and exponential decay are two of the most common applications of exponential functions.
Systems that exhibit exponential growth follow a model of the form $y={y}_{0}{e}^{kt}.$
In exponential growth, the rate of growth is proportional to the quantity present. In other words, ${y}^{'}=ky.$
Systems that exhibit exponential growth have a constant doubling time, which is given by $(\text{ln}\ 2)\text{/}k.$
Systems that exhibit exponential decay follow a model of the form $y={y}_{0}{e}^{\text{-}kt}.$
Systems that exhibit exponential decay have a constant half-life, which is given by $(\text{ln}\ 2)\text{/}k.$

Exponential Growth and Decay

True or False? If true, prove it. If false, find the true answer.

For the following exercises, use $y={y}_{0}{e}^{kt}.$

For the next set of exercises, use the following table, which features the world population by decade.

Years since 1950	Population (millions)
$0$	$2,556$
$10$	$3,039$
$20$	$3,706$
$30$	$4,453$
$40$	$5,279$
$50$	$6,083$
$60$	$6,849$

For the next set of exercises, use the following table, which shows the population of San Francisco during the 19th century.

Years since 1850	Population (thousands)
$0$	$21.00$
$10$	$56.80$
$20$	$149.5$
$30$	$234.0$

Exponential Growth Model

\[{y}^{'}=k{y}_{0}{e}^{kt}=ky.\]

Time (min)	Population Size (no. of bacteria)
$10$	$244$
$20$	$298$
$30$	$364$
$40$	$445$
$50$	$544$
$60$	$664$
$70$	$811$
$80$	$991$
$90$	$1210$
$100$	$1478$
$110$	$1805$
$120$	$2205$

Example

Try it.

Solution

We have $f(t)=200{e}^{0.02t}.$ Then

\[f(300)=200{e}^{0.02(300)}\approx 80,686.\]

There are $80,686$ bacteria in the population after $5$ hours.

To find when the population reaches $100,000$ bacteria, we solve the equation

\[\begin{array}{lll}100,000 & = & 200{e}^{0.02t} \\ 500 & = & {e}^{0.02t} \\ \text{ln}\ 500 & = & 0.02t \\ t & = & \frac{\text{ln}\ 500}{0.02}\approx 310.73.\end{array}\]

The population reaches $100,000$ bacteria after $310.73$ minutes.

\[1000(1+0.02)=\text{\$}1020.\]\[\text{Balance}=P{e}^{rt}.\]

Condensed — the full section is in OpenStax Calculus Volume 1.

Exponential Decay Model

As with exponential growth, there is a differential equation associated with exponential decay. We have

\[{y}^{'}=\text{-}k{y}_{0}{e}^{\text{-}kt}=\text{-}ky.\]

The following figure shows a graph of a representative exponential decay function.

\[{T}^{'}=\text{-}k(T-{T}_{a}).\]

\[{y}^{'}=\text{-}ky.\]

From our previous work, we know this relationship between y and its derivative leads to exponential decay. Thus,

\[y={y}_{0}{e}^{\text{-}kt},\]

and we see that

Condensed — the full section is in OpenStax Calculus Volume 1.

Key Concepts

Exponential growth and exponential decay are two of the most common applications of exponential functions.
Systems that exhibit exponential growth follow a model of the form $y={y}_{0}{e}^{kt}.$
In exponential growth, the rate of growth is proportional to the quantity present. In other words, ${y}^{'}=ky.$
Systems that exhibit exponential growth have a constant doubling time, which is given by $(\text{ln}\ 2)\text{/}k.$
Systems that exhibit exponential decay follow a model of the form $y={y}_{0}{e}^{\text{-}kt}.$
Systems that exhibit exponential decay have a constant half-life, which is given by $(\text{ln}\ 2)\text{/}k.$

Exponential Growth and Decay

True or False? If true, prove it. If false, find the true answer.

For the following exercises, use $y={y}_{0}{e}^{kt}.$

For the next set of exercises, use the following table, which features the world population by decade.

Years since 1950	Population (millions)
$0$	$2,556$
$10$	$3,039$
$20$	$3,706$
$30$	$4,453$
$40$	$5,279$
$50$	$6,083$
$60$	$6,849$

For the next set of exercises, use the following table, which shows the population of San Francisco during the 19th century.

Years since 1850	Population (thousands)
$0$	$21.00$
$10$	$56.80$
$20$	$149.5$
$30$	$234.0$

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Consider the population of bacteria described earlier. This population grows according to the function $f(t)=200{e}^{0.02t},$ where t is measured in minutes. How many bacteria are present in the population after $5$ hours $(300$ minutes)? When does the population reach $100,000$ bacteria?

答えを明らかにしろ

We have $f(t)=200{e}^{0.02t}.$ Then
\[f(300)=200{e}^{0.02(300)}\approx 80,686.\]
There are $80,686$ bacteria in the population after $5$ hours.
To find when the population reaches $100,000$ bacteria, we solve the equation
\[\begin{array}{lll}100,000 & = & 200{e}^{0.02t} \\ 500 & = & {e}^{0.02t} \\ \text{ln}\ 500 & = & 0.02t \\ t & = & \frac{\text{ln}\ 500}{0.02}\approx 310.73.\end{array}\]
The population reaches $100,000$ bacteria after $310.73$ minutes.
Consider a population of bacteria that grows according to the function $f(t)=500{e}^{0.05t},$ where $t$ is measured in minutes. How many bacteria are present in the population after 4 hours? When does the population reach $100$ million bacteria?

答えを明らかにしろ

There are $81,377,396$ bacteria in the population after $4$ hours. The population reaches $100$ million bacteria after $244.12$ minutes.
A 25-year-old student is offered an opportunity to invest some money in a retirement account that pays $5\text{\%}$ annual interest compounded continuously. How much does the student need to invest today to have $\text{\$}1$ million when she retires at age $65?$ What if she could earn $6\text{\%}$ annual interest compounded continuously instead?

答えを明らかにしろ

We have
\[\begin{array}{lll}1,000,000 & = & P{e}^{0.05(40)} \\ P & = & 135,335.28.\end{array}\]
She must invest $\text{\$}135,335.28$ at $5\text{\%}$ interest.
If, instead, she is able to earn $6\text{\%},$ then the equation becomes
\[\begin{array}{lll}1,000,000 & = & P{e}^{0.06(40)} \\ P & = & 90,717.95.\end{array}\]
In this case, she needs to invest only $\text{\$}90,717.95.$ This is roughly two-thirds the amount she needs to invest at $5\text{\%}.$ The fact that the interest is compounded continuously greatly magnifies the effect of the $1\text{\%}$ increase in interest rate.
Suppose instead of investing at age $25$, the student waits until age $35.$ How much would she have to invest at $5\text{\%}?$ At $6\text{\%}?$

答えを明らかにしろ

At $5\text{\%}$ interest, she must invest $\text{\$}223,130.16.$ At $6\text{\%}$ interest, she must invest $\text{\$}165,298.89.$
Assume a population of fish grows exponentially. A pond is stocked initially with $500$ fish. After $6$ months, there are $1000$ fish in the pond. The owner will allow his friends and neighbors to fish on his pond after the fish population reaches $10,000.$ When will the owner’s friends be allowed to fish?

答えを明らかにしろ

We know it takes the population of fish $6$ months to double in size. So, if t represents time in months, by the doubling-time formula, we have $6=(\text{ln}\ 2)\text{/}k.$ Then, $k=(\text{ln}\ 2)\text{/}6.$ Thus, the population is given by $y=500{e}^{((\text{ln}\ 2)\text{/}6)t}.$ To figure out when the population reaches $10,000$ fish, we must solve the following equation:
\[\begin{array}{lll}10,000 & = & 500{e}^{(\text{ln}\ 2\text{/}6)t} \\ 20 & = & {e}^{(\text{ln}\ 2\text{/}6)t} \\ \text{ln}\ 20 & = & (\frac{\text{ln}\ 2}{6})t \\ t & = & \frac{6(\text{ln}\ 20)}{\text{ln}\ 2}\approx 25.93.\end{array}\]
The owner’s friends have to wait $25.93$ months (a little more than $2$ years) to fish in the pond.
Suppose it takes $9$ months for the fish population in to reach $1000$ fish. Under these circumstances, how long do the owner’s friends have to wait?

答えを明らかにしろ

$38.90$ months
According to experienced baristas, the optimal temperature to serve coffee is between $155\text{^{\circ}}\text{F}$ and $175\text{^{\circ}}\text{F}.$ Suppose coffee is poured at a temperature of $200\text{^{\circ}}\text{F},$ and after $2$ minutes in a $70\text{^{\circ}}\text{F}$ room it has cooled to $180\text{^{\circ}}\text{F}.$ When is the coffee first cool enough to serve? When is the coffee too cold to serve? Round answers to the nearest half minute.

答えを明らかにしろ

We have
\[\begin{array}{lll}T & = & ({T}_{0}-{T}_{a}){e}^{\text{-}kt}+{T}_{a} \\ 180 & = & (200-70){e}^{\text{-}k(2)}+70 \\ 110 & = & 130{e}^{-2k} \\ \frac{11}{13} & = & {e}^{-2k} \\ \text{ln}\ \frac{11}{13} & = & -2k \\ \text{ln}\ 11-\text{ln}\ 13 & = & -2k \\ k & = & \frac{\text{ln}\ 13-\text{ln}\ 11}{2}.\end{array}\]
Then, the model is
\[T=130{e}^{(\frac{\text{ln}\ 11-\text{ln}\ 13}{2})t}+70.\]
The coffee reaches $175\text{^{\circ}}\text{F}$ when
\[\begin{array}{lll}175 & = & 130{e}^{(\frac{\text{ln}\ 11-\text{ln}\ 13}{2})t}+70 \\ 105 & = & 130{e}^{(\frac{\text{ln}\ 11-\text{ln}\ 13}{2})t} \\ \frac{21}{26} & = & {e}^{(\frac{\text{ln}\ 11-\text{ln}\ 13}{2})t} \\ \text{ln}\ \frac{21}{26} & = & \frac{\text{ln}\ 11-\text{ln}\ 13}{2}t \\ \text{ln}\ 21-\text{ln}\ 26 & = & \frac{\text{ln}\ 11-\text{ln}\ 13}{2}t \\ t & = & \frac{2(\text{ln}\ 21-\text{ln}\ 26)}{\text{ln}\ 11-\text{ln}\ 13}\approx 2.56.\end{array}\]
The coffee can be served about $2.5$ minutes after it is poured. The coffee reaches $155\text{^{\circ}}\text{F}$ at
\[\begin{array}{lll}155 & = & 130{e}^{(\frac{\text{ln}\ 11-\text{ln}\ 13}{2})t}+70 \\ 85 & = & 130{e}^{(\frac{\text{ln}\ 11-\text{ln}\ 13}{2})t} \\ \frac{17}{26} & = & {e}^{(\frac{\text{ln}\ 11-\text{ln}\ 13}{2})t} \\ \text{ln}\ 17-\text{ln}\ 26 & = & (\frac{\text{ln}\ 11-\text{ln}\ 13}{2})t \\ t & = & \frac{2(\text{ln}\ 17-\text{ln}\ 26)}{\text{ln}\ 11-\text{ln}\ 13}\approx 5.09.\end{array}\]
The coffee is too cold to be served about $5$ minutes after it is poured.
Suppose the room is warmer $(75\text{^{\circ}}\text{F})$ and, after $2$ minutes, the coffee has cooled only to $185\text{^{\circ}}\text{F}.$ When is the coffee first cool enough to serve? When is the coffee be too cold to serve? Round answers to the nearest half minute.

答えを明らかにしろ

The coffee is first cool enough to serve about $3.5$ minutes after it is poured. The coffee is too cold to serve about $7$ minutes after it is poured.
One of the most common applications of an exponential decay model is carbon dating. $\text{Carbon-}14$ decays (emits a radioactive particle) at a regular and consistent exponential rate. Therefore, if we know how much carbon was originally present in an object and how much carbon remains, we can determine the age of the object. The half-life of $\text{carbon-}14$ is approximately $5730$ years—meaning, after that many years, half the material has converted from the original $\text{carbon-}14$ to the new nonradioactive $\text{nitrogen-}14.$ If we have $100$ g $\text{carbon-}14$ today, how much is left in $50$ years? If an artifact that originally contained $100$ g of carbon now contains $10$ g of carbon, how old is it? Round the answer to the nearest hundred years.

答えを明らかにしろ

We have
\[\begin{array}{lll}5730 & = & \frac{\text{ln}\ 2}{k} \\ k & = & \frac{\text{ln}\ 2}{5730}.\end{array}\]
So, the model says
\[y=100{e}^{\text{-}(\text{ln}\ 2\text{/}5730)t}.\]
In $50$ years, we have
\[\begin{array}{lll}y & = & 100{e}^{\text{-}(\text{ln}\ 2\text{/}5730)(50)} \\ & \approx & 99.40.\end{array}\]
Therefore, in $50$ years, $99.40$ g of $\text{carbon-}14$ remains.
To determine the age of the artifact, we must solve
\[\begin{array}{lll}10 & = & 100{e}^{\text{-}(\text{ln}\ 2\text{/}5730)t} \\ \frac{1}{10} & = & {e}^{\text{-}(\text{ln}\ 2\text{/}5730)t} \\ t & \approx & 19035.\end{array}\]
The artifact is about $19,000$ years old.
If we have $100$ g of $\text{carbon-}14,$ how much is left after $500$ years? If an artifact that originally contained $100$ g of carbon now contains $20g$ of carbon, how old is it? Round the answer to the nearest hundred years.

答えを明らかにしろ

A total of $94.13$ g of carbon remains. The artifact is approximately $13,300$ years old.
The doubling time for $y={e}^{ct}$ is $(\text{ln}\ (2))\text{/}(\text{ln}\ (c)).$
If you invest $\text{\$}500,$ an annual rate of interest of $3\text{\%}$ yields more money in the first year than a $2.5\text{\%}$ continuous rate of interest.

答えを明らかにしろ

True
If you leave a $100\text{^{\circ}}\text{C}$ pot of tea at room temperature $(25\text{^{\circ}}\text{C})$ and an identical pot in the refrigerator $(5\text{^{\circ}}\text{C}),$ with $k=0.02,$ the tea in the refrigerator reaches a drinkable temperature $(70\text{^{\circ}}\text{C})$ more than $5$ minutes before the tea at room temperature.
If given a half-life of t years, the constant $k$ for $y={e}^{kt}$ is calculated by $k=\text{ln}\ (1\text{/}2)\text{/}t.$

答えを明らかにしろ

True; $k=\frac{\text{ln}\ (2)}{t}$
If a culture of bacteria doubles in $3$ hours, how many hours does it take to multiply by $10?$
If bacteria increase by a factor of $10$ in $10$ hours, how many hours does it take to increase by $100?$

答えを明らかにしろ

$20$ hours
How old is a skull that contains one-fifth as much radiocarbon as a modern skull? Note that the half-life of radiocarbon is $5730$ years.
If a relic contains $90\text{\%}$ as much radiocarbon as new material, can it have come from the time of Christ (approximately $2000$ years ago)? Note that the half-life of radiocarbon is $5730$ years.

答えを明らかにしろ

No. The relic is approximately $871$ years old.
The population of Cairo grew from $5$ million to $10$ million in $20$ years. Use an exponential model to find when the population was $8$ million.
The populations of New York and Los Angeles are growing at $1\text{\%}$ and $1.4\text{\%}$ a year, respectively. Starting from $8$ million (New York) and $6$ million (Los Angeles), when are the populations equal? Round your answer to a whole number of years.

答えを明らかにしろ

$73$ years
Suppose the value of $\text{\$}1$ in Japanese yen decreases at $2\text{\%}$ per year. Starting from $\text{\$}1=\text{¥}250,$ when will $\text{\$}1=\text{¥}1?$
The effect of advertising decays exponentially. If $40\text{\%}$ of the population remembers a new product after $3$ days, how long will $20\text{\%}$ remember it?

答えを明らかにしろ

$5$ days $6$ hours $27$ minutes
If $y=1000$ at $t=3$ and $y=3000$ at $t=4,$ what was ${y}_{0}$ at $t=0?$
If $y=100$ at $t=4$ and $y=10$ at $t=8,$ when does $y=1?$

答えを明らかにしろ

$12$
If a bank offers annual interest of $7.5\text{\%}$ or continuous interest of $7.25\text{\%},$ which has a better annual yield?
What continuous interest rate has the same yield as an annual rate of $9\text{\%}?$

答えを明らかにしろ

$8.618\text{\%}$
If you deposit $\text{\$}5000$ at $8\text{\%}$ annual interest, how many years can you withdraw $\text{\$}500$ (starting after the first year) without running out of money?
You are trying to save $\text{\$}50,000$ in $20$ years for college tuition for your child. If interest is a continuous $10\text{\%},$ how much do you need to invest initially?

答えを明らかにしろ

$\text{\$}6766.76$
You are cooling a turkey that was taken out of the oven with an internal temperature of $165\text{^{\circ}}\text{F}.$ After $10$ minutes of resting the turkey in a $70\text{^{\circ}}\text{F}$ apartment, the temperature has reached $155\text{^{\circ}}\text{F}\text{.}$ What is the temperature of the turkey $20$ minutes after taking it out of the oven?
You are trying to thaw some vegetables that are at a temperature of $1\text{^{\circ}}\text{F}\text{.}$ To thaw vegetables safely, you must put them in the refrigerator, which has an ambient temperature of $44\text{^{\circ}}\text{F}.$ You check on your vegetables $2$ hours after putting them in the refrigerator to find that they are now $12\text{^{\circ}}\text{F}\text{.}$ Plot the resulting temperature curve and use it to determine when the vegetables reach $33\text{^{\circ}}\text{F}\text{.}$

答えを明らかにしろ

$9$ hours $13$ minutes
You are an archaeologist and are given a bone that is claimed to be from a Tyrannosaurus Rex. You know these dinosaurs lived during the Cretaceous Era $(146$ million years to $65$ million years ago), and you find by radiocarbon dating that there is $0.000001\text{\%}$ the amount of radiocarbon. Is this bone from the Cretaceous?
The spent fuel of a nuclear reactor contains plutonium-239, which has a half-life of $24,000$ years. If $1$ barrel containing $10\ \text{kg}$ of plutonium-239 is sealed, how many years must pass until only $10g$ of plutonium-239 is left?

答えを明らかにしろ

$239,179$ years
[T] The best-fit exponential curve to the data of the form $P(t)=a{e}^{bt}$ is given by $P(t)=2686{e}^{0.01604t}.$ Use a graphing calculator to graph the data and the exponential curve together.
[T] Find and graph the derivative ${y}^{'}$ of your equation. Where is it increasing and what is the meaning of this increase?

答えを明らかにしろ

$P'(t)=43{e}^{0.01604t}.$ The population is always increasing.
[T] Find and graph the second derivative of your equation. Where is it increasing and what is the meaning of this increase?
[T] Find the predicted date when the population reaches $10$ billion. Using your previous answers about the first and second derivatives, explain why exponential growth is unsuccessful in predicting the future.

答えを明らかにしろ

The population reaches $10$ billion people in $2032.$
[T] The best-fit exponential curve to the data of the form $P(t)=a{e}^{bt}$ is given by $P(t)=35.26{e}^{0.06407t}.$ Use a graphing calculator to graph the data and the exponential curve together.
[T] Find and graph the derivative ${y}^{'}$ of your equation. Where is it increasing? What is the meaning of this increase? Is there a value where the increase is maximal?

答えを明らかにしろ

$P'(t)=2.259{e}^{0.06407t}.$ The population is always increasing.
[T] Find and graph the second derivative of your equation. Where is it increasing? What is the meaning of this increase?
Consider the population of bacteria described earlier. This population grows according to the function $f(t)=200{e}^{0.02t},$ where t is measured in minutes. How many bacteria are present in the population after $5$ hours $(300$ minutes)? When does the population reach $100,000$ bacteria?

答えを明らかにしろ

We have $f(t)=200{e}^{0.02t}.$ Then
\[f(300)=200{e}^{0.02(300)}\approx 80,686.\]
There are $80,686$ bacteria in the population after $5$ hours.
To find when the population reaches $100,000$ bacteria, we solve the equation
\[\begin{array}{lll}100,000 & = & 200{e}^{0.02t} \\ 500 & = & {e}^{0.02t} \\ \text{ln}\ 500 & = & 0.02t \\ t & = & \frac{\text{ln}\ 500}{0.02}\approx 310.73.\end{array}\]
The population reaches $100,000$ bacteria after $310.73$ minutes.

Symbols used here

^\circ

degrees
1/360 of a full turn. 180° = π radians.

\approx

approximately equal
Equal to the precision shown, not exactly.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

Euler's number
2.71828…, the base whose exponential is its own derivative.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: Exponential Growth and Decay

Use the exponential growth model in applications, including population growth and compound interest.
Explain the concept of doubling time.
Use the exponential decay model in applications, including radioactive decay and Newton’s law of cooling.
Explain the concept of half-life.
Exponential growth and exponential decay are two of the most common applications of exponential functions.
Systems that exhibit exponential growth follow a model of the form
In exponential growth, the rate of growth is proportional to the quantity present. In other words,
Systems that exhibit exponential growth have a constant doubling time, which is given by

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

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Parts of this page are adapted from OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0), OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

ここに Calculus

Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests

Time (min)	Population Size (no. of bacteria)
\(10\)	\(244\)
\(20\)	\(298\)
\(30\)	\(364\)
\(40\)	\(445\)
\(50\)	\(544\)
\(60\)	\(664\)
\(70\)	\(811\)
\(80\)	\(991\)
\(90\)	\(1210\)
\(100\)	\(1478\)
\(110\)	\(1805\)
\(120\)	\(2205\)

Years since 1950	Population (millions)
\(0\)	\(2,556\)
\(10\)	\(3,039\)
\(20\)	\(3,706\)
\(30\)	\(4,453\)
\(40\)	\(5,279\)
\(50\)	\(6,083\)
\(60\)	\(6,849\)

Years since 1850	Population (thousands)
\(0\)	\(21.00\)
\(10\)	\(56.80\)
\(20\)	\(149.5\)
\(30\)	\(234.0\)

Time (min)	Population Size (no. of bacteria)
\(10\)	\(244\)
\(20\)	\(298\)
\(30\)	\(364\)
\(40\)	\(445\)
\(50\)	\(544\)
\(60\)	\(664\)
\(70\)	\(811\)
\(80\)	\(991\)
\(90\)	\(1210\)
\(100\)	\(1478\)
\(110\)	\(1805\)
\(120\)	\(2205\)

Years since 1950	Population (millions)
\(0\)	\(2,556\)
\(10\)	\(3,039\)
\(20\)	\(3,706\)
\(30\)	\(4,453\)
\(40\)	\(5,279\)
\(50\)	\(6,083\)
\(60\)	\(6,849\)