# 6.8 Exponential growth and decay  (Page 3/9)

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If a quantity grows exponentially, the time it takes for the quantity to double remains constant. In other words, it takes the same amount of time for a population of bacteria to grow from $100$ to $200$ bacteria as it does to grow from $10,000$ to $20,000$ bacteria. This time is called the doubling time. To calculate the doubling time, we want to know when the quantity reaches twice its original size. So we have

$\begin{array}{ccc}\hfill 2{y}_{0}& =\hfill & {y}_{0}{e}^{kt}\hfill \\ \hfill 2& =\hfill & {e}^{kt}\hfill \\ \hfill \text{ln}\phantom{\rule{0.2em}{0ex}}2& =\hfill & kt\hfill \\ \hfill t& =\hfill & \frac{\text{ln}\phantom{\rule{0.2em}{0ex}}2}{k}.\hfill \end{array}$

## Definition

If a quantity grows exponentially, the doubling time    is the amount of time it takes the quantity to double. It is given by

$\text{Doubling time}\phantom{\rule{0.2em}{0ex}}=\frac{\text{ln}\phantom{\rule{0.2em}{0ex}}2}{k}.$

## Using the doubling time

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?

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 $6=\left(\text{ln}\phantom{\rule{0.2em}{0ex}}2\right)\text{/}k.$ Then, $k=\left(\text{ln}\phantom{\rule{0.2em}{0ex}}2\right)\text{/}6.$ Thus, the population is given by $y=500{e}^{\left(\left(\text{ln}\phantom{\rule{0.2em}{0ex}}2\right)\text{/}6\right)t}.$ To figure out when the population reaches $10,000$ fish, we must solve the following equation:

$\begin{array}{ccc}\hfill 10,000& =\hfill & 500{e}^{\left(\text{ln}\phantom{\rule{0.2em}{0ex}}2\text{/}6\right)t}\hfill \\ \hfill 20& =\hfill & {e}^{\left(\text{ln}\phantom{\rule{0.2em}{0ex}}2\text{/}6\right)t}\hfill \\ \hfill \text{ln}\phantom{\rule{0.2em}{0ex}}20& =\hfill & \left(\frac{\text{ln}\phantom{\rule{0.2em}{0ex}}2}{6}\right)t\hfill \\ \hfill t& =\hfill & \frac{6\left(\text{ln}\phantom{\rule{0.2em}{0ex}}20\right)}{\text{ln}\phantom{\rule{0.2em}{0ex}}2}\approx 25.93.\hfill \end{array}$

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 [link] to reach $1000$ fish. Under these circumstances, how long do the owner’s friends have to wait?

$38.90$ months

## 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

${y}^{\prime }=\text{−}k{y}_{0}{e}^{\text{−}kt}=\text{−}ky.$

## Rule: exponential decay model

Systems that exhibit exponential decay    behave according to the model

$y={y}_{0}{e}^{\text{−}kt},$

where ${y}_{0}$ represents the initial state of the system and $k>0$ is a constant, called the decay constant .

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

${T}^{\prime }=\text{−}k\left(T-{T}_{a}\right).$

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 $y\left(t\right)=T\left(t\right)-{T}_{a}.$ Then ${y}^{\prime }\left(t\right)={T}^{\prime }\left(t\right)-0={T}^{\prime }\left(t\right),$ and our equation becomes

${y}^{\prime }=\text{−}ky.$

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

questions solve y=sin x
Solve it for what?
Tim
you have to apply the function arcsin in both sides and you get arcsin y = acrsin (sin x) the the function arcsin and function sin cancel each other so the ecuation becomes arcsin y = x you can also write x= arcsin y
Ioana
what is the question ? what is the answer?
Suman
there is an equation that should be solve for x
Ioana
ok solve it
Suman
are you saying y is of sin(x) y=sin(x)/sin of both sides to solve for x... therefore y/sin =x
Tyron
or solve for sin(x) via the unit circle
Tyron
what is unit circle
Suman
a circle whose radius is 1.
Darnell
the unit circle is covered in pre cal...and or trigonometry. it is the multipcation table of upper level mathematics.
Tyron
what is function?
A set of points in which every x value (domain) corresponds to exactly one y value (range)
Tim
what is lim (x,y)~(0,0) (x/y)
limited of x,y at 0,0 is nt defined
Alswell
But using L'Hopitals rule is x=1 is defined
Alswell
Could U explain better boss?
emmanuel
value of (x/y) as (x,y) tends to (0,0) also whats the value of (x+y)/(x^2+y^2) as (x,y) tends to (0,0)
NIKI
can we apply l hospitals rule for function of two variables
NIKI
why n does not equal -1
Andrew
I agree with Andrew
Bg
f (x) = a is a function. It's a constant function.
proof the formula integration of udv=uv-integration of vdu.?
Find derivative (2x^3+6xy-4y^2)^2
no x=2 is not a function, as there is nothing that's changing.
are you sure sir? please make it sure and reply please. thanks a lot sir I'm grateful.
The
i mean can we replace the roles of x and y and call x=2 as function
The
if x =y and x = 800 what is y
y=800
800
Bg
how do u factor the numerator?
Nonsense, you factor numbers
Antonio
You can factorize the numerator of an expression. What's the problem there? here's an example. f(x)=((x^2)-(y^2))/2 Then numerator is x squared minus y squared. It's factorized as (x+y)(x-y). so the overall function becomes : ((x+y)(x-y))/2
The
The problem is the question, is not a problem where it is, but what it is
Antonio
I think you should first know the basics man: PS
Vishal
Yes, what factorization is
Antonio
Antonio bro is x=2 a function?
The
Yes, and no.... Its a function if for every x, y=2.... If not is a single value constant
Antonio
you could define it as a constant function if you wanted where a function of "y" defines x f(y) = 2 no real use to doing that though
zach
Why y, if domain its usually defined as x, bro, so you creates confusion
Antonio
Its f(x) =y=2 for every x
Antonio
Yes but he said could you put x = 2 as a function you put y = 2 as a function
zach
F(y) in this case is not a function since for every value of y you have not a single point but many ones, so there is not f(y)
Antonio
x = 2 defined as a function of f(y) = 2 says for every y x will equal 2 this silly creates a vertical line and is equivalent to saying x = 2 just in a function notation as the user above asked. you put f(x) = 2 this means for every x y is 2 this creates a horizontal line and is not equivalent
zach
The said x=2 and that 2 is y
Antonio
that 2 is not y, y is a variable 2 is a constant
zach
So 2 is defined as f(x) =2
Antonio
No y its constant =2
Antonio
what variable does that function define
zach
the function f(x) =2 takes every input of x within it's domain and gives 2 if for instance f:x -> y then for every x, y =2 giving a horizontal line this is NOT equivalent to the expression x = 2
zach
Yes true, y=2 its a constant, so a line parallel to y axix as function of y
Antonio
Sorry x=2
Antonio
And you are right, but os not a function of x, its a function of y
Antonio
As function of x is meaningless, is not a finction
Antonio
yeah you mean what I said in my first post, smh
zach
I mean (0xY) +x = 2 so y can be as you want, the result its 2 every time
Antonio
OK you can call this "function" on a set {2}, but its a single value function, a constant
Antonio
well as long as you got there eventually
zach
2x^3+6xy-4y^2)^2 solve this
femi
moe
volume between cone z=√(x^2+y^2) and plane z=2
Fatima
It's an integral easy
Antonio
V=1/3 h π (R^2+r2+ r*R(
Antonio
How do we find the horizontal asymptote of a function using limits?
Easy lim f(x) x-->~ =c
Antonio
solutions for combining functions
what is a function? f(x)
one that is one to one, one that passes the vertical line test
Andrew
It's a law f() that to every point (x) on the Domain gives a single point in the codomain f(x)=y
Antonio
is x=2 a function?
The
restate the problem. and I will look. ty
is x=2 a function?
The