# 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,

find the domain and range of f(x)= 4x-7/x²-6x+8
find the range of f(x)=(x+1)(x+4)
-1, -4
Marcia
That's domain. The range is [-9/4,+infinity)
Jacob
If you're using calculus to find the range, you have to find the extrema through the first derivative test and then substitute the x-value for the extrema back into the original equation.
Jacob
Good morning,,, how are you
d/dx{1/y - lny + X^3.Y^5}
How to identify domain and range
hello
Akpevwe
He,,
Harrieta
hi
Dr
hello
velocity
I only talk to girls
Dr
women are smart then guys
Dr
Smarter
sorry
Dr
Dr
:(
Shun
was up
Dr
hello
is it chatting app?.. I do not see any calculus here. lol
Find the arc length of the graph of f(x) = In (sinx) on the interval [Π/4, Π/2].
Sand falling freely from a lorry form a conical shape whose height is always equal to one-third the radius of the base. a. How fast is the volume increasing when the radius of the base is (1m) and increasing at the rate of 1/4cm/sec Pls help me solve
show that lim f(x) + lim g(x)=m+l
list the basic elementary differentials
Differentiation and integration
yes
Damien
proper definition of derivative
the maximum rate of change of one variable with respect to another variable
terms of an AP is 1/v and the vth term is 1/u show that the sum of uv terms is 1/2(uv+1)
what is calculus?
calculus is math that studies the change in math, such as the rate and distance,
Tamarcus
what are the topics in calculus
Augustine
what is limit of a function?
what is x and how x=9.1 take?
what is f(x)
the function at x
Marc
also known as the y value so I could say y=2x or f(x)= 2x same thing just using functional notation your next question is what is dependent and independent variables. I am Dyslexic but know math and which is which confuses me. but one can vary the x value while y depends on which x you use. also
Marc
up domain and range
Marc
enjoy your work and good luck
Marc
I actually wanted to ask another questions on sets if u dont mind please?
Inembo
I have so many questions on set and I really love dis app I never believed u would reply
Inembo
Hmm go ahead and ask you got me curious too much conversation here
am sorry for disturbing I really want to know math that's why *I want to know the meaning of those symbols in sets* e.g n,U,A', etc pls I want to know it and how to solve its problems
Inembo
and how can i solve a question like dis *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
next questions what do dy mean by (A' n B^c)^c'
Inembo
The sets help you to define the function. The function is like a magic box where you put inside stuff(numbers or sets) and you get out the stuff but in different shapes (forms).
I dont understand what you wanna say by (A' n B^c)^c'
(A' n B (rise to the power of c)) all rise to the power of c
Inembo
Aaaahh
Ok so the set is formed by vectors and not numbers
A vector of length n
But you can make a set out of matrixes as well
I I don't even understand sets I wat to know d meaning of all d symbolsnon sets
Inembo
High-school?
yes
Inembo
am having big problem understanding sets more than other math topics
Inembo
So f:R->R means that the function takes real numbers and provides real numer. For ex. If f(x) =2x this means if you give to your function a real number like 2,it gives you also a real number 2times2=4
pls answer this question *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
If you have f:R^n->R^n you give to your function a vector of length n like (a1,a2,...an) where all a1,.. an are reals and gives you also a vector of length n... I don't know if i answering your question. Otherwise on YouTube you havr many videos where they explain it in a simple way
I would say 24
Offer both
Sorry 20
Actually you have 40 - 4 =36 who offer maths or physics or both.
I know its 20 but how to prove it
Inembo
You have 32+24=56who offer courses
56-36=20 who give both courses... I would say that
solution: In a question involving sets and Venn diagram, the sum of the members of set A + set B - the joint members of both set A and B + the members that are not in sets A or B = the total members of the set. In symbolic form n(A U B) = n(A) + n (B) - n (A and B) + n (A U B)'.
Mckenzie
In the case of sets A and B use the letters m and p to represent the sets and we have: n (M U P) = 40; n (M) = 24; n (P) = 32; n (M and P) = unknown; n (M U P)' = 4
Mckenzie
Now substitute the numerical values for the symbolic representation 40 = 24 + 32 - n(M and P) + 4 Now solve for the unknown using algebra: 40 = 24 + 32+ 4 - n(M and P) 40 = 60 - n(M and P) Add n(M and P), as well, subtract 40 from both sides of the equation to find the answer.
Mckenzie
40 - 40 + n(M and P) = 60 - 40 - n(M and P) + n(M and P) Solution: n(M and P) = 20
Mckenzie
thanks
Inembo
Simpler form: Add the sums of set M, set P and the complement of the union of sets M and P then subtract the number of students from the total.
Mckenzie
n(M and P) = (32 + 24 + 4) - 40 = 60 - 40 = 20
Mckenzie