rate of growth/decay in exponential function

Hannibal lecter · 2021-12-12 17:24:36

Hi, please see this example talking about converting exponential function form from

into form

it's easy and understandable as you see in the following picture :-

as you can see the growth rate here is (1 - 1.65= 65%) and it's increasing rate because 1.65 > 1
and in the other function is (1-0.819 = 18.1%) and it's decreasing rate because 0 < 0.819 < 1
and I tested them all,

my problem is I found another function I wonder if I can convert it like the others too and the function is because I'm unable to find the rate while I convert it to the classic form (without e) :-

lets say k=1,t≥0
I want to convert it too so I can see the initial value and the growth rate
I just want to convert it because I want to compare between growth/decay in the form of

and growth/decay in the second from of

the reason that I want to convert it to classic form because in the book as highlighted in yellow "..the quantity increases at a decreasing rate"

again directly it said "This is realistic because as the quantity of the drug in the body increases, so does the rate at which
the body excretes the drug..."

so now I'm confused! once it say the rate is decreasing and again it said the rate is increasing, I want to know the rate in the classic form to check it by myself if it's decreasing or if it's not ( I mean the rate change not the function)

also it's strange equation it's without initial value!

I tried to convert it with my own way like this :

by assuming t is 1 and k is 1 too
so it's :-

but here when finding the rate is (1 - 0.367879441171 = 0.63212055882) and it's mean 63.2%
first it's not like k I chosed k = 1 it has to be 1%!! I choses it myself
second it's not a correct rate it has to be around 36.7% because when I try these values with the two forms of this functions it's identical
but in 63% rate it's not correct,
and this is my another problem question why this time I minus (1-0.367879441171) the rate is wrong? I minus 1 always as you can see above and it's correct why this time in this function is not correct

Last edited by Hannibal lecter (2021-12-12 18:06:28)

Bob · 2021-12-13 00:38:13

The graph of S against t shows a curve that is going up (means increasing) but with a gradient that is getting more horizontal (ie. a decreasing rate of change)

I used the function grapher to experiment with functions and found 1-1.65^(-x) gives roughly the right curve.

You can get a more accurate value than 1.65 by substituting x=1 into the first equation to find y.

Bob

Hannibal lecter · 2021-12-13 13:45:52

yes what is that gradient value of -k?
what you mean when it's decreasing rate what you mean you mean it become for example -5 % then -4% then -3%
is it in that function decreasing and getting horizontal like that? or how, is it a constant value decreasing rate for example -50%

and what is value of k represented in that mentioned function is it rate of the medicine in the blood of patient being decreased
or it's the rate of change in the saturation level will get to its end, or what is it I I know S meaning saturation level and t mean time but what is k represent in that example

Hannibal lecter · 2021-12-13 14:08:46

and please I have another question confused me again also :

why we say function P growth rate is continuous and function y growth rate is just increasing
isn't should be continuous too as in P??? I graph it in graph calculator it looks continuous
I see the manual solution it's also state the P is has continuous rate of change, but in Y he just said "increasing or decay"

note that I found this problem from an exercise and here it is :

I learned in math that if the graph is a line it's continuous if it's not a separate points!!

Last edited by Hannibal lecter (2021-12-13 14:15:10)

Bob · 2021-12-13 22:23:45

yes what is that gradient value of -k?

The question doesn't set a value to this. If you look at the S-t graph, the S axis doesn't have a scale, so k could be anything.

-5%, then -4%, then -3% is NOT decreasing. It is getting more positive so it is increasing.

It is continuous and this is true with either equation. It would have to be since they both represent the same thing.

I hope I have answered all your questions here. I found it difficult to keep scrolling back to two separate posts. I would rather you kept them separate, thanks.

Bob

Hannibal lecter · 2021-12-14 10:02:31

Bob wrote:

yes what is that gradient value of -k?
I hope I have answered all your questions here. I found it difficult to keep scrolling back to two separate posts. I would rather you kept them separate, thanks.
Bob

yes sorry I'm trying to use the site in better ways

so branch 7,8 continues growth of rate

and

what about branch 5,6 :

can we say it's continuous growth of rate too?

Mathegocart · 2021-12-14 12:04:21

Hannibal lecter wrote:

Bob wrote:
yes what is that gradient value of -k?
I hope I have answered all your questions here. I found it difficult to keep scrolling back to two separate posts. I would rather you kept them separate, thanks.
Bob
yes sorry I'm trying to use the site in better ways
so branch 7,8 continues growth of rate

and
what about branch 5,6 :

can we say it's continuous growth of rate too?

No, because . Continuous growth is modeled with the equation

, where is the ending value, is the initial value, is Euler's constant, is the continuous growth rate, and is the time that has passed.

Diregard what I wrote above; as zetafunc helpfully noted,

could be represented as .

Last edited by Mathegocart (2021-12-15 09:38:10)

zetafunc · 2021-12-14 12:39:49

Mathegocart wrote:

No, because continuous growth is modeled with the equation
, where
is the ending value,
is the initial value,
is Euler's constant,
is the continuous growth rate, and
is the time that has passed.

What's stopping you from taking k = log(1.07)?

Mathegocart · 2021-12-15 09:36:18

zetafunc wrote:

Mathegocart wrote:
No, because continuous growth is modeled with the equation
, where
is the ending value,
is the initial value,
is Euler's constant,
is the continuous growth rate, and
is the time that has passed.
What's stopping you from taking k = log(1.07)?

I was a little fatigued when I wrote that up; that's true.

Hannibal lecter · 2021-12-15 12:16:44

Mathegocart wrote:

Hannibal lecter wrote:
Bob wrote:
I hope I have answered all your questions here. I found it difficult to keep scrolling back to two separate posts. I would rather you kept them separate, thanks.
Bob
yes sorry I'm trying to use the site in better ways
so branch 7,8 continues growth of rate

and
what about branch 5,6 :

can we say it's continuous growth of rate too?
No, because . Continuous growth is modeled with the equation
, where
is the ending value,
is the initial value,
is Euler's constant,
is the continuous growth rate, and
is the time that has passed.
Diregard what I wrote above; as zetafunc helpfully noted,
could be represented as
.

I'm talking about e and k and exponential form I'm talking about this form :

is the growth rate 7% here is called continuous or we can't call it this

Mathegocart · 2021-12-15 13:03:13

Hannibal lecter wrote:

Mathegocart wrote:
Hannibal lecter wrote:
yes sorry I'm trying to use the site in better ways
so branch 7,8 continues growth of rate

and
what about branch 5,6 :

can we say it's continuous growth of rate too?
No, because . Continuous growth is modeled with the equation
, where
is the ending value,
is the initial value,
is Euler's constant,
is the continuous growth rate, and
is the time that has passed.
Diregard what I wrote above; as zetafunc helpfully noted,
could be represented as
.
I'm talking about e and k and exponential form I'm talking about this form :
is the growth rate 7% here is called continuous or we can't call it this

Yes, the equation does represent continuous growth. As noted, you can convert that equation into the form of the continuous standard equation

with the substitution mentioned above.

Hannibal lecter · 2021-12-15 14:53:28

Mathegocart wrote:

Yes, the equation does represent continuous growth. As noted, you can convert that equation into the form of the continuous standard equation
with the substitution mentioned above.

I think it's not right, because I found the answer today
the form model without e and k like this :

is called non-continuous exponential models functions
because in case of increasing (a>1) while in case of model contain e, and k, if k is positive it's increasing continuously rate (because any positive number is acceptable and in case of base a there is numbers less then 1 are positive but not included)
and in case of decreasing (0<a<1) while in case of e, and k, here k is has to be negative <0 so it's continuously decreasing rate

so now

its growth rate is not continuous,

am I right?

by the way I learned this today from a YouTube channel but I still don't understand why we can't called the base a a continuous if it's >1
I knew because it doesn't included the numbers that less than 1 and bigger than 0
but why? why this is a reasons that we can't call it continuously, what is meaning continuously in this case anyway? because every graph I draw in form of non-continuous exponential models I see them clearly in my naked eyes that they are continuous, I draw these random examples as in example I see them all goes beyond without end

Last edited by Hannibal lecter (2021-12-15 15:05:16)

Mathegocart · 2021-12-15 17:44:42

Hannibal lecter wrote:

Mathegocart wrote:
Yes, the equation does represent continuous growth. As noted, you can convert that equation into the form of the continuous standard equation
with the substitution mentioned above.
I think it's not right, because I found the answer today
the form model without e and k like this :

is called non-continuous exponential models functions
because in case of increasing (a>1) while in case of model contain e, and k, if k is positive it's increasing continuously rate (because any positive number is acceptable and in case of base a there is numbers less then 1 are positive but not included)
and in case of decreasing (0<a<1) while in case of e, and k, here k is has to be negative <0 so it's continuously decreasing rate
so now

its growth rate is not continuous,
am I right?

by the way I learned this today from a YouTube channel but I still don't understand why we can't called the base a a continuous if it's >1
I knew because it doesn't included the numbers that less than 1 and bigger than 0
but why? why this is a reasons that we can't call it continuously, what is meaning continuously in this case anyway? because every graph I draw in form of non-continuous exponential models I see them clearly in my naked eyes that they are continuous, I draw these random examples as in example I see them all goes beyond without end
https://i.ibb.co/bvG98qw/2021-12-16-060158.png

How so? As zetafunc mentions above, the function

is equivalent to the function you mentioned, and the equation is continuous.

Last edited by Mathegocart (2021-12-15 17:44:57)

Hannibal lecter · 2021-12-16 04:38:26

Mathegocart wrote:

How so? As zetafunc mentions above, the function
is equivalent to the function you mentioned, and the equation is continuous.

can you please see this YouTube link it talked about it and it said there is differences :-

https://www.youtube.com/watch?v=VmxESGI4zdA&t=18s

also see this sheet pls :

Last edited by Hannibal lecter (2021-12-16 04:45:39)

Mathegocart · 2021-12-16 11:42:47

Hannibal lecter wrote:

Mathegocart wrote:
How so? As zetafunc mentions above, the function
is equivalent to the function you mentioned, and the equation is continuous.

can you please see this YouTube link it talked about it and it said there is differences :-
https://www.youtube.com/watch?v=VmxESGI4zdA&t=18s

also see this sheet pls :
https://i.ibb.co/2F92yjQ/2021-12-16-194132.png

As stated, one can be converted into the other. So it is continuous. The current form isn't, but it can be converted into an equivalent continuous form.

Last edited by Mathegocart (2021-12-16 11:43:48)

Bob · 2021-12-16 21:00:39

I'm not following this discussion about continuous functions.

The book says P = 300(1.05^x) is non-continuous.

I've just plotted it on the function grapher and it definitely is continuous. So is its gradient function. So why is it described as non-continuous.

Surely if you have two different equations for a function and they are fully the same function, then they must both be continuous or neither. I thought continuity was a property of the graph.

Help!

Bob

zetafunc · 2021-12-17 00:25:10

Hi Bob,

I think 'continuous' in this context is referring to a growth rate which is 'convertible continuously'.

For example, suppose you have £100 in a savings account which grows at an effective rate of interest of 5% pa (not likely these days but who knows, given current market conditions!). The bank isn't likely to give you the 5% at the end of each year, though -- so they might offer a rate convertible monthly.

So if the savings account offered you a return on your £100 deposit at an effective interest rate of 5% pa offering monthly interest payments, then we want to find the interest rate such that:

The bank can then claim to offer you an interest rate of convertible monthly. So every month, you get about on top of your savings. That's the same as getting an effective interest rate of over a year, except you get the equivalent every month.

In this case we subdivided our annual effective interest rate into 12 intervals of equal length (months). But there's no reason why we can't continue subdividing into even smaller intervals, which is where the phrase 'convertible continuously' comes from (where the intervals become infinitesimally small). So suppose we have an interest rate convertible pthly and let p go to infinity. We have:

where we've used the limit definition of and writing Here, is the nominal rate of interest (growth) per unit time convertible continuously (sometimes called the 'force of interest'). That's what I think is meant by 'continuous' here. In this case we see that and in particular we can accumulate a payment of 100 from time 0 to time t like this:

In Hannibal lecter's example the emphasis should probably be on the fact that the rate is continuous, i.e. a continuous growth rate of 5% (i.e. ) which is slightly different to a growth rate of pa (i.e. ).

In other words, if someone told you to accumulate 300 at a rate of 5% pa for 10 years you would rightly do

but at a continuous rate of 5% per unit time you would do

Bob · 2021-12-18 22:54:02

hi zetafunc,

Thanks for clearing that up.

Bob

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#1 2021-12-12 17:24:36

rate of growth/decay in exponential function

#2 2021-12-13 00:38:13

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#3 2021-12-13 13:45:52

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