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YAMHT: How do you find the function of an exponential growth function.

Ricemarine

Ricemarine

Lifer
Ok, um yeah, I've asked everybody, and tried resources... I hope for a quicker answer (while googling)...

1. For a science experiment in biology, Binh is observing the growth of bacteria in the colony. At 2 P.M., he estimates that there are 1000 bacteria. When he returns to check at 5 P.M., there are 2200.

a) If the bacteria continue to reproduce at this rate, how many will be there at midnight?
b) Find a function that will tell you how many bacteria will be present at any given time.
Click to expand...

I know its exponential growth... But... how do you find the function?... (the rest of the homework is like this..)

Um yeah... integrated math III FTL. 😛 (pre calc next year heh...)
 
y = c*e^(at)

At t = 0 (i.e. 2 PM) y = 1000.

At t = 3, y = 2200

Solve for a.

Edit: Units for a are in [1/hours]
 
Originally posted by: hypn0tik
y = c*e^(at)

At t = 0 (i.e. 2 PM) y = 1000.

At t = 3, y = 2200

Solve for a.

Edit: Units for a are in [1/hours]
Click to expand...

OK uhh seriously... We have never ever... used that equation... But hey my sister solved it for me with that equation... I just don't know how to put it in terms of a*b^x

A and B are set numbers... So uh... help? 🙂. Like how do you solve it in terms of a*b^x?

 
Well, at 2pm, you start the stopwatch. Thus, time at 2pm = 0.
Other than that, it's simple "plug it in!"
 
All the book told us is that determining special numbers are called parameters...

Is this what the book is telling us to do?!... Plug and guess and check?

instead of doing like the equation hypn0tik gave?
 
Originally posted by: Ricemarine
Originally posted by: hypn0tik
y = c*e^(at)

At t = 0 (i.e. 2 PM) y = 1000.

At t = 3, y = 2200

Solve for a.

Edit: Units for a are in [1/hours]
Click to expand...

OK uhh seriously... We have never ever... used that equation... But hey my sister solved it for me with that equation... I just don't know how to put it in terms of a*b^x

A and B are set numbers... So uh... help? 🙂. Like how do you solve it in terms of a*b^x?
Click to expand...

It's not that hard to change c*e^(at) into the form of a*b^t. Hint: e^(at) = (e^a)^t
 
Originally posted by: Ricemarine
Originally posted by: hypn0tik
y = c*e^(at)

At t = 0 (i.e. 2 PM) y = 1000.

At t = 3, y = 2200

Solve for a.

Edit: Units for a are in [1/hours]
Click to expand...

OK uhh seriously... We have never ever... used that equation... But hey my sister solved it for me with that equation... I just don't know how to put it in terms of a*b^x

A and B are set numbers... So uh... help? 🙂. Like how do you solve it in terms of a*b^x?
Click to expand...
Basic exponential growth P = P(0) * e^(kt), where P(0) = population at time 0 (2pm, 1000); and k is a rate constant.

ln [P / P(0)] = kt, therefore k = ln [P/P(0)] / t <= Plug in t=3, P = 2200, P(0) = 1000 to get k.

Doubling time is when P = 2 P(0), therefore e^(kt) = 2, solve for t.

Edit: once you have k, you know the amount of bacteria at any given time. For example at midnight the population will be P = 1000 e^(k*10)
 
Originally posted by: Ricemarine
All the book told us is that determining special numbers are called parameters...

Is this what the book is telling us to do?!... Plug and guess and check?

instead of doing like the equation hypn0tik gave?
Click to expand...

Guess and check will probably take forever. If they said it's an exponential then you start with a basic exponential equation with variables or 'parameters'. Just like if they say it's a 2nd order polynomial, you start with y = a*x^2 + bx + c and start solving.
 
Originally posted by: TuxDave
Originally posted by: Ricemarine
Originally posted by: hypn0tik
y = c*e^(at)

At t = 0 (i.e. 2 PM) y = 1000.

At t = 3, y = 2200

Solve for a.

Edit: Units for a are in [1/hours]
Click to expand...

OK uhh seriously... We have never ever... used that equation... But hey my sister solved it for me with that equation... I just don't know how to put it in terms of a*b^x

A and B are set numbers... So uh... help? 🙂. Like how do you solve it in terms of a*b^x?
Click to expand...

It's not that hard to change c*e^(at) into the form of a*b^t. Hint: e^(at) = (e^a)^t
Click to expand...

Ok let's start over shall we...
We never ever learned y=C*e^(at), never used natural logarithms...

So I'm trying to solve "the simple dumb" way... uh yeah.
 
Originally posted by: Ricemarine

Ok let's start over shall we...
We never ever learned y=C*e^(at), never used natural logarithms...

So I'm trying to solve "the simple dumb" way... uh yeah.
Click to expand...

y= a*b^t

(stolen the rest from above)
At t = 0 (i.e. 2 PM) y = 1000. Solve for a.

At t = 3, y = 2200. Solve for b.

You're allowed to use logs right? Otherwise... I guess you have no choice but to guess and check on solving for b.
 
Originally posted by: TuxDave
Originally posted by: Ricemarine

Ok let's start over shall we...
We never ever learned y=C*e^(at), never used natural logarithms...

So I'm trying to solve "the simple dumb" way... uh yeah.
Click to expand...

y= a*b^t

(stolen the rest from above)
At t = 0 (i.e. 2 PM) y = 1000. Solve for a.

At t = 3, y = 2200. Solve for b.

You're allowed to use logs right? Otherwise... there's probably a 2nd method which I'm not aware of.
Click to expand...

We've never used logs, so probably not... 🙁... Maybe there is a 2nd method... Somewhere....

and most of my homework is doing this 🙁
 
Hmmm. Maybe it isn't exponential growth considering you have never dealt with logarithms (natural or otherwise).

Perhaps is it a simple linear function with a constant growth rate.

You have 2 points, which is all you need to find the equation of a line:

(0,1000) and (3,2200).

Consider y = mt + b
where y = Population
t = time.

Slope = m = (2200-1000)/(3-0) = 1200/3 = 400.

y = 400t + b
substitute (0,1000)

-> y = 400t + 1000.

If that doesn't work I really don't know what else to do.

Edit: At midnight, t = 10 --> y = 5000.
 
Originally posted by: Ricemarine
Originally posted by: TuxDave
Originally posted by: Ricemarine

Ok let's start over shall we...
We never ever learned y=C*e^(at), never used natural logarithms...

So I'm trying to solve "the simple dumb" way... uh yeah.
Click to expand...

y= a*b^t

(stolen the rest from above)
At t = 0 (i.e. 2 PM) y = 1000. Solve for a.

At t = 3, y = 2200. Solve for b.

You're allowed to use logs right? Otherwise... there's probably a 2nd method which I'm not aware of.
Click to expand...

We've never used logs, so probably not... 🙁... Maybe there is a 2nd method... Somewhere....

and most of my homework is doing this 🙁
Click to expand...
What about P = P(0) * 2^[t / t(d)], where t(d) is the doubling time.

In this case I would say that t(d) ~ 3 hrs or 2.5 hrs if you assume linear growth (which doesn't apply because it's exponential growth). I'd say use 3 hrs for t(d). That should give you a rough approximation.
 
Originally posted by: Ricemarine

Ok let's start over shall we...
We never ever learned y=C*e^(at), never used natural logarithms...

So I'm trying to solve "the simple dumb" way... uh yeah.
Click to expand...

You don't have to do it in base e. The difference between bases for logarithmic or exponential relationships is just a constant. That is, if you solve for the equation y = C_1*e^(\tau_1*t), and solve for the equation y = C_2*10^(\tau_2*t), the two equations will give you the same exponential curve. So all you have to do is choose some arbitrary base. It doesn't matter what it is as long as you solve the relationship correctly. That's why all they tell you is that the relationship is exponential, there are an infinite number of ways of expressing the relationship.

So you do not need to have any knowledge of natural number or natural logarithm to do this. There is no other way to do this without logarithms. Well, I guess you could do it using Taylor's expansion.

1000 = C*(1+x+1/2*x^2+1/6*x^3+1/24*x^4+....) where x=\tau*0
--> C = 1000
2200 = 1000*(1+x+1/2*x^2+1/6*x^3+1/24*x^4+....) where x=\tau*3

So you could solve for the leading terms of the Taylor's expansion to find \tau and then approximate the actual \tau from that. Solving for \tau to the fourth order will give you \tau to almost 10^-3 error.
 
What about plugging in x=3 y=2200 into the exponential growth equation A*B^X...

So then
2200=1000*B^3
2.2=B^3
~1.3=B

? lol.
 
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