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Exponential growth

  • Thread starter Banana
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Can anyone help me with a problem for my physics lab? We are studying capacitance and I'm not even sure what this problem has to do with it. I think you're supposed to use a log or ln, but I never had calculus, so I don't think you're supposed to do it that way. Any help would be appreciated!

The yearly growth rate of the US is 1 percent. Assuming this growth rate to be constant, in how many years will the population double?
 

Answers and Replies

chroot
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The equation you need is this one:

[tex]N = N_0 e^{kt}[/tex]

Where [itex]N[/itex] is the population at time t, [itex]N_0[/itex] is the population a time 0, [itex]k[/itex] is the "growth constant," and [itex]t[/itex] is time.

The first step in getting a handle on these kinds of problems is finding k. Once you find k, you can easily find the population at any time t.

To find k, use the condition you already know: in one year, the population grows by one percent. That is, when t = 1, [itex]N = 1.01 N_0[/itex].

The equation is thus:

[tex]1.01 N_0 = N_0 e^{k \cdot 1}[/tex]

From which you should be able to readily calculate k.

To determine when the population doubles, use your known value of k and set the equation up in the following way:

[tex]2 N_0 = N_0 e^{kt}[/tex]

And solve for t.

- Warren
 
HallsofIvy
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Here's how I would do that problem: Saying that the annual growth rate is 1% means that the population, P, increases by .01P each year or to 1.01P each year: the population is multiplies by 1.01 each year so the formula is P(t)= P0(1.01)t. The population will have doubled when P(t)= 2P0= P0(1.01)t or 2= (1.01)t.

Taking logs of both sides, log(2)=tlog(1.01) or t=log(2)/log(1.01).

In Chroot's method, k= ln(1.01) so this is, in fact, exactly the same answer. (I used log instead of ln because any base will do.)
 

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