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0313phd
06-03-2011, 11:54 AM
This is an exponential growth problem that I used to be able to figure out, but I did not keep notes on it, and in review I am stumped:

A colony of bacteria starts with two bacteria at noon. If the number of bacteria triples every thirty minutes, how many bacteria will be present at 3:00pm the same day.

The equation looks like: Q at 3:00pm= Qat 12:00 to the Kt power
t is the time or 6half-hour intervals. K is the constant that would triple the population or 3. The answer I am getting is incorrect.

JeffM
06-03-2011, 01:26 PM
This is an exponential growth problem that I used to be able to figure out, but I did not keep notes on it, and in review I am stumped:

A colony of bacteria starts with two bacteria at noon. If the number of bacteria triples every thirty minutes, how many bacteria will be present at 3:00pm the same day.

The equation looks like: Q at 3:00pm= Qat 12:00 to the Kt power
t is the time or 6half-hour intervals. K is the constant that would triple the population or 3. The answer I am getting is incorrect.
Memorizing is not necessary although of course it saves time.

When you can't remember, induction is one technique that substitutes for memory whenever a problem involves a variable expressed solely in integers. You already know that k = 3 should appear somewhere, and the initial value of 2 should also be relevant as should the value of t So let's look for a pattern that has 2, 3 and t in it.
Let t = time measured in units of 30 minutes with t = 0 at noon.
Let b be the number of bugs.
t = 0, b = 2 = 2 * 1 = 2 * 3[sup:343v29og]0[/sup:343v29og]. Remember in induction that you can always replace 1 with z[sup:343v29og]0[/sup:343v29og], where z is any real, or (2z - 1), where z = 1.
t = 1, b = 3 * 2 = 6 = 2 * 3 = 2 * 3[sup:343v29og]1[/sup:343v29og].
t = 2, b = 3 * 6 = 18 = 2 * 9 = 2 * 3[sup:343v29og]2[/sup:343v29og].

It seems that: b = 2 * (3[sup:343v29og]t[/sup:343v29og]). If that does not jog your memory, you can validate it for yourself by mathematical induction (which despite its name is a deductive argument).

SO b = 2 * (3[sup:343v29og]6[/sup:343v29og]) = 1,458.

Generalizing, y[sub:343v29og]t[/sub:343v29og] = x[sub:343v29og]0[/sub:343v29og] * (k[sup:343v29og]t[/sup:343v29og])
Or log(y[sub:343v29og]t[/sub:343v29og]) = log(x[sub:343v29og]0[/sub:343v29og]) + [t * log(k)]

Need more review or do you have it solid now?

0313phd
06-03-2011, 02:17 PM
Thank you, but I have one question. How did you know to multiply 3 to the 6th power by 2. I know 2 is what we start out with, but how did you know to multiply by 2. 0313

Denis
06-03-2011, 03:32 PM
Look up geometric series: formula for nth term
Here's a site: http://www.mathguide.com/lessons/SequenceGeometric.html

JeffM
06-03-2011, 04:57 PM
Thank you, but I have one question. How did you know to multiply 3 to the 6th power by 2. I know 2 is what we start out with, but how did you know to multiply by 2. 0313

I knew because (a) the initial value (what I called x[sub:3dw0u2b3]0[/sub:3dw0u2b3] in my generalization and what I said was b when t = 0) has to influence the final result. If the initial value had been 5 instead of 2, don't you think the final answer would have been different? I also knew because it grew out of the induction. Let's go back and look at the induction again.

Let b[sub:3dw0u2b3]t[/sub:3dw0u2b3] be the number of bugs at time t.

b[sub:3dw0u2b3]0[/sub:3dw0u2b3] = 2 (that is GIVEN).
The increase per unit of time = k = 3. (That too is GIVEN).
So, b[sub:3dw0u2b3]1[/sub:3dw0u2b3] = k * b[sub:3dw0u2b3]0[/sub:3dw0u2b3] = 3 * 2 = 6.
And b[sub:3dw0u2b3]2[/sub:3dw0u2b3] = k * b[sub:3dw0u2b3]1[/sub:3dw0u2b3] = 3 * 6 = 18.

Now I want a formula that contains t, k, and b[sub:3dw0u2b3]0[/sub:3dw0u2b3] because a change in any one of them will presumably change the answer.

Let's start with b[sub:3dw0u2b3]0[/sub:3dw0u2b3] = 2.

b[sub:3dw0u2b3]0[/sub:3dw0u2b3] = 2.
b[sub:3dw0u2b3]1[/sub:3dw0u2b3] = 6 = 2 * 3.
b[sub:3dw0u2b3]2[/sub:3dw0u2b3] = 18 = 2 * 9.
There are other possible ways to use 2 but this is a very natural one.

Now how about using k = 3.

b[sub:3dw0u2b3]2[/sub:3dw0u2b3] = 2 * 9 = 2 * 3 * 3.
b[sub:3dw0u2b3]1[/sub:3dw0u2b3] = 2 * 3.
b[sub:3dw0u2b3]0[/sub:3dw0u2b3] = 2 * 1, which is not 3, but I know 3[sup:3dw0u2b3]0[/sup:3dw0u2b3] = 1.
So b[sub:3dw0u2b3]0[/sub:3dw0u2b3] = 2 * 3[sup:3dw0u2b3]0[/sup:3dw0u2b3].

Still have not used t except in the case of b[sub:3dw0u2b3]0[/sub:3dw0u2b3] = 2 * 3[sup:3dw0u2b3]0[/sup:3dw0u2b3], but let's see if the pattern extends.
b[sub:3dw0u2b3]1[/sub:3dw0u2b3] = 2 * 3 = 2 * 3[sup:3dw0u2b3]1[/sup:3dw0u2b3].
b[sub:3dw0u2b3]2[/sub:3dw0u2b3] = 2 * 3 * 3 = 2 * 3[sup:3dw0u2b3]2[/sup:3dw0u2b3].

Now, the pattern b[sub:3dw0u2b3]0[/sub:3dw0u2b3] * k[sup:3dw0u2b3]t[/sup:3dw0u2b3] = 2 * 3[sup:3dw0u2b3]t[/sup:3dw0u2b3] jumps out. Furthermore, it is a common pattern in math, the geometric series, as Denis points out. Furthermore, its applicability to this case can be formally demonstrated quickly by mathematical induction if you still feel a qualm.

Clearer?

0313phd
06-04-2011, 07:38 PM
Thank you. Believe it or not I was going over geometric sequences today, and I thought that is exactly what the bacteria problem was. THANKS 0313