# Growth Rate Problem

1. Oct 11, 2008

### Swerting

The problem states that the number of bacteria in a culture is proportional to the number present. The culture has 10,000 bacteria initially, 20,000 at time t1 minutes, and 100,000 bacteria at time (t1+10) minutes.
I am asked to find the number of bacteria at time t in terms of t only.
The rate of growth equation is $$y=ce^{kt}$$.

I understand c to be the initial population, 10,000, and y to be the population at time t, which makes sense because according to this logic, y=10,000 at time 0. Over 10 minutes the population seems to grow 80,000, so I divided 80,000 by 10 and got 8,000 as a rate. I plugged this in as my 'k' value and solved for t1. When I plugged this into the formula when y=100,000, the two values weren't equal (the value on the right side of the equation was so big I couldn't even compute it). I have been trying to figure out other ways to calcuate 'k', but am unable to since the time t1 was not specified. I appreciate any help.

2. Oct 11, 2008

### symbolipoint

You seem on the correct track. Start this as y=20000 at a new start time, and that y=100000 at ten minutes from that start time (meaning when t=10).

3. Oct 11, 2008

### jacksonpeeble

This has the same format as the popular Pert formula for continuous compound interest. It appears as though you know how to use it, so I take it the issue is finding the value for variable k.

As k is in the exponent, the way you went about solving for it is absolutely incorrect. Remember that exponents impact the answer in a big way - had the equation been linear, it would be a different story.

The problem is certainly solvable. Try making a table and comparing the differences, then the differences of the differences. The "differences of the differences" should be the same.

I'll check in after a while to see how you're doing. I assume you have until Monday?

4. Oct 11, 2008

### Swerting

Thank you all very much! I got it now!
With your help, I was able to get the idea of substitution into my mind, putting 'k' into terms of t1. From this, I was able to plug it into the 100,000 equation, solve for t1, and then solve for k. When I plugged these values into BOTH equations, the 100,000 and 20,000 equation (where y equals those values), it worked perfectly!
Again, thanks for the help! I appreciate it!

5. Oct 12, 2008

### Mentallic

Actually this doesn't work for exponents. The differences of the differences of the diff... etc. are always increasing at the rate the exponential increases.

For linear patterns, n, the difference is always equal.
For quadratic patterns, n2, the difference of the difference is always equal.
For cubic patterns, n3, the diff. of the diff. of the diff. is always equal... etc.