1. ## integral word problem

A population of bacteria is changing at a rate of dp/dt = (3000/e^3+3t) , where t is the time in minutes. If the initial population is 13000, find the population when t=5 minutes.

so i have to use substitution on this...i tried let u = 3t, but the problem didnt work out right on the quiz and he corrected it saying it should have been u = e^3 + 3t, which differentiates into du=3dt => (1/3)du=dt ? what happened to the e^3? doesnt e^3 = e^3 after you differentiate it?

2. Originally Posted by jeph
A population of bacteria is changing at a rate of dp/dt = (3000/e^3+3t) , where t is the time in minutes. If the initial population is 13000, find the population when t=5 minutes.

so i have to use substitution on this...i tried let u = 3t, but the problem didnt work out right on the quiz and he corrected it saying it should have been u = e^3 + 3t, which differentiates into du=3dt => (1/3)du=dt ? what happened to the e^3? doesnt e^3 = e^3 after you differentiate it?
That differencial equation makes no sense.
I think you are missing something.
Because you have,
$e^3$ maybe you mean,
$e^{3t}$

3. thats what it says....i took a picture

i just dont get how you make du/dt = e^3 + 3t into (1/3)du = dx. the e^3 just disappears...he even substituted the whole bottom line (e^3 + 3t) with u

4. What the grader wrote in red is absolutely correct.
Perhaps what you have misunderstood is that $e^3$ is a constant!
The derivative of $e^3 + 3t$ is just $3$.

5. oh...

that constant thing keeps getting me. this wasnt the first time i made this mistake =T