# integral word problem

• Dec 18th 2006, 11:34 AM
jeph
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?
• Dec 18th 2006, 01:43 PM
ThePerfectHacker
Quote:

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,
\$\displaystyle e^3\$ maybe you mean,
\$\displaystyle e^{3t}\$
• Dec 18th 2006, 02:45 PM
jeph
thats what it says....i took a picture

http://img352.imageshack.us/img352/7...0082kt3.th.jpg

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
• Dec 18th 2006, 03:16 PM
Plato
What the grader wrote in red is absolutely correct.
Perhaps what you have misunderstood is that \$\displaystyle e^3\$ is a constant!
The derivative of \$\displaystyle e^3 + 3t\$ is just \$\displaystyle 3\$.
• Dec 18th 2006, 03:43 PM
jeph
oh...

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