# calculus 2 word problem

• Aug 28th 2007, 08:33 PM
chocolatelover
calculus 2 word problem
Hi everyone,

Can you please tell me if this is correct?

A bacteria population starts with 400 bacteria and grows at a rate of r(t)=(450.268)e1.2567t bacteria per hour. How many bacteria will there be after three hours?

I did:

r(t)=ln450.268e1.2567(3)
=(1.2567)(3)ln450.268
=3.7701(6.10984261)

r(t)=23.034 bacteria

I'm not sure if I did it right. This seems too easy.

Thank you
• Aug 28th 2007, 10:46 PM
Rebesques
Quote:

This seems too easy.
Yeap, I use the same criterion to see what's right :o

Here you are given the rate of growth, and not the actual population. Call this N(t). The problem is of the form

$N'(t)=ce^{\lambda t}$

and sorry but I'm just too lazy to write down the constants :o So integrating this gives:

$N(t)-N(0)=\frac{ce^{\lambda t}}{\lambda}$

where N(0) is the initial population.
• Aug 28th 2007, 10:46 PM
CaptainBlack
Quote:

Originally Posted by chocolatelover
Hi everyone,

Can you please tell me if this is correct?

A bacteria population starts with 400 bacteria and grows at a rate of r(t)=(450.268)e1.2567t bacteria per hour. How many bacteria will there be after three hours?

I did:

r(t)=ln450.268e1.2567(3)
=(1.2567)(3)ln450.268
=3.7701(6.10984261)

r(t)=23.034 bacteria

I'm not sure if I did it right. This seems too easy.

Thank you

Let the number of bacteria at time $t$ be $n(t)$, then the question tells you that:

$
\frac{dn}{dt}=450.268 ~e^{1.2567~t}
$

so:

$
n(t)=\frac{450.268}{1.2567} ~e^{1.2567~t}+C
$

and as at $t=0\ \ n=400$ we have:

$
C=400-\frac{450.268}{1.2567}
$

RonL
• Aug 28th 2007, 10:48 PM
CaptainBlack
Quote:

Originally Posted by chocolatelover
Hi everyone,

Can you please tell me if this is correct?

A bacteria population starts with 400 bacteria and grows at a rate of r(t)=(450.268)e1.2567t bacteria per hour. How many bacteria will there be after three hours?

I did:

r(t)=ln450.268e1.2567(3)
=(1.2567)(3)ln450.268
=3.7701(6.10984261)

r(t)=23.034 bacteria

I'm not sure if I did it right. This seems too easy.

Thank you

As the rate of growth in numbers of bacteria is positive there must be