# Word problem help

• Nov 20th 2012, 11:57 AM
nubshat
Word problem help
One hundred grams of a certain chemical is removed from its airtight container and
immediately placed in a 20 gram dish on a sensitive scale. The scale reads 120 grams. The
chemical absorbs water from the air at a rate of 15e^-t grams per minute, where t is the
time in minutes that the chemical has been exposed to air.

a) What is the reading on the scale after t minutes?
b) What is the reading on the scale after a very long time has passed?

I am not asking you to do this for me, I just have no idea how to do it an help would be appreciated
• Nov 20th 2012, 12:09 PM
MarkFL
Re: Word problem help
Let $R(t)$ be the reading on the scale in grams after $t$ minutes. From the information provided, we may state the IVP:

$\frac{dR}{dt}=15e^{-t}$ where $R(0)=120$

Can you proceed?
• Nov 21st 2012, 03:50 PM
nubshat
Re: Word problem help
Thank you for the help
I got R(t)= 15e^-t + 135 for part a and 135 for part b
Can someone please confirm that this is the answer or help me figure out what I did wrong?
Thanks
• Nov 21st 2012, 04:10 PM
MarkFL
Re: Word problem help
You can confirm whether or not your solution to the IVP is correct by differentiating it to see if you get back to the differential equation, or at least check to see if the initial condition is met.
• Nov 21st 2012, 04:13 PM
skeeter
Re: Word problem help
$R(t) = 135 - 15e^{-t}$

$\lim_{t \to \infty} R(t) = 135$
• Nov 21st 2012, 04:24 PM
MarkFL
Re: Word problem help
Yes, using the boundaries and introducing dummy variables, we find:

$\int_{120}^{R(t)}\,du=15\int_0^te^{-v}\,dv$

$R(t)-120=-15(e^{-t}-1)$

$R(t)=135-15e^{-t}=15(9-e^{-t})$