There is no snow on Janet's driveway when snow begins to fall at midnight. From midnight to 9 am, snow accumulates on the driveway at a rate modeled by f(t)=7t(e^cost) cubic feet per hour, where t is measured in hours since midnight. Janet starts removing snow at 6 am (t=6). The rate g(t), in cubic feet per hour at which Janet removes snow from the driveway at time t hours after midnight is modeled by:
g(t) =
0 0≤t

Question

There is no snow on Janet's driveway when snow begins to fall at midnight. From midnight to 9 am, snow accumulates on the driveway at a rate modeled by f(t)=7t(e^cost) cubic feet per hour, where t is measured in hours since midnight. Janet starts removing snow at 6 am (t=6). The rate g(t), in cubic feet per hour at which Janet removes snow from the driveway at time t hours after midnight is modeled by: 
g(t) = 
0          0≤t<6
125      6≤t<7
108      7≤t≤9
A) Find the rate of change of the volume of snow on the driveway at 8am?
B) How many cubic feet of snow are on the driveway at 9am?

anonymous · Answer

Okay let's start with A).

anonymous · Answer

The rate of change of the volume os snow is going to be the rate is falling $f(t)$, minus the rate it is being removed $g(t)$.   So they want $f(t)-g(t)$ but at the time 8 am, $t=8$.

anonymous · Answer

So A) is a pretty simple one.  Just plug in the numbers.

anonymous · Answer

B) seems to just be asking for $$ \Large 
\int_0^9 f(t)-g(t) dt
$$

anonymous · Answer

but also rememberthat g(t) is a piece-wise function...

anonymous · Answer

So it's an improper integral. $$ \Large
\int_0^6 f(t)-g(t)dt + \int_6^7f(t)-g(t)dt + \int_7^9f(t)-g(t)dt
$$