1. Part (a)
# of gallons entering tank 7
100t2 sin t dt =
8,264 gallons

2. This occurs on the intervals 0  t  1.617 and 3  t  5.076
Part (b)
1.617
5.076
f(t)
g(t)
The amount of H2O in the tank is decreasing when g(t)  f(t).
This occurs on the intervals 0  t  and 3  t  5.076

3. There are only 3 possible answers:
Part (c)
There are only 3 possible answers:
t=3
f(t)
g(t)
t=0
t=7
@ t=0, because the amount of water begins to decrease immediately.
@ t=3, because that’s when the graph of f(t) goes back below g(t) again.
@ t=7, because it’s the right endpoint, and f(t)  g(t) from to 7.

4. There are only 3 possible answers:
Part (c)
There are only 3 possible answers:
t=3
f(t)
5,000 gallons
g(t)
t=0
t=7
@ t=0, there are 5,000 gallons of water in the tank.

5. There are only 3 possible answers:
Part (c)
5,127 gallons
t=3
f(t)
5,000
5,000 gallons
g(t)
t=0
t=7
5, t=3
Now we’ll look at each interval:
1.617
(250 – 100t2 sin t ) dx
gallons lost
1.617 3
(100t2 sin t - 250) dx
gallons gained

6. There are only 3 possible answers:
Part (c)
5,127 gallons
t=3
f(t) 5,
-1,
5,000 gallons
g(t)
t=0
4,514 gallons
t=7
4, t=7
Now we’ll look at each interval: 3
5.076
(2000 – 100t2 sin t ) dx
1, gallons lost
5.076 7
(100t2 sin t ) dx
gallons gained