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

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  1.617 and 3  t  5.076

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 5.076 to 7.

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.

There are only 3 possible answers: Part (c) 5,127 gallons t=3 f(t) 5,000 -280.777 +407.368 5,000 gallons g(t) t=0 t=7 5,126.591 gallons @ t=3 Now we’ll look at each interval: 1.617 (250 – 100t2 sin t ) dx 280.777 gallons lost 1.617 3 (100t2 sin t - 250) dx 407.368 gallons gained

There are only 3 possible answers: Part (c) 5,127 gallons t=3 f(t) 5,126.591 -1,109.710 +496.926 5,000 gallons g(t) t=0 4,514 gallons t=7 4,513.807 gallons @ t=7 Now we’ll look at each interval: 3 5.076 (2000 – 100t2 sin t ) dx 1,109.710 gallons lost 5.076 7 (100t2 sin t - 2000) dx 496.926 gallons gained