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Total water drained via Net Change Theorem

Water drains from a tank at a rate of r(t)=500−8tr(t) = 500 - 8t gallons per minute, where tt is measured in minutes since draining began.

Assuming r(t)≥0r(t) \geq 0 for all tt in the interval [0,60][0, 60], use the Net Change Theorem to find the total amount of water drained from the tank between t=0t = 0 and t=60t = 60 minutes. Give your answer in gallons.

Hint

The rate function r(t)r(t) gives gallons per minute, so the total amount drained over the interval is ∫060r(t) dt\int_0^{60} r(t)\, dt. Evaluate this definite integral using the Fundamental Theorem of Calculus.

Step-by-step solution

By the Net Change Theorem, the total water drained equals ∫060r(t) dt=∫060(500−8t) dt\int_0^{60} r(t)\, dt = \int_0^{60} (500 - 8t)\, dt, since the rate function accumulates to give total change (here, total volume drained). First find an antiderivative: ∫(500−8t) dt=500t−4t2+C\int (500 - 8t)\, dt = 500t - 4t^2 + C.

Now evaluate this antiderivative at the bounds: [500t−4t2]060=(500(60)−4(60)2)−(500(0)−4(0)2)[500t - 4t^2]_0^{60} = (500(60) - 4(60)^2) - (500(0) - 4(0)^2). This gives (30000−14400)−0=15600(30000 - 14400) - 0 = 15600. So a total of 15,60015{,}600 gallons of water drained from the tank.

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