Water flows from the bottom of a storage tank at a rate of r(t) = 300 − 6t liters per minute, where 0 ≤ t ≤ 50. Find the amount of water that flows from the tank during the first 30 minutes.
Question
Water flows from the bottom of a storage tank at a rate of r(t) = 300 − 6t liters per minute, where 0 ≤ t ≤ 50. Find the amount of water that flows from the tank during the first 30 minutes.
Solution
To find the amount of water that flows from the tank during the first 30 minutes, we need to integrate the rate function from 0 to 30.
The integral of r(t) from 0 to 30 gives the total amount of water that flows out of the tank during the first 30 minutes.
The integral ∫ from 0 to 30 of (300 - 6t) dt represents the total amount of water that flows out of the tank from time t=0 to t=30.
To solve the integral, we first find the antiderivative of 300 - 6t. The antiderivative of a constant is the constant times the variable, and the antiderivative of a t is (1/2)t^2. So, the antiderivative of 300 - 6t is 300t - 3t^2.
We evaluate this from 0 to 30:
[300(30) - 3(30)^2] - [300(0) - 3(0)^2] = 9000 - 2700 = 6300 liters.
So, 6300 liters of water flow out of the tank during the first 30 minutes.
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