Filling a tank A 2000-liter cistern is empty when water begins flowing into it (at t=0 at a rate (in L/min) given by Q′(t) = 3√t, where t is measured in minutes.


a. How much water flows into the cistern in 1 hour?

Verified step by step guidance

1. Identify the given rate of water flow into the cistern as a function of time: \(Q'(t) = 3\sqrt{t}\) liters per minute, where \(t\) is in minutes.
2. Recognize that \(Q'(t)\) represents the derivative of the volume of water \(Q(t)\) with respect to time, so to find the total volume of water that has flowed in by time \(t\), you need to integrate \(Q'(t)\) over the interval from 0 to \(t\).
3. Set up the definite integral to find the total volume of water that has flowed into the cistern in 1 hour (which is 60 minutes): \(\displaystyle Q(60) = \int_0^{60} 3\sqrt{t} \, dt\)
4. Rewrite the integrand \(3\sqrt{t}\) as \$3t^{1/2}$ to make integration straightforward.
5. Integrate \$3t^{1/2}\( with respect to \)t$ using the power rule for integration: \(\int 3t^{1/2} \, dt = 3 \cdot \frac{t^{3/2}}{\frac{3}{2}} + C = 2t^{3/2} + C\). Then evaluate this antiderivative from 0 to 60 to find the total volume of water that has flowed into the cistern in 1 hour.

Key Concepts

Definite Integral

A definite integral calculates the total accumulation of a quantity over an interval. In this problem, integrating the rate function Q′(t) from 0 to 60 minutes gives the total volume of water that has flowed into the cistern during that time.

Rate of Change and Accumulation

The rate function Q′(t) represents how fast water flows into the tank at any time t. Understanding that integrating this rate over time accumulates the total amount of water is essential to solving the problem.

Units and Time Conversion

Since the rate is given in liters per minute and time is in minutes, it is important to convert the time interval correctly (1 hour = 60 minutes) to ensure the integral limits match the units and yield a meaningful total volume.