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.


b. Find the function that gives the amount of water in the tank at any time t≥0.

Verified step by step guidance

1. Identify the given rate of change of the amount of water in the tank, which is the derivative of the volume function: \(Q'(t) = 3\sqrt{t} = 3t^{1/2}\) liters per minute.
2. Recognize that to find the amount of water in the tank at any time \(t\), denoted by \(Q(t)\), you need to integrate the rate function \(Q'(t)\) with respect to \(t\).
3. Set up the integral: \(Q(t) = \int Q'(t) \, dt = \int 3t^{1/2} \, dt\).
4. Perform the integration using the power rule for integrals: \(\int t^{n} \, dt = \frac{t^{n+1}}{n+1} + C\), where \(C\) is the constant of integration.
5. Apply the initial condition that the tank is empty at \(t=0\), so \(Q(0) = 0\), to solve for the constant \(C\) and write the final expression for \(Q(t)\).

Key Concepts

Understanding the Rate of Change

The rate of change function Q′(t) represents how quickly the amount of water in the tank changes over time. In this problem, Q′(t) = 3√t gives the instantaneous flow rate in liters per minute, which is essential for determining the total volume of water at any time.

Definite and Indefinite Integration

Integration is used to find the total accumulated quantity from a rate function. By integrating Q′(t) with respect to time, we obtain the function Q(t) that represents the total amount of water in the tank at time t, starting from zero at t=0.

Initial Conditions and Constants of Integration

When integrating a rate function, an arbitrary constant appears. The initial condition, here that the tank is empty at t=0, allows us to solve for this constant, ensuring the solution accurately reflects the physical situation.