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.


c. When will the tank be full?

Verified step by step guidance

1. Identify the rate of change of the volume of water in the tank, which is given by the function \(Q'(t) = 3\sqrt{t}\). This represents the inflow rate in liters per minute at time \(t\) minutes.
2. To find the total volume of water \(Q(t)\) that has flowed into the tank by time \(t\), integrate the rate function \(Q'(t)\) with respect to \(t\): \[Q(t) = \int 3\sqrt{t} \, dt = \int 3t^{1/2} \, dt.\]
3. Perform the integration by applying the power rule for integrals: \[Q(t) = 3 \times \frac{2}{3} t^{3/2} + C = 2 t^{3/2} + C,\] where \(C\) is the constant of integration.
4. Use the initial condition that the tank is empty at \(t=0\), so \(Q(0) = 0\). Substitute \(t=0\) into the integrated function to solve for \(C\): \[0 = 2 \times 0^{3/2} + C \implies C = 0.\] Thus, the volume function simplifies to \[Q(t) = 2 t^{3/2}.\]
5. Set the volume function equal to the tank's capacity to find when the tank is full: \[2000 = 2 t^{3/2}.\] Solve this equation for \(t\) to determine the time at which the tank reaches 2000 liters.

Key Concepts

Definite Integral as Accumulated Quantity

The definite integral of a rate function over time gives the total accumulated quantity. Here, integrating the flow rate Q′(t) from 0 to t will yield the total volume of water that has entered the tank by time t.

Solving for Time Using Integral Equations

To find when the tank is full, set the integral of the flow rate equal to the tank's capacity (2000 liters) and solve for t. This involves evaluating the integral and then isolating t in the resulting equation.

Integration of Power Functions

The flow rate Q′(t) = 3√t can be rewritten as 3t^(1/2). Integrating power functions involves increasing the exponent by one and dividing by the new exponent, which is essential to find the volume function Q(t).