Question

A cylindrical tank is full at time \(t=0\) when a valve in the bottom of the tank is opened. By Torricelli's law, the volume of water in the tank after \(t\) hours is \(V=100(200-t)^{2}\) measured in cubic meters. a. Graph the volume function. What is the volume of water in the tank before the valve is opened? b. How long does it take for the tank to empty? c. Find the rate at which water flows from the tank and plot the flow rate function. d. At what time is the magnitude of the flow rate a minimum? A maximum?

Short answer

b. How long does it take for the tank to empty completely? c. What is the flow rate function? d. At what time does the flow rate have a maximum and minimum magnitude? a) The initial volume of the tank is \(4 \times 10^6\) cubic meters. b) It takes 200 hours for the tank to empty completely. c) The flow rate function is \(\frac{dV}{dt} = -400(200-t)\). d) The flow rate has a maximum magnitude at \(t = 0\) and a minimum magnitude at \(t = 200\) hours.

Step-by-step solution

a. Graph the Volume Function and Initial Volume

The volume function is given by: \(V(t) = 100(200 - t)^2\) To find the initial volume of the tank at \(t = 0\): \(V(0) = 100(200 - 0)^2\) \(V(0) = 100(200)^2 = 4 \times 10^6\) cubic meters. Now to graph the function, we can choose some points, such as \(t = 50, 100,\) and \(150\) to see the behavior of the function: \(V(50) = 100(150)^2 = 2.25 \times 10^6\) cubic meters. \(V(100) = 100(100)^2 = 10^6\) cubic meters. \(V(150) = 100(50)^2 = 2.5 \times 10^5\) cubic meters. Plotting these points on a graph will give us a parabolic curve of the volume function where the volume decreases as time increases.

b. Time taken to empty the tank

To find out how long it takes for the tank to empty, we need to find the time when the volume function becomes zero. We set the volume function \(V(t)\) equal to zero and solve for \(t\): \(100(200 - t)^2 = 0\) \((200 - t)^2 = 0\) \(200 - t = 0\) \(t = 200\) So, it takes 200 hours to empty the tank completely.

c. Flow rate function

To find the flow rate function, we need to calculate the derivative of the volume function \(V(t)\) with respect to time \(t\): \(\frac{dV}{dt} = -400(200-t)\) Now we can plot the flow rate function by choosing some points: \(\frac{dV}{dt}(50) = -400(150) = -60000 \; \text{m}^3\text{/hr}\). \(\frac{dV}{dt}(100) = -400(100) = -40000 \; \text{m}^3\text{/hr}\). \(\frac{dV}{dt}(150) = -400(50) = -20000 \; \text{m}^3\text{/hr}\). These points show that the flow rate function is a linear function, decreasing with time.

d. Minimum and Maximum Flow Rate Magnitude

Since the flow rate function is a linear function of time, the magnitude of the flow rate function has a minimum when the flow rate is 0, which corresponds to the completely empty tank at \(t=200\) hours. The maximum magnitude of the flow rate occurs when the valve is opened at \(t = 0\), and its value is: \(\frac{dV}{dt}(0) = -400(200-0) = -80000 \; \text{m}^3\text{/hr}\). Thus, the flow rate has a maximum magnitude at \(t = 0\) and a minimum magnitude at \(t = 200\) hours.

Key concepts

Cylindrical Tank

A cylindrical tank is essentially a three-dimensional shape with two parallel circular bases connected by a curved surface at a constant distance, known as the height or depth of the cylinder. In practical applications, cylindrical tanks are often used for storing liquids because their shape allows for a uniform distribution of pressure. In this problem, we assume the tank is completely full at the start (time \( t = 0 \)).

The problem also references Torricelli's law, which relates to the fluid outflow under gravity from an orifice in such a tank. This principle allows us to examine how the water drains over time, rather than staying constant. The tank's design impacts how we calculate the volume and flow rate, leading us to think about the geometry when applying formulas—whether calculating capacity or evaluating outflow dynamics.

Volume Function

In calculus and physics, the volume function \( V(t) \) represents the volume of liquid remaining in a tank over time. In this scenario, the volume function is given by \( V(t) = 100(200 - t)^2 \). The initial volume—representing a full tank—is found by plugging in \( t = 0 \), which yields \( V(0) = 4 \times 10^6 \) cubic meters.

Graphically, \( V(t) \) is a parabolic function that decreases over time, indicating a reduction in volume as water drains out. You can imagine plotting a curve that starts at a peak (the full tank) and gradually declines until the tank is empty. The trajectory of this graph gives us important insights about real-life scenarios like how quickly the volume decreases over time, which relates directly to the rate at which the fluid exits the tank.

Flow Rate

The flow rate describes how fast the volume of liquid changes in the tank, essentially measuring the speed at which water exits. By finding the derivative of the volume function \( V(t) \), we obtain the flow rate function \( \frac{dV}{dt} = -400(200-t) \). This derivative tells us how quickly the water level decreases with respect to time.

As such, the flow rate changes linearly with time, as indicated by the linear expression. The negative sign shows the reduction of water, indicating that water is being lost. This function helps identify key points like when the flow rate is fastest—a critical insight for predicting when the tank will empty or just monitoring flow for operational needs.

• At \( t = 0 \): Flow rate is at maximum, \( -80000 \text{ m}^3/\text{hr} \)
• At \( t = 200 \): Flow rate is zero—the tank is empty

These points illustrate the flow rate's maximal and minimal values right before the tank is depleted.

Calculus

Calculus provides us the mathematical tools to investigate changes described in the problem, such as the variable flow rate for our cylindrical tank. It primarily involves differentiation, where we examine how functions change, allowing us to determine the flow rate from the volume function. Understanding derivatives is crucial in this context.

Differentiation of the volume function \( V(t) \) produces the flow rate function \( \frac{dV}{dt} \). This relates directly to how the water level changes over time. Calculus answers functional questions beyond simple arithmetic by enabling models of real-world scenarios regarding rates of change and growth of functions.

• Derivatives help us understand instantaneous rates, in this case, the tank's emptying rate.
• It provides insight into identifying times of peak and minimal flow during drainage.

As such, these concepts allow us to predict, analyze, and optimize liquid dynamics under certain conditions, such as those depicted in this exercise.