Question

Torricelli's law states that for a water tank with a hole in the bottom that has a cross-section of \(A\) and with a height of water \(h\) above the bottom of the tank, the rate of change of volume of water flowing from the tank is proportional to the square root of the height of water, according to \(\frac{d V}{d t}=-A \sqrt{2 g h}\), where \(g\) is the acceleration due to gravity. Note that \(\frac{d V}{d t}=A \frac{d h}{d t} .\) Solve the resulting initial-value problem for the height of the water, assuming a tank with a hole of radius \(2 \mathrm{ft}\). The initial height of water is \(100 \mathrm{ft}\)

Short answer

The height of water is \( h(t) = \left(-\frac{\sqrt{2g}}{2} t + 10\right)^2 \).

Step-by-step solution

Express Unknowns and Known Quantities

Given the problem, we need to solve for the height of the water, \( h(t) \), in the tank over time. We know that \( \frac{dV}{dt} = -A \sqrt{2gh} \) and \( \frac{dV}{dt} = A \frac{dh}{dt} \). The radius of the hole is \(2 \text{ ft}\), so the cross-sectional area \(A\) of the hole is \( A = \pi (2)^2 = 4\pi \). The initial height, \(h_0\), is \(100 \text{ ft}\).

Set Up Differential Equation

Using Torricelli's law and the relationship \( \frac{dV}{dt} = A \frac{dh}{dt} \), substitute \( -A \sqrt{2gh} \) into the second equation: \[ A \frac{dh}{dt} = -A \sqrt{2gh} \] This simplifies to \( \frac{dh}{dt} = -\sqrt{2gh} \).

Separate Variables

To solve the differential equation \( \frac{dh}{dt} = -\sqrt{2gh} \), we first separate the variables: \[ \frac{dh}{\sqrt{h}} = -\sqrt{2g} dt \]

Integrate Both Sides

Integrate \( \frac{dh}{\sqrt{h}} = -\sqrt{2g} dt \):\[\int \frac{dh}{\sqrt{h}} = \int -\sqrt{2g} \, dt\]The integral of \( \frac{1}{\sqrt{h}} \) is \( 2\sqrt{h} \), so we get:\[2\sqrt{h} = -\sqrt{2g} t + C\] where \(C\) is a constant of integration.

Solve for Constant of Integration

Use the initial condition that the initial height \(h(0) = 100\) ft. Substitute into the equation:\[2\sqrt{100} = -\sqrt{2g} (0) + C \]Simplifies to: \[ C = 20 \]

Solve for Height as a Function of Time

Substitute \(C = 20\) back into the equation:\[2\sqrt{h} = -\sqrt{2g} t + 20\]Solve for \(h\):\[\sqrt{h} = -\frac{\sqrt{2g}}{2} t + 10\]\(h = \left(-\frac{\sqrt{2g}}{2} t + 10\right)^2\)

Key concepts

Water Flow

In the context of Torricelli's Law, understanding water flow is essential. Water flow refers to the movement of water out of a container, such as a tank through a hole, and it is a phenomenon influenced by various forces, notably gravity.
Gravity plays a crucial role in water flow by pulling the water downwards, accelerating its movement through the hole. In this particular exercise, the flow rate or the rate at which water is flowing out is dictated by Torricelli's Law. This specific law indicates that the flow rate is proportional to the square root of the height of water above the hole.

• The flow rate decreases as the height of water decreases because less gravitational potential energy is converted to kinetic energy as water height lowers.
• Water flow can be affected by the size and shape of the hole, velocity of exit, and pressure.

When analyzing a water flow situation, like the one stated in the original exercise, the hole's radius and the initial height of water are key parameters. The larger the radius, the faster the flow, given that more water can exit at the same time.

Differential Equations

Differential equations are mathematical equations that involve derivatives, which represent rates of change. In this exercise, the differential equation is used to describe how the height of water in the tank changes over time with reference to Torricelli's Law.
Mathematically, this is conveyed through the equation: \( \frac{dh}{dt} = -\sqrt{2gh} \)This is a first-order ordinary differential equation (ODE) because it involves only the first derivative of the height function, \( h(t) \). Differential equations like this one, not only express the relationship between the changes in height and time, but also allow us to predict the behavior of physical systems such as fluid motion.

• They are broadly used in physics and engineering to model real-world situations.
• Solve to find unknown functions that describe a particular behavior or process over time.
• Often need initial conditions (starting facts) to find specific solutions, such as the water’s initial height being 100 feet here.

Through understanding differential equations, we can delve deeper into analyzing dynamic systems and predicting their future states.

Initial Value Problem

The initial value problem (IVP) is a type of differential equation that, in addition to the equation itself, provides specific information about the value of the unknown function at a certain point, usually at the initial time. This forms a complete problem setup enabling us to find a unique solution.
In this scenario, we are tasked with determining how the height of the water, \( h(t) \), varies over time given the initial height of 100 feet. We phrase this through:

• The differential equation: \( \frac{dh}{dt} = -\sqrt{2gh} \)
• The initial condition: \( h(0) = 100 \text{ ft} \)

The initial condition is critical as it anchors our solution, ensuring that the function describing the water height starts from the known value and evolves accurately over time according to the governing differential equation.
Solving an initial value problem involves finding the function that satisfies both the differential equation and the initial condition, providing insights into how the system behaves initially and subsequently.

Separation of Variables

Separation of variables is a method for solving differential equations. It involves rearranging an equation so that each variable and its differentials appear on opposite sides of the equation. Once separated, each side can be independently integrated to find a function relating the variables.
For the exercise at hand, we are given the differential equation:\( \frac{dh}{dt} = -\sqrt{2gh} \)To use separation of variables here, we write:\( \frac{dh}{\sqrt{h}} = -\sqrt{2g} \, dt \)This separation allows us to integrate each side individually:

• The left side with respect to \( h \)
• The right side with respect to \( t \)

Integration provides a solution linking \( h \) and \( t \) with the constant of integration, determined by the initial condition. In our specific example, this yields the equation:\( 2\sqrt{h} = -\sqrt{2g}t + C \)Separation of variables is a fundamental technique, especially useful for solving ordinary differential equations where the equation can be manipulated into a separable form.