Problem 64 - Calculus Early Transcendentals

Question

An open cylindrical tank initially filled with water drains through a hole in the bottom of the tank according to Torricelli's Law (see figure). If \(h(t)\) is the depth of water in the tank for \(t \geq 0,\) then Torricelli's Law implies \(h^{\prime}(t)=2 k \sqrt{h}\), where \(k\) is a constant that includes the acceleration due to gravity, the radius of the tank, and the radius of the drain. Assume that the initial depth of the water is \(h(0)=H\). a. Find the general solution of the equation. b. Find the solution in the case that \(k=0.1\) and \(H=0.5 \mathrm{m}\). c. In general, how long does it take the tank to drain in terms of \(k\) and \(H ?\)

Step-by-Step Solution

Answer

Answer: The general solution is \(h(t) = 2kt + H\). For the case when \(k=0.1\) and \(H=0.5m\), the height function is \(h(t) = 0.2t + 0.5\), and the time required for the tank to drain completely is \(t = \frac{-H}{2k} = 2.5s\).

Step 1: Rewrite the differential equation as a separable equation

To rewrite the differential equation as a separable equation, we want to have \(h\) and its derivatives on one side of the equation and \(t\) and its derivatives on the other side. Divide both sides by \(\sqrt{h}\) to get the separable equation: \(\frac{dh}{dt} = \frac{2k\sqrt{h}}{\sqrt{h}} \Rightarrow \frac{dh}{dt} = 2k\)

Step 2: Integrate both sides with respect to \(t\)

Integrate both sides of \(\frac{dh}{dt} = 2k\) with respect to \(t\) to get: \(\int \frac{dh}{dt} dt = \int 2k dt \Rightarrow h(t) = 2kt + C\)

Step 3: Find the general solution

Using the initial condition \(h(0) = H\), substitute \(t=0\) into the expression for \(h(t)\): \(H = 2k(0) + C \Rightarrow C = H\). The general solution is \(h(t) = 2kt + H\).

Step 4: Find the solution for given values of \(k\) and \(H\)

For the case when \(k=0.1\) and \(H=0.5m\), substitute these values into the general solution: \(h(t) = 2(0.1)t + 0.5 \Rightarrow h(t) = 0.2t + 0.5\).

Step 5: Find the time it takes for the tank to drain

To find the time required for the tank to empty, set \(h(t) = 0\): \(0 = 2kt + H\). Solving for \(t\), we get: \(t = \frac{-H}{2k}\) as the time required for the tank to drain completely in terms of \(k\) and \(H\).

Key Concepts

Differential Equations

Differential equations are powerful tools used to describe various physical phenomena, like the flow of water from a tank, which involves rates of change. Specifically, a differential equation is an equation that relates a function with one or more of its derivatives.

In the context of Torricelli's Law, we deal with a first-order ordinary differential equation (ODE) since it involves the first derivative of the water height, denoted as \( h'(t) \). The equation \( h^{\textprime}(t) = 2k\textbackslash\textbackslash\textbackslash\textbackslash\textsquare{\textbackslash\textbackslash\textbackslash\textbackslash\textsubscript{h}} \) models the rate at which the height of the water in the tank decreases over time. In this case, the independent variable is time \( t \), and the dependent variable is the water height \( h \). Understanding differential equations is crucial because it allows us to predict how systems evolve over time.

Separable Equations

Separable equations, a subset of differential equations, are those that can be manipulated algebraically to express the equation in a form where one side depends only on \( x \) (or the independent variable, like \( t \) in our case), and the other side depends only on \( y \) (or the dependent variable, like \( h \)).

By doing so, as shown in the solution provided, the equation becomes \( \frac{dh}{dt} = 2k \), allowing the variables to be separated and integrated individually. This separation simplifies the process of solving the differential equation, transforming it into a format where standard calculus integration techniques apply. Solving a separable equation often leads to an explicit function or a general solution, defining the relationship between the variables.

Calculus Integration

Calculus integration is the process of finding the integral of a function, essentially determining the quantity where the rate of change (the derivative) is known. When solving differential equations, we often integrate to find a function that describes the system in question.

In the exercise, we integrated both sides of the separable equation with respect to \( t \) to solve for the function \( h(t) \). This process is akin to 'adding up' the small changes in height over time to find the total change in height. It involves finding the antiderivative, which, in the case of constant rate \( 2k \), is simply \( 2kt+C \), where \( C \) is the constant of integration reflecting the initial condition of the system.

Initial Value Problems

Initial value problems (IVPs) are a specific type of differential equation where the solution is required to satisfy given preliminary conditions. In physics and engineering, these conditions typically represent the state of a system at the onset of observation or the beginning of the experiment.

In our exercise, the initial condition is stated as \( h(0) = H \), where \( H \) is the initial depth of water. By applying this condition to the general solution \( h(t) = 2kt + C \), we can solve for the constant \( C \). As it turns out, \( C = H \), which then fully defines the function \( h(t) \) over time. IVPs are essential for predicting future states of a system, allowing us to utilize general solutions in concrete scenarios.