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Chapter 8: Problem 34

Consider the tank problem in Example \(6 .\) For the following parameter values, find the water height function. Then determine the approximate time at which the tank is first empty and graph the solution. $$H=2.25 \mathrm{m}, A=2 \mathrm{m}^{2}, a=0.5 \mathrm{m}^{2}$$

Short Answer

Expert verified
Answer: The tank will become empty at approximately 12 hours.

Step by step solution

01

Formulate the differential equation

To set up the differential equation, we can use the conservation of mass principle. The rate of change of the volume of water in the tank must be equal to the rate at which the water flows out. Therefore, we have: $$A \frac{dH}{dt}=-a\sqrt{H}$$ Where \(A\) is the area of the tank, \(H\) is the height of the water, \(a\) is the area of the outlet, and \(t\) is the time. For the given parameter values, \(A = 2 \mathrm{m}^{2}\), \(a = 0.5 \mathrm{m}^{2}\), and the initial height \(H_0 = 2.25 \mathrm{m}\).
02

Solve the differential equation

Separate the variables as follows: $$\frac{dH}{\sqrt{H}}=-\frac{a}{A}dt$$ Integrate both sides: $$\int \frac{1}{\sqrt{H}} dH=-\frac{a}{A}\int dt$$ Using substitution, let \(u=\sqrt{H}\), then \(du=\frac{1}{2\sqrt{H}}dH\). The integral becomes: $$\int 2u du =-\frac{a}{A}\int dt$$ Now integrate: $$u^2+C_1=-\frac{at}{A}+C_2$$ Replace \(u=\sqrt{H}\) back: $$H(t)=\left(-\frac{at}{2A}+C\right)^2$$ Retrieve the constants using initial value \(H_0=2.25\mathrm{m}\) at \(t=0\): $$2.25=\left(C\right)^2$$ We get \(C=\pm 1.5\mathrm{m}\). Since the height of the tank cannot be negative, we use the positive value: $$H(t)=\left(-\frac{at}{2A}+1.5\right)^2$$ Substitute the values of \(a\) and \(A\): $$H(t)=\left(-\frac{0.5t}{4}+1.5\right)^2$$
03

Find the time when the tank is empty

To find the time when the tank is empty, set the height function equal to zero: $$0=\left(-\frac{0.5t}{4}+1.5\right)^2$$ Solve for t: $$t=\frac{6}{0.5}$$ $$t=12 \,\text{hours}$$ The tank becomes empty at approximately 12 hours.
04

Graph the solution

To graph the solution, plot the height function \(H(t)=\left(-\frac{0.5t}{4}+1.5\right)^2\) over the interval [0, 12]. The graph will show how the height of the water in the tank decreases over time until it reaches zero at approximately 12 hours.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Separation of Variables
The technique of separation of variables is a fundamental method used to solve differential equations. It involves rearranging the equation so that each side contains a different variable. This allows us to integrate both sides separately, making it easier to find the solution. In the context of the tank problem, we started with a differential equation representing the rate of change of water height in the tank:
  • Initial equation: \(A \frac{dH}{dt} = -a\sqrt{H}\)
  • Rearranged for separation: \(\frac{dH}{\sqrt{H}} = -\frac{a}{A} dt\)
By separating variables, you've positioned all terms involving the water height \(H\) on one side and the time \(t\) on the other. This makes it feasible to perform integration on each side separately. Integrating both sides gives us a relationship between \(H\) and \(t\). The separation of variables is particularly useful here because it simplifies complex integrals, making the task manageable.
Initial Value Problem
An initial value problem involves solving a differential equation starting from a specified initial condition. In this case, the initial condition given was the height of the water in the tank at \(t=0\).
  • Initial height: \(H_0 = 2.25\, \mathrm{m}\)
To find the solution to a differential equation that satisfies this initial condition, after integration, we determine the particular constants using the initial values.
In the solution, after integrating both sides of the separated equation, we had:
  • \(H(t)=\left(-\frac{at}{2A}+C\right)^2\).
Replacing \(t=0\) with the initial height \(H_0=2.25\), we find the constant \(C\). The calculated \(C\) helps define the specific solution that satisfies our initial condition, ensuring it reflects the tank's physical state at the start.
Conservation of Mass
The principle of the conservation of mass is crucial in formulating the differential equation for the tank problem. It implies that the mass entering a system must be equal to the mass exiting, plus any accumulation in the system over time.
For the tank, the rate at which water flows out is equal to the change in height over time. Since no water is entering and only exiting the tank through the outlet, this principle simplifies to:
  • Rate of decrease in water volume = rate of flow out through the outlet
  • Expressed mathematically: \(A \frac{dH}{dt} = -a\sqrt{H}\)
This equation directly results from ensuring that mass (or, here, volume) is conserved. By using the areas of the tank and outlet, the differential equation represents the change in height due to outflow, illustrating the practical application of conservation laws in everyday scenarios.

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Most popular questions from this chapter

Determine whether the following equations are separable. If so, solve the initial value problem. $$y^{\prime}(t)=\frac{e^{t}}{2 y}, y(\ln 2)=1$$

Consider the first-order initial value problem \(y^{\prime}(t)=a y+b, y(0)=A,\) for \(t \geq 0,\) where \(a, b,\) and \(A\) are real numbers. a. Explain why \(y=-b / a\) is an equilibrium solution and corresponds to a horizontal line in the direction field. b. Draw a representative direction field in the case that \(a>0\) Show that if \(A>-b / a,\) then the solution increases for \(t \geq 0\) and if \(A<-b / a,\) then the solution decreases for \(t \geq 0\). c. Draw a representative direction field in the case that \(a<0\) Show that if \(A>-b / a,\) then the solution decreases for \(t \geq 0\) and if \(A<-b / a,\) then the solution increases for \(t \geq 0\).

Suppose Euler's method is applied to the initial value problem \(y^{\prime}(t)=a y, y(0)=1,\) which has the exact solution \(y(t)=e^{a t} .\) For this exercise, let \(h\) denote the time step (rather than \(\Delta t\) ). The grid points are then given by \(t_{k}=k h .\) We let \(u_{k}\) be the Euler approximation to the exact solution \(y\left(t_{k}\right),\) for \(k=0,1,2, \ldots\) a. Show that Euler's method applied to this problem can be written \(u_{0}=1, u_{k+1}=(1+a h) u_{k},\) for \(k=0,1,2, \ldots\) b. Show by substitution that \(u_{k}=(1+a h)^{k}\) is a solution of the equations in part (a), for \(k=0,1,2, \dots\) c. \(\lim _{h \rightarrow 0}(1+a h)^{1 / h}=e^{a} .\) Use this fact to show that as the time step goes to zero \(\left(h \rightarrow 0, \text { with } t_{k}=k h \text { fixed }\right),\) the approximations given by Euler's method approach the exact solution of the initial value problem; that is, \(\lim _{h \rightarrow 0} u_{k}=\lim _{h \rightarrow 0}(1+a h)^{k}=y\left(t_{k}\right)=e^{a t_{k}}\).

A special class of first-order linear equations have the form \(a(t) y^{\prime}(t)+a^{\prime}(t) y(t)=f(t),\) where \(a\) and \(f\) are given functions of \(t.\) Notice that the left side of this equation can be written as the derivative of a product, so the equation has the form $$a(t) y^{\prime}(t)+a^{\prime}(t) y(t)=\frac{d}{d t}(a(t) y(t))=f(t)$$ Therefore, the equation can be solved by integrating both sides with respect to \(t.\) Use this idea to solve the following initial value problems. $$e^{-t} y^{\prime}(t)-e^{-t} y=e^{2 t}, y(0)=4$$

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