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Chapter 9: Problem 48

Draining tanks Consider the tank problem in Example 7. 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
Based on the given parameters (initial height H = 2.25 m, cross-sectional area A = 2 m虏, and draining area a = 0.5 m虏), the height function of the water in the tank over time is: $$h(t)=(1-\frac{a\sqrt{2g}}{2A}t)^{2}$$ It takes approximately 2.89 seconds for the tank to empty completely. The graph of the height function vs. time would be a parabolic curve starting at the initial height of 2.25 meters and decreasing until it reaches 0 at around 2.89 seconds.

Step by step solution

01

Find the rate of height change using Torricelli's law

The tank's draining speed can be modeled using Torricelli's law, which states that the water's outflow speed (\(v\)) is related to the height at that instant (\(h\)) and acceleration due to gravity (g) as follows: $$v=\sqrt{2gh}$$ Since the area of the draining hole is a, the volume flow rate is: $$Q=a\times v$$ For our tank, the new height function will be given by the time rate of change of the volume of water inside the tank. To understand this, the change in volume inside is given by \(A\times dh\), which should be equal to the volume flow rate \(Q\) times the time interval dt. So, $$A\,dh = a\,v\,dt$$ Replace v with Torricelli's law formula: $$A\,dh=a\times\sqrt{2gh}\,dt$$
02

Separation of Variables

Now, given the differential equation (\(A\,dh = a\,v\,dt\)), we need to separate the variables: $$\frac{dh}{\sqrt{h}}=- \frac{a}{A}\sqrt{2g}\, dt$$
03

Integrating both sides

Now we need to integrate both sides with respect to their respective variables. To do this, we will integrate with respect to h on the left side and with respect to t on the right side: $$\int\frac{dh}{\sqrt{h}} = - \frac{a}{A}\sqrt{2g}\int dt$$ Evaluating the integrals: $$2\sqrt{h}+C_{1}=-\frac{a}{A}\sqrt{2g}\,t+C_{2}$$ Let's solve for \(C_{1}\) and \(C_{2}\) using the initial condition: when \(t=0\), \(h=H\). So, $$2\sqrt{H}=C_{1}$$ So our equation is now: $$2\sqrt{h}= -\frac{a}{A}\sqrt{2g}\,t+2\sqrt{H}$$
04

Solve for h as a function of t

Now we will solve for height h as a function of time t: $$\sqrt{h}=1-\frac{a\sqrt{2g}}{2A}t$$ Now square both sides: $$h(t)=(1-\frac{a\sqrt{2g}}{2A}t)^{2}$$ This is the height function of the water in the tank over time, using the given parameters.
05

Find the time when the tank is empty

To find when the tank is empty, set the height h(t) to 0 and solve for t: $$0=(1-\frac{a\sqrt{2g}}{2A}t)^{2}$$ Taking the square root of both sides: $$0=1-\frac{a\sqrt{2g}}{2A}t$$ Now solve for t: $$t = \frac{2A}{a\sqrt{2g}}$$ Now, substitute the given values of A, a, and g (g = 9.81 \(m/s^2\)) into the equation: $$t = \frac{2(2)}{0.5\sqrt{2(9.81)}}$$ $$t\approx2.89 \,seconds$$
06

Graphing the solution

To graph the solution, we will plot the height function h(t) vs. time t: Using the derived height function: $$h(t)=(1-\frac{a\sqrt{2g}}{2A}t)^{2}$$ And the given values for A, a, and g, we can plot the height of the water in the tank over time. The graph will start at the initial height (\(h = 2.25\,m\)) and decrease as time goes on until it reaches 0 at \(t\approx2.89\,s\). The graph should be a parabolic curve that touches the time-axis, or t-axis, at around 2.89 seconds, illustrating the water height in the tank over time as it drains.

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Key Concepts

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

Differential Equations
Differential equations are mathematical equations that involve the rates at which quantities change. These are fundamental in expressing the laws of nature, many of which involve rates at which changes occur 鈥� such as the velocity of an object or the growth rate of a population. In the context of our tank draining problem, we use a differential equation to describe the rate at which the height of the water in a tank changes over time.

Understanding how to solve such an equation is crucial for predicting future states of the system it models 鈥� in our case, determining when the tank will be empty. The differential equation comes from Torricelli's law which relates the outflow speed of water to the water height in the tank.
Separation of Variables
Separation of variables is a method used to solve differential equations, where the equation is rearranged so that each variable is on a separate side of the equation. This highly effective technique simplifies complex differential equations into a form where both sides can be integrated individually. In the provided solution, we separate the variable representing height, h, and the variable representing time, t, onto opposite sides of an equation to facilitate the process of integration.

This method not only separates variables but also helps us visualize the relationship between them by comparing the integrals on either side of the equation. Once the variables are separated, we can proceed to solve the differential equation step by step.
Rate of Height Change
In physics and engineering, understanding the rate of height change is essential for modeling and controlling systems involving liquids. Torricelli's law is applied to determine the rate of height change of water in a tank. This is expressed in the differential equation, \(A dh = a\times\sqrt{2gh} dt\), which links the change in height (dh) to the change in time (dt).

The negative sign tells us that the water level is decreasing over time. By analyzing how quickly the height of water in a tank decreases, we can not only predict when the tank will be empty but also devise measures to regulate the flow if necessary.
Integration

Integration in Calculus

Integral calculus, or simply integration, is the inverse process of differentiation, and it's used to find functions when their rates of change are known. When we integrate, we are essentially summing up small pieces to find a total quantity, like area under a curve or the accumulated change in a function.

In our case, we integrate both sides of the separated differential equation to find the relationship between the water height and time. This process is pivotal because it transforms the rate of change into a functional form which we can graph, analyze, and use to make concrete predictions about the system's behavior.

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Most popular questions from this chapter

In this section, several models are presented and the solution of the associated differential equation is given. Later in the chapter, we present methods for solving these differential equations. The delivery of a drug (such as an antibiotic) through an intravenous line may be modeled by the differential equation \(m^{\prime}(t)+k m(t)=I,\) where \(m(t)\) is the mass of the drug in the blood at time \(t \geq 0, k\) is a constant that describes the rate at which the drug is absorbed, and \(I\) is the infusion rate. a. Show by substitution that if the initial mass of drug in the blood is zero \((m(0)=0),\) then the solution of the initial value problem is \(m(t)=\frac{I}{k}\left(1-e^{-k t}\right)\) b. Graph the solution for \(I=10 \mathrm{mg} / \mathrm{hr}\) and \(k=0.05 \mathrm{hr}^{-1}\) c. Evaluate \(\lim _{t \rightarrow \infty} m(t),\) the steady-state drug level, and verify the result using the graph in part (b).

Chemical rate equations Let \(y(t)\) be the concentration of a substance in a chemical reaction (typical units are moles/liter). The change in the concentration, under appropriate conditions, is modeled by the equation \(\frac{d y}{d t}=-k y^{n},\) for \(t \geq 0,\) where \(k>0\) is a rate constant and the positive integer \(n\) is the order of the reaction. a. Show that for a first-order reaction \((n=1),\) the concentration obeys an exponential decay law. b. Solve the initial value problem for a second-order reaction \((n=2)\) assuming \(y(0)=y_{0}\). c. Graph the concentration for a first-order and second-order reaction with \(k=0.1\) and \(y_{0}=1\).

Solve the differential equation for Newton's Law of Cooling to find the temperature function in the following cases. Then answer any additional questions. A cup of coffee has a temperature of \(90^{\circ} \mathrm{C}\) when it is poured and allowed to cool in a room with a temperature of \(25^{\circ} \mathrm{C}\). One minute after the coffee is poured, its temperature is \(85^{\circ} \mathrm{C}\). How long must you wait until the coffee is cool enough to drink, say \(30^{\circ} \mathrm{C} ?\)

Find the equilibrium solution of the following equations, make a sketch of the direction field, for \(t \geq 0,\) and determine whether the equilibrium solution is stable. The direction field needs to indicate only whether solutions are increasing or decreasing on either side of the equilibrium solution. $$y^{\prime}(t)=12 y-18$$

Logistic equation for spread of rumors Sociologists model the spread of rumors using logistic equations. The key assumption is that at any given time, a fraction \(y\) of the population, where \(0 \leq y \leq 1,\) knows the rumor, while the remaining fraction \(1-y\) does not. Furthermore, the rumor spreads by interactions between those who know the rumor and those who do not. The number of such interactions is proportional to \(y(1-y) .\) Therefore, the equation that describes the spread of the rumor is \(y^{\prime}(t)=k y(1-y)\) for \(t \geq 0,\) where \(k\) is a positive real number and \(t\) is measured in weeks. The number of people who initially know the rumor is \(y(0)=y_{0},\) where \(0 \leq y_{0} \leq 1\). a. Solve this initial value problem and give the solution in terms of \(k\) and \(y_{0}.\) b. Assume \(k=0.3\) weeks \(^{-1}\) and graph the solution for \(y_{0}=0.1\) and \(y_{0}=0.7.\) c. Describe and interpret the long-term behavior of the rumor function, for any \(0 \leq y_{0} \leq 1\).
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