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Chapter 8: Problem 33

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. $$H=1.96 \mathrm{m}, A=1.5 \mathrm{m}^{2}, a=0.3 \mathrm{m}^{2}$$

Short Answer

Expert verified
Question: Determine the water height function, the approximate time at which the tank is first empty, and describe the graph based on the given solution. Answer: The water height function is given by \(h(t) = (-0.0051\, t + 1.4)^{2}\). The approximate time at which the tank is first empty is 274.509 seconds. The graph demonstrates how the water height decreases over time, with the water height (h(t)) on the y-axis and time (t) on the x-axis, showing the tank to be empty at around t = 274.509 seconds.

Step by step solution

01

Set up the differential equation for the water height function

We know that the rate at which the water drains from the tank is proportional to the height of the water in the tank. This can be represented as a differential equation: $$\frac{dh}{dt} = -k\sqrt{h}$$ where \(h(t)\) is the water height function, and \(k\) is a constant. To find the constant \(k\), we can use the initial conditions: when the tank is full, the height is \(H=1.96 \mathrm{m}\), and the rate at which it drains is \(a=0.3 \mathrm{m}^{2}\)/s. Therefore, we have $$-\frac{0.3}{1.5}=\sqrt{1.96}\cdot k$$ Solving for \(k\): $$k = -\frac{0.3}{1.96\cdot 1.5} = 0.0102 \mathrm{s}^{-1}$$ Now the differential equation is: $$\frac{dh}{dt} = -0.0102\sqrt{h}$$
02

Solve the differential equation for the water height function

We'll solve for the water height function \(h(t)\): $$\frac{dh}{\sqrt{h}} = -0.0102\, dt$$ Integrating both sides, we get: $$2\sqrt{h} = -0.0102\, t + C $$ Now we need to find the integration constant \(C\). We know that at time \(t=0\), the height is \(1.96 \mathrm{m}\), so we can plug these values in to get: $$2\sqrt{1.96}=-0.0102 \cdot 0 + C$$ Solving for \(C\), we obtain \(C=2.8\). Thus, our water height function is given by: $$2\sqrt{h} = -0.0102\, t + 2.8 $$
03

Determine the approximate time when the tank is empty

To find when the tank is empty, we want the height to be zero, so we can set \(h=0\) in our water height equation: $$2\sqrt{0} = -0.0102\, t + 2.8 $$ $$0=-0.0102\, t + 2.8$$ Solving for \(t\), we get: $$t = \frac{2.8}{0.0102} \approx 274.509\,\mathrm{s}$$ So, the approximate time when the tank is empty is \(274.509\) seconds.
04

Graph the solution

To graph the solution, we can solve for \(h(t)\) explicitly: $$\sqrt{h} = -0.0051\, t + 1.4 $$ $$h(t) = (-0.0051\, t + 1.4)^{2}$$ Now we can plot this function on a graph, with \(t\) on the x-axis and \(h(t)\) on the y-axis, illustrating how the water height decreases over time as the tank empties. The graph will show that the tank is empty at around \(t \approx 274.509\,\mathrm{s}\).

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Key Concepts

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

Tank Drainage
In a tank drainage problem, we're primarily dealing with how the water level in a tank decreases over time due to water being released. This is a classic example of applying differential equations in real-world scenarios. The tank is initially filled to a certain height, and water drains out through an outlet at the bottom.

The flow rate, or how quickly the water leaves the tank, is often proportional to the square root of the water height. This concept stems from Torricelli鈥檚 law, which describes fluid flow under the force of gravity, suggesting that the velocity of efflux is related to the height of the liquid surface above the outlet.
  • As water drains, the water height decreases, thus reducing the flow rate.
  • This dynamic change is modeled using a differential equation, which provides a powerful tool to predict when the tank will be completely drained.
Being able to define and solve this differential equation is crucial for understanding how long it will take for the tank to empty entirely.
Rate of Change
The rate of change in the context of this problem refers to how rapidly the water height in the tank decreases over time. Specifically, the equation for this change is given by \[\frac{dh}{dt} = -k\sqrt{h}\] Here, \(dh/dt\) is the rate of change of the height. The negative sign indicates that the height is decreasing. The term \(k\sqrt{h}\) reflects how the height of the water affects this rate.

In simple terms:
  • The higher the water, the faster it empties because of increased pressure at the outlet.
  • Over time, as the water level drops, the rate of change (drainage speed) decreases.
From a calculus perspective, this differential relationship provides essential insights: it implies that as time progresses, the contribution of \(h\) becomes less impactful, which is why tanks do not empty instantly but rather at a diminishing pace. This concept is vital to control and manage fluid dynamics in engineering and environmental applications.
Water Height Function
The water height function describes how the water height in the tank changes over time as the tank drains. Solving the differential equation gives us this function, which is central to predicting when the tank will be empty. We begin with: \[2\sqrt{h} = -0.0102\, t + 2.8 \]This equation is derived from integrating the rate of change equation. Upon solving, it provides a \(h(t)\), the height of the water at any time \(t\). To find this function:
  • We express \(h\) as a function of \(t\) by squaring both sides:
\[h(t) = (-0.0051\, t + 1.4)^2\]
  • This quadratic form allows us to easily see how height decreases as time increases.
  • By setting \(h(t) = 0\), we solve for \(t\) to determine when the tank will be empty, confirming our calculations of approximately 274.5 seconds.
Graphing this function beautifully illustrates the gradual decline in water height, offering a visual representation of the drainage process. It's through this function that engineers and scientists can simulate, predict, and analyze fluid behaviors in various scenarios, ensuring practical and theoretical fluency in handling differential equations.

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Most popular questions from this chapter

One possible model that describes the free fall of an object in a gravitational field subject to air resistance uses the equation \(v^{\prime}(t)=g-b v,\) where \(v(t)\) is the velocity of the object for \(t \geq 0\), \(g=9.8 \mathrm{m} / \mathrm{s}^{2}\) is the acceleration due to gravity, and \(b>0\) is a constant that involves the mass of the object and the air resistance. a. Verify by substitution that a solution of the equation, subject to the initial condition \(v(0)=0,\) is \(v(t)=\frac{g}{b}\left(1-e^{-b t}\right)\). b. Graph the solution with \(b=0.1 s^{-1}\). c. Using the graph in part (c), estimate the terminal velocity \(\lim _{t \rightarrow \infty} v(t)\).

U.S. population projections According to the U.S. Census Bureau, the nation's population (to the nearest million) was 281 million in 2000 and 310 million in \(2010 .\) The Bureau also projects a 2050 population of 439 million. To construct a logistic model, both the growth rate and the carrying capacity must be estimated. There are several ways to estimate these parameters. Here is one approach: a. Assume that \(t=0\) corresponds to 2000 and that the population growth is exponential for the first ten years; that is, between 2000 and \(2010,\) the population is given by \(P(t)=P(0) e^{n}\) Estimate the growth rate \(r\) using this assumption. b. Write the solution of the logistic equation with the value of \(r\) found in part (a). Use the projected value \(P(50)=439 \mathrm{mil}\) lion to find a value of the carrying capacity \(K\) c. According to the logistic model determined in parts (a) and (b), when will the U.S. population reach \(95 \%\) of its carrying capacity? d. Estimations of this kind must be made and interpreted carefully. Suppose the projected population for 2050 is 450 million rather than 439 million. What is the value of the carrying capacity in this case? e. Repeat part (d) assuming the projected population for 2050 is 430 million rather than 439 million. What is the value of the carrying capacity in this case? f. Comment on the sensitivity of the carrying capacity to the 40-year population projection.

Write a logistic equation with the following parameter values. Then solve the initial value problem and graph the solution. Let \(r\) be the natural growth rate, \(K\) the carrying capacity, and \(P_{0}\) the initial population. $$r=0.4, K=5500, P_{0}=100$$

Consider the initial value problem \(y^{\prime}(t)=y^{n+1}, y(0)=y_{0},\) where \(n\) is a positive integer. a. Solve the initial value problem with \(n=1\) and \(y_{0}=1\) b. Solve the initial value problem with \(n=2\) and \(y_{0}=\frac{1}{\sqrt{2}}\) c. Solve the problem for positive integers \(n\) and \(y_{0}=n^{-1 / n}\) How do solutions behave as \(t \rightarrow 1^{-2}\)

Widely used models for population growth involve the logistic equation \(P^{\prime}(t)=r P\left(1-\frac{P}{K}\right),\) where \(P(t)\) is the population, for \(t \geq 0,\) and \(r>0\) and \(K>0\) are given constants. a. Verify by substitution that the general solution of the equation is \(P(t)=\frac{K}{1+C e^{-n}},\) where \(C\) is an arbitrary constant. b. Find that value of \(C\) that corresponds to the initial condition \(P(0)=50\). c. Graph the solution for \(P(0)=50, r=0.1,\) and \(K=300\). d. Find \(\lim _{t \rightarrow \infty} P(t)\) and check that the result is consistent with the graph in part (c).
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