Question

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

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

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: $$\frac{dh}{dt} = -0.0102\sqrt{h}$$

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: $$2\sqrt{h} = -0.0102\, t + 2.8 $$

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.

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 \(t \approx 274.509\,\mathrm{s}\).

Key concepts

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’s 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.