Problem 11
Question
(a) The initial-value problem is \(d h / d t=-8 A_{h} \sqrt{h} / A_{w}, h(0)=H\) Separating variables and integrating we have \\[ \frac{d h}{\sqrt{h}}=-\frac{8 A_{h}}{A_{w}} d t \quad \text { and } \quad 2 \sqrt{h}=-\frac{8 A_{h}}{A_{w}} t+c \\] Using \(h(0)=H\) we find \(c=2 \sqrt{H},\) so the solution of the initial-value problem is \(\sqrt{h(t)}=\left(A_{w} \sqrt{H}-4 A_{h} t\right) / A_{w},\) where \(A_{w} \sqrt{H}-4 A_{h} t \geq 0 .\) Thus, \\[ h(t)=\left(A_{w} \sqrt{H}-4 A_{h} t\right)^{2} / A_{w}^{2} \quad \text { for } \quad 0 \leq t \leq A_{w} H / 4 A_{h} \\] (b) Identifying \(H=10, A_{w}=4 \pi,\) and \(A_{h}=\pi / 576\) we have \(h(t)=t^{2} / 331,776-(\sqrt{5 / 2} / 144) t+10 .\) Solving \(h(t)=0\) we see that the tank empties in \(576 \sqrt{10}\) seconds or 30.36 minutes.
Step-by-Step Solution
Verified Answer
The tank empties in 576√10 seconds, or about 30.36 minutes.
1Step 1: Setting up the Equation for Integration
Given the differential equation \( \frac{dh}{dt} = -\frac{8A_h \sqrt{h}}{A_w} \), we will perform separation of variables by rearranging the terms. Multiply both sides by \( \frac{dt}{\sqrt{h}} \) to isolate the variables: \( \frac{dh}{\sqrt{h}} = -\frac{8A_h}{A_w} \, dt \).
2Step 2: Integrating Both Sides
Integrate both sides with respect to their variables: \[ \int \frac{dh}{\sqrt{h}} = -\int \frac{8A_h}{A_w} \, dt \] This results in the antiderivative expressions \[ 2\sqrt{h} = -\frac{8A_h}{A_w}t + c. \]
3Step 3: Applying Initial Condition
Use the initial condition \( h(0) = H \) to find \( c \). Substitute \( t = 0 \) and \( h = H \) into the equation: \[ 2\sqrt{H} = c. \] Thus, \( c = 2\sqrt{H}. \)
4Step 4: Solving for \( h(t) \)
Substitute \( c = 2\sqrt{H} \) back into the integrated equation: \[ 2\sqrt{h} = -\frac{8A_h}{A_w}t + 2\sqrt{H}. \] This gives the expression \[ \sqrt{h(t)} = \frac{A_w \sqrt{H} - 4A_ht}{A_w}. \]
5Step 5: Obtaining the Final Expression for \( h(t) \)
Square both sides to obtain \( h(t) \):\[ h(t) = \left(\frac{A_w \sqrt{H} - 4A_h t}{A_w}\right)^2 = \frac{(A_w \sqrt{H} - 4A_h t)^2}{A_w^2}. \]
6Step 6: Establishing the Time Interval
The condition \( A_w \sqrt{H} - 4A_h t \geq 0 \) implies \[ t \leq \frac{A_w H}{4A_h}. \] This defines the valid range for \( t \).
7Step 7: Substituting Known Values
Identify \( H = 10, A_w = 4\pi, \) and \( A_h = \frac{\pi}{576} \), and substitute these into the expression for \( h(t) \): \[ h(t) = \left(4\pi \sqrt{10} - \frac{4 \cdot \pi}{576}t\right)^2 / (4\pi)^2. \]
8Step 8: Solving \( h(t) = 0 \)
Solve the equation \( h(t) = 0 \) for \( t \):\[ \left(4\pi \sqrt{10} - \frac{4\pi}{576}t\right) = 0. \] Solving for \( t \) gives \[ t = 576\sqrt{10}. \]
Key Concepts
Initial-Value ProblemSeparation of VariablesIntegration in CalculusTank Draining Problem
Initial-Value Problem
An initial-value problem in differential equations involves finding a function that satisfies a given differential equation and fulfills a predetermined condition, called the initial condition. This initial condition provides specific information about the state of the function at a starting point, generally time equals zero.
In the tank draining problem, the initial-value problem is set up with the equation \( \frac{dh}{dt} = -\frac{8A_h \sqrt{h}}{A_w} \), and the initial condition is \( h(0) = H \). This initial condition stipulates the height \( h \) of the liquid in the tank at the initial time \( t = 0 \).
By solving this initial-value problem, we obtain a function \( h(t) \) that describes how the height of the liquid changes over time, starting from the initial height \( H \). This process combines the concepts of differential equations with specific initial constraints to model real-world scenarios, like how quickly a tank drains.
In the tank draining problem, the initial-value problem is set up with the equation \( \frac{dh}{dt} = -\frac{8A_h \sqrt{h}}{A_w} \), and the initial condition is \( h(0) = H \). This initial condition stipulates the height \( h \) of the liquid in the tank at the initial time \( t = 0 \).
By solving this initial-value problem, we obtain a function \( h(t) \) that describes how the height of the liquid changes over time, starting from the initial height \( H \). This process combines the concepts of differential equations with specific initial constraints to model real-world scenarios, like how quickly a tank drains.
Separation of Variables
Separation of variables is a technique used to solve differential equations. The method involves rearranging an equation so that each side contains only one of the variables.
For the tank draining problem, we start with the differential equation \( \frac{dh}{dt} = -\frac{8A_h \sqrt{h}}{A_w} \). To separate the variables, we multiply both sides by \( \frac{dt}{\sqrt{h}} \). This results in an equation where the variables \( h \) and \( t \) are separated: \( \frac{dh}{\sqrt{h}} = -\frac{8A_h}{A_w} dt \).
By separating the variables, we can independently integrate each side with respect to its variable. This ultimately helps us express the relationship between \( h \) and \( t \) in terms of an integrated equation, paving the way for solving the initial-value problem.
For the tank draining problem, we start with the differential equation \( \frac{dh}{dt} = -\frac{8A_h \sqrt{h}}{A_w} \). To separate the variables, we multiply both sides by \( \frac{dt}{\sqrt{h}} \). This results in an equation where the variables \( h \) and \( t \) are separated: \( \frac{dh}{\sqrt{h}} = -\frac{8A_h}{A_w} dt \).
By separating the variables, we can independently integrate each side with respect to its variable. This ultimately helps us express the relationship between \( h \) and \( t \) in terms of an integrated equation, paving the way for solving the initial-value problem.
Integration in Calculus
Integration is a fundamental concept in calculus used to find the antiderivative or the area under a curve. When solving differential equations, integration allows us to recover the original function from its derivative.
In the context of the tank draining problem, after separating variables, the next step involves integrating both sides of the equation. We integrate \( \int \frac{dh}{\sqrt{h}} \) and \(-\int \frac{8A_h}{A_w} dt \).
The integral of \( \frac{1}{\sqrt{h}} \) with respect to \( h \) results in \( 2\sqrt{h} \). The integral of a constant with respect to \( t \) is straightforward, yielding \(-\frac{8A_h}{A_w}t + c \), where \( c \) is the constant of integration. This step gives structure to our solution and requires an initial condition to identify \( c \).
Through this process, integration helps convert our separated differential equation into an equation that can be solved for \( h(t) \).
In the context of the tank draining problem, after separating variables, the next step involves integrating both sides of the equation. We integrate \( \int \frac{dh}{\sqrt{h}} \) and \(-\int \frac{8A_h}{A_w} dt \).
The integral of \( \frac{1}{\sqrt{h}} \) with respect to \( h \) results in \( 2\sqrt{h} \). The integral of a constant with respect to \( t \) is straightforward, yielding \(-\frac{8A_h}{A_w}t + c \), where \( c \) is the constant of integration. This step gives structure to our solution and requires an initial condition to identify \( c \).
Through this process, integration helps convert our separated differential equation into an equation that can be solved for \( h(t) \).
Tank Draining Problem
The tank draining problem is a classic example used to apply differential equation methods like separation of variables and integration. It models how liquid flows out of a tank over time due to gravity, making it a practical application in physics and engineering.
In this problem, we determine how the height of the liquid, \( h(t) \), changes with time as the liquid drains. The differential equation \( \frac{dh}{dt} = -\frac{8A_h \sqrt{h}}{A_w} \) helps model this process, where \( A_h \) is the cross-sectional area of the hole, \( A_w \) is the cross-sectional area of the top of the tank, and \( \sqrt{h} \) relates to the velocity of the draining liquid.
By solving this equation within the bounds of the initial condition \( h(0) = H \), we can deduce the function \( h(t) \) and predict the time the tank will be empty. The function demonstrates the rate at which the tank empties, which plays a vital role in engineering scenarios requiring fluid dynamics analysis, like water treatment facilities and chemical processing plants.
In this problem, we determine how the height of the liquid, \( h(t) \), changes with time as the liquid drains. The differential equation \( \frac{dh}{dt} = -\frac{8A_h \sqrt{h}}{A_w} \) helps model this process, where \( A_h \) is the cross-sectional area of the hole, \( A_w \) is the cross-sectional area of the top of the tank, and \( \sqrt{h} \) relates to the velocity of the draining liquid.
By solving this equation within the bounds of the initial condition \( h(0) = H \), we can deduce the function \( h(t) \) and predict the time the tank will be empty. The function demonstrates the rate at which the tank empties, which plays a vital role in engineering scenarios requiring fluid dynamics analysis, like water treatment facilities and chemical processing plants.
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