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Chapter 3: Problem 32

Find the equilibrium solution to \(d v / d t=\) \(-g-(c / m) v^{2}\). What is the limiting velocity?

Short Answer

Expert verified
The limiting velocity is \( v = \text{sqrt} \frac{mg}{c} \.

Step by step solution

01

- Understand the problem

To find the equilibrium solution to the differential equation \(\frac{d v}{d t} = -g - \frac{c}{m} v^{2}\), identify the conditions required for equilibrium.
02

- Set the derivative to zero

For equilibrium, the velocity must be constant, so set \( \frac{d v}{d t} = 0 \). This gives: \ 0 = -g - \frac{c}{m} v^{2} \.
03

- Solve for v

Rearrange the equation to solve for \( v \): \ \frac{c}{m} v^{2} = -g \. Then, multiply both sides by \( -1 \) to get: \ \frac{c}{m} v^{2} = g \.
04

- Isolate \( v \)

Isolate \( v \) by multiplying both sides by \( \frac{m}{c} \): \ v^{2} = \frac{mg}{c} \.
05

- Take the square root

Take the square root of both sides to solve for \( v \): \ v = \frac{\text{sqrt}(mg)}{\text{sqrt}(c)} \ or \ v = \text{sqrt} \frac{mg}{c}\.
06

- Identify the limiting velocity

The limiting velocity is the equilibrium solution, which is the value of \( v \) found in the previous step.

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Key Concepts

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

limiting velocity
Limiting velocity is the maximum speed an object can reach when considering forces like gravity and air resistance. When an object falls, gravity pulls it downward. But air resistance opposes this motion which increases with speed. Eventually, the forces balance out, leading to a constant velocity. This is known as the limiting or terminal velocity. The velocity stops increasing as the downward force of gravity is equal to the upward force of air resistance. Hence, an equilibrium is reached. In our problem, the limiting velocity is found by solving the equation when the acceleration is zero. This gives us \( v = \text{sqrt} \frac{mg}{c}\).
constant velocity
Constant velocity means an object's speed and direction do not change over time. This implies no acceleration. In our equation, constant velocity is achieved when the derivative of velocity with respect to time \( \frac{d v}{d t} = 0 \). At this point, forces acting on the object are balanced. In practical scenarios like free fall with air resistance, reaching constant velocity means the object has hit its limiting velocity. For differential equations, setting the rate of change to zero is a common method to identify steady-state or equilibrium solutions. Thus, constant velocity is a state where there is neither speeding up nor slowing down.
solving differential equations
Solving differential equations involves finding a function that satisfies the equation's condition. In our case, we start from the differential equation \(\frac{d v}{d t} = -g - \frac{c}{m} v^{2}\). First, determine the equilibrium state where \(\frac{d v}{d t} = 0\). This simplifies to \(0 = -g - \frac{c}{m} v^{2}\). We solve for \(v\) to find the steady-state solution. By isolating \(v\) and taking the square root, we get \(v = \text{sqrt} \frac{mg}{c}\). This method is typical for first-order ordinary differential equations. Equilibrium solutions often reveal important behavior about the system's dynamics over time.

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Most popular questions from this chapter

(Tumor and Organism Growth) Ludwig von Bertalanffy (1901-1972) made valuable contributions to the study of organism growth, including the growth of tumors, based on the relationship between body size of the organism and the metabolic rate. He theorized that weight is directly proportional to volume and that the metabolic rate is proportional to surface area. This indicates that surface area is proportional to \(V^{2 / 3}\) because \(V\) is measured in cubic units and \(S\) in square units. Therefore, Bertalanffy studied the IVP \(d V / d t=a V^{2 / 3}-b V, V(0)=V_{0}\). (a) Solve this initial value problem to show that volume is given by \(V(t)=\) \(\left[\frac{a}{b}-\left(\frac{a}{b}-V_{0}^{1 / 3}\right) e^{-b t / 3}\right]^{3}\). (b) Find \(\lim _{t \rightarrow \infty} V(t)\) and explain what the limit represents. (c) This initial value problem is similar to that involving the Logistic equation \(d V / d t=a V^{2}-b V, V(0)=V_{0}\). Compare the limiting volume between the Logistic equation and the model proposed by Bertalanffy in the cases when \(a / b>1\) and when \(a / b<1\).

(Harvesting) If we wish to model a population of size \(P(t)\) at time \(t\) and consider a constant harvest rate \(h\) (like hunting, fishing, or disease), then we might modify the logistic equation and use the equation \(P^{\prime}=\) \(r P-a P^{2}-h\) to model the population under consideration. Assume that \(h \geq r^{2} /(4 a)\). (a) Show that if \(h \geq r^{2} /(4 a)\), a general solution of \(P^{\prime}=r P-a P^{2}-h\) is $$ \begin{aligned} P(t) &=\frac{1}{2 a}\left[r+\sqrt{4 a h-r^{2}}\right.\\\ &\left.\times \tan \left(\frac{1}{2 a}(C-a t) \sqrt{4 a h-r^{2}}\right)\right] . \end{aligned} $$ (b) Suppose that for a certain species it is found that \(r=0.03, a=0.0001, h=2.26\), and \(C=-1501.85\). At what time will the species become extinct? (c) If \(r=0.03, a=0.0001\), and \(P(0)=5.3\), graph \(P(t)\) if \(h=0,0.5,1.0,1.5,2.0,2.25\) and 2.5. (d) What is the maximum allowable harvest rate to assure that the species survives? (e) Generalize your result from (d). For arbitrary \(a\) and \(r\), what is the maximum allowable harvest rate that ensures survival of the species?

Find the amount of salt \(y(t)\) in a tank with initial volume \(V_{0}\) gallons of liquid and \(y_{0}\) pounds of salt using the given conditions. (a) \(R_{1}=4 \mathrm{gal} / \mathrm{min}, R_{2}=4 \mathrm{gal} / \mathrm{min}, S_{1}=2\) \(\mathrm{lb} / \mathrm{gal}, V_{0}=400, y_{0}=0\). (b) \(R_{1}=4 \mathrm{gal} / \mathrm{min}, R_{2}=2 \mathrm{gal} / \mathrm{min}, S_{1}=1\) \(\mathrm{lb} / \mathrm{gal}, V_{0}=500, y_{0}=20\). (c) \(R_{1}=4 \mathrm{gal} / \mathrm{min}, R_{2}=6 \mathrm{gal} / \mathrm{min}, S_{1}=1\) \(\mathrm{lb} / \mathrm{gal}, V_{0}=600, y_{0}=100\). Describe how the values of \(R_{1}\) and \(R_{2}\) affect the volume of liquid in the tank.

A tank contains \(200 \mathrm{gal}\) of a brine solution in which \(10 \mathrm{lb}\) of salt is initially dissolved. A brine solution with concentration \(2 \mathrm{lb} / \mathrm{gal}\) is then allowed to flow into the tank at a rate of \(4 \mathrm{gal} / \mathrm{min}\) and the well-stirred mixture flows out of the tank at a rate of \(3 \mathrm{gal} / \mathrm{min}\). Determine the amount of salt \(y(t)\) at any time \(t\). If the tank can hold a maximum of \(400 \mathrm{gal}\), what is the concentration of the brine solution in the tank when the volume reaches this maximum?

What is the equilibrium solution of \(d v / d t=\) \(32-v\) ? How does this relate to the solution \(v=32+C e^{-t}\) ?
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