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Chapter 2: Problem 39

(a) Writing the differential equation in the form $$\frac{d v}{d t}=\frac{k}{m}\left(\frac{m g}{k}-v\right)$$ we see that a critical point is \(m g / k\). From the phase portrait we see that \(m g / k\) is an asymptotically stable critical point. Thus, \(\lim _{t \rightarrow \infty} v=m g / k\). (b) Writing the differential equation in the form $$\frac{d v}{d t}=\frac{k}{m}\left(\frac{m g}{k}-v^{2}\right)=\frac{k}{m}(\sqrt{\frac{m g}{k}}-v)(\sqrt{\frac{m g}{k}}+v)$$ we see that the only physically meaningful critical point is \(\sqrt{m g / k}\). From the phase portrait we see that \(\sqrt{m g / k}\) is an asymptotically stable critical point. Thus, \(\lim _{t \rightarrow \infty} v=\sqrt{m g / k}\).

Short Answer

Expert verified
(a) \( \lim_{t \rightarrow \infty} v = \frac{mg}{k} \), (b) \( \lim_{t \rightarrow \infty} v = \sqrt{\frac{mg}{k}} \).

Step by step solution

01

Identify the Differential Equation Form

We are given two different forms of the differential equation. The first form is \( \frac{d v}{d t}=\frac{k}{m}\left(\frac{mg}{k}-v\right) \). This form indicates how the velocity \( v \) changes over time.
02

Determine the Critical Point for First Form

For the equation \( \frac{d v}{d t}=\frac{k}{m}\left(\frac{mg}{k}-v\right) \), the critical point is where \( \frac{d v}{d t}=0 \). Solving \( \frac{mg}{k}-v=0 \) gives the critical point \( v=\frac{mg}{k} \).
03

Analyze Stability from Phase Portrait (First Form)

The problem states that \( \frac{mg}{k} \) is an asymptotically stable critical point. This means that as time \( t \) approaches infinity, the velocity \( v \) will approach \( \frac{mg}{k} \).
04

Rewrite the Differential Equation for Second Form

The second form of the equation given is \( \frac{d v}{d t}=\frac{k}{m}\left(\frac{mg}{k}-v^2\right) \). This indicates a new dynamic in the change of velocity over time as it involves \( v^2 \).
05

Factor and Determine Critical Points for Second Form

Factoring the equation \( \frac{d v}{d t}=\frac{k}{m}\left(\sqrt{\frac{mg}{k}}-v\right)\left(\sqrt{\frac{mg}{k}}+v\right) \), we set \( \sqrt{\frac{mg}{k}}-v=0 \) to find the critical point \( v=\sqrt{\frac{mg}{k}} \).
06

Analyze Stability from Phase Portrait (Second Form)

The phase portrait indicates \( \sqrt{\frac{mg}{k}} \) is asymptotically stable, which implies \( \lim_{t \rightarrow \infty} v = \sqrt{\frac{mg}{k}} \).
07

Conclude with Long-Term Behavior Comparison

In part (a), the limit of velocity as time goes to infinity is \( \frac{mg}{k} \). In part (b), it is \( \sqrt{\frac{mg}{k}} \). These are the values \( v \) approaches as time goes to infinity for each form of the differential equation.

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Key Concepts

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

Asymptotically Stable Critical Points
In differential equations, a critical point occurs where the rate of change equals zero, meaning that the system is at a point of equilibrium.
An "asymptotically stable" critical point is a type of equilibrium where solutions that start near it tend to approach it as time progresses. In simple terms, it's like a ball in a bowl; if you slightly push the ball, it will roll back to the bottom.
For the given equation \( \frac{d v}{d t} = \frac{k}{m} \left( \frac{mg}{k} - v \right) \), the critical point is \( \frac{mg}{k} \). Analyzing its stability, we find that disturbances in the system tend to vanish, causing the velocity \( v \) to stabilize at this critical point over time.
Critical points, like \( \frac{mg}{k} \) and \( \sqrt{mg/k} \), are not only significant for determining where a system settles but also for understanding the overall behavior of a dynamical system. Recognizing and calculating them helps us predict long-term behavior of the system.
Phase Portrait
A phase portrait is a graphical representation of the trajectories of a dynamic system in the phase plane. It helps in visualizing how the state of a system evolves over time.
In the context of our differential equations, a phase portrait can show us whether a critical point is stable or unstable by illustrating the direction of the trajectories.
For example, in the phase portrait for the equation \( \frac{d v}{d t} = \frac{k}{m} \left( \frac{mg}{k} - v \right) \), you would see arrows that point towards \( \frac{mg}{k} \), showing it is an attractor or asymptotically stable point. This means trajectories lead towards it as time increases.
Understanding phase portraits is crucial because they offer insights that aren't immediately apparent from algebraic solutions alone. By visualizing how solutions move, you can better understand potential behavior for various initial conditions.
Long-Term Behavior Analysis
Long-term behavior analysis in differential equations involves understanding where and how a system behaves as time approaches infinity. This kind of analysis tells us the end-state or equilibrium point of a system.
For both forms of differential equations given, identifying the limits helps in understanding this behavior.
In the first case, with equation \( \frac{d v}{d t} = \frac{k}{m} \left( \frac{mg}{k} - v \right) \), the velocity \( v \) approaches the value \( \frac{mg}{k} \) as time goes to infinity.
In the second equation form \( \frac{d v}{d t} = \frac{k}{m} \left( \frac{mg}{k} - v^2 \right) \), the velocity approaches \( \sqrt{mg/k} \) over the long term.
Such analysis is pivotal, especially in fields like engineering and physics, where predicting the final state of a system is essential for application, design, and understanding broader phenomena.

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Most popular questions from this chapter

(a) From \(m d v / d t=m g-k v\) we obtain \(v=m g / k+c e^{-k t / m} .\) If \(v(0)=v_{0}\) then \(c=v_{0}-m g / k\) and the solution of the initial-value problem is $$v(t)=\frac{m g}{k}+\left(v_{0}-\frac{m g}{k}\right) e^{-k t / m}$$. (b) As \(t \rightarrow \infty\) the limiting velocity is \(m g / k\) (c) From \(d s / d t=v\) and \(s(0)=0\) we obtain $$s(t)=\frac{m g}{k} t-\frac{m}{k}\left(v_{0}-\frac{m g}{k}\right) e^{-k t / m}+\frac{m}{k}\left(v_{0}-\frac{m g}{k}\right)$$.

When the height of the water is \(h,\) the radius of the top of the water is \(\frac{2}{5}(20-h)\) and \(A_{w}=4 \pi(20-h)^{2} / 25 .\) The differential equation is \\[ \frac{d h}{d t}=-c \frac{A_{h}}{A_{w}} \sqrt{2 g h}=-0.6 \frac{\pi(2 / 12)^{2}}{4 \pi(20-h)^{2} / 25} \sqrt{64 h}=-\frac{5}{6} \frac{\sqrt{h}}{(20-h)^{2}} \\] Separating variables and integrating we have \\[ \frac{(20-h)^{2}}{\sqrt{h}} d h=-\frac{5}{6} d t \quad \text { and } \quad 800 \sqrt{h}-\frac{80}{3} h^{3 / 2}+\frac{2}{5} h^{5 / 2}=-\frac{5}{6} t+c \\] Using \(h(0)=20\) we find \(c=2560 \sqrt{5} / 3,\) so an implicit solution of the initial-value problem is \\[ 800 \sqrt{h}-\frac{80}{3} h^{3 / 2}+\frac{2}{5} h^{5 / 2}=-\frac{5}{6} t+\frac{2560 \sqrt{5}}{3} \\] To find the time it takes the tank to empty we set \(h=0\) and solve for \(t .\) The tank empties in \(1024 \sqrt{5}\) seconds or 38.16 minutes. Thus, the tank empties more slowly when the base of the cone is on the bottom.

Writing the differential equation as \(\frac{d E}{d t}+\frac{1}{R C} E=0\) we see that an integrating factor is \(e^{t / R C}\). Then $$\begin{aligned} \frac{d}{d t}\left[e^{t / R C} E\right] &=0 \\ e^{t / R C} E &=c \\ E &=c e^{-t / R C} \end{aligned}$$ From \(E(4)=c e^{-4 / R C}=E_{0}\) we find \(c=E_{0} e^{4 / R C} .\) Thus, the solution of the initial-value problem is $$E=E_{0} e^{4 / R C} e^{-t / R C}=E_{0} e^{-(t-4) / R C}$$

The differential equation for the first container is \(d T_{1} / d t=k_{1}\left(T_{1}-0\right)=k_{1} T_{1},\) whose solution is \(T_{1}(t)=c_{1} e^{k_{1} t}\) since \(T_{1}(0)=100\) (the initial temperature of the metal bar), we have \(100=c_{1}\) and \(T_{1}(t)=100 e^{k_{1} t}\). After 1 minute, \(T_{1}(1)=100 e^{k_{1}}=90^{\circ} \mathrm{C},\) so \(k_{1}=\ln 0.9\) and \(T_{1}(t)=100 e^{t \ln 0.9} .\) After 2 minutes, \(T_{1}(2)=100 e^{2 \ln 0.9}=\) \(100(0.9)^{2}=81^{\circ} \mathrm{C}\). The differential equation for the second container is \(d T_{2} / d t=k_{2}\left(T_{2}-100\right),\) whose solution is \(T_{2}(t)=\) \(100+c_{2} e^{k_{2} t} .\) When the metal bar is immersed in the second container, its initial temperature is \(T_{2}(0)=81,\) so $$T_{2}(0)=100+c_{2} e^{k_{2}(0)}=100+c_{2}=81$$ and \(c_{2}=-19 .\) Thus, \(T_{2}(t)=100-19 e^{k_{2} t} .\) After 1 minute in the second tank, the temperature of the metal bar is \(91^{\circ} \mathrm{C},\) so $$\begin{aligned} T_{2}(1) &=100-19 e^{k_{2}}=91 \\ e^{k_{2}} &=\frac{9}{19} \\ k_{2} &=\ln \frac{9}{19} \end{aligned}$$ and \(T_{2}(t)=100-19 e^{t \ln (9 / 19)} .\) Setting \(T_{2}(t)=99.9\) we have $$\begin{aligned} 100-19 e^{t \ln (9 / 19)} &=99.9 \\ e^{t \ln (9 / 19)} &=\frac{0.1}{19} \\ t &=\frac{\ln (0.1 / 19)}{\ln (9 / 19)} \approx 7.02 \end{aligned}$$ Thus, from the start of the "double dipping" process, the total time until the bar reaches \(99.9^{\circ} \mathrm{C}\) in the second container is approximately 9.02 minutes.

(a) Solving \(r-k x=0\) for \(x\) we find the equilibrium solution \(x=r / k\). When \(x0\) and when \(x>r / k, d x / d t<0 .\) From the phase portrait we see that \(\lim _{t \rightarrow \infty} x(t)=r / k\). (b) From \(d x / d t=r-k x\) and \(x(0)=0\) we obtain \(x=r / k-(r / k) e^{-k t}\) so that \(x \rightarrow r / k\) as \(t \rightarrow \infty .\) If \(x(T)=r / 2 k\) then \(T=(\ln 2) / k\).
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