Science:Math Exam Resources/Courses/MATH102/December 2016/Question B 04 (b)

MATH102 December 2016

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Question B 04 (b)
The velocity of a particle is determined by its position so that the position function $x(t)$ satisfies the equation $x'=rx(1-x^{2})$ where $r>0$. The particle is at $x(0)=3$ initially. (b) Sketch the phase line for the equation and determine what happens to the position of the particle as $t\to \infty$?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Draw the graph of $f(x)=rx(1-x^{2})$ and find interval of $x'=f(x)>0$ and $x'=f(x)<0$. Based on this, sketch the phase line.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The below is the phase line for the given equation. As you see, based on the graph of $f(x)=rx(1-x^{2})$, we can easily get $x'=f(x)>0$ for $x<-1$ and $0<x<1$, while $x'=f(x)<0$ for $-1<x<0$ and $x>1$. Therefore, on $(-\infty ,-1)\cup (0,1)$, $x$ is increasing (i.e, move to the right on the graph), while on $(-1,0)\cup (1,\infty )$, $x$ is decreasing (i.e., move to the left). This gives the phase line as you see in the picture above. In particular, if we start at the point $x=3$, since it lies on $(1,\infty )$ it moves to the left and approaches to $x=1$ as $t\to \infty$. Answer: $\color {blue}\lim _{t\to \infty }x(t)=1$