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

MATH102 December 2016

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Question B 04 (c)

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. (c) Suppose that at $t=1s$, the particle is at $x(1)=2$. Another particle whose position $x_{2}(t)$ is determined by the same equation is initially at $x_{2}(0)=-3$. What is the location of this particle at $t=1s$? Hint: Look at the phase line carefully. Explain briefly.

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
Observe that the phase line has the reflectional symmetry with respect to the point $x=0$.

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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. As we mentioned in the Hint, the phase line has the reflectional symmetry with respect to the point $x=0$. This implies that if we start at $x_{2}(0)=-x(0)=-3$, then in time $t(s)$, it moves the same amount but to the opposite direction. In particular, in $1s$, we have $\color {blue}x_{2}(1)=-x(1)=-2$.