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

MATH102 December 2016

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Question B 04 (a)
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. (a) What is the speed of the particle at $t=0$?

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
For given position function $x(t)$ of some object, the speed of it at $t$ means $|x'(t)|=|v(t)|$.

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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. First, we find $x'(0)$. Since we have the differential equation $x'(t)=rx(t)(1-(x(t))^{2})$, plugging $t=0$ and using $x(0)=3$, we get $x'(0)=rx(0)(1-(x(0))^{2})=3r(1-3^{2})=-24r$ Then, by its definition in the Hint, we get the speed of the particle at $t=0$ as $\color {blue}|x'(0)|=24r$.