Science:Math Exam Resources/Courses/MATH102/December 2017/Question 07 (c)

MATH102 December 2017

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Question 07 (c)

New words are constantly being introduced into a language. The usage of one particular new word is thought to grow according to a logistic equation $$u'=2u(1-u)-au$$ where ${\textstyle u}$ is the fraction of the population using the word. The usage increases at a rate proportional to the product of the fraction of people who use the word and the fraction that don’t use it and decreases at a rate proportional to the number of users. (c) When the word does catch on and spread, what is the steady state fraction of the population that uses the word? A. ${\textstyle 0}$ B. ${\textstyle {\dfrac {a}{2}}}$ C. ${\textstyle 1-{\dfrac {a}{2}}}$ D. ${\textstyle 2-a}$ E. ${\textstyle 1}$

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
At the steady state, the rate of change (derivative) is zero.

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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. By the Hint, we need to solve for ${\textstyle u'=0}$ in the case ${\textstyle u>0}$: $$2u(1-u)-au=0,$$ $$[2(1-u)-a]u=0.$$ Since ${\textstyle u>0}$ (the word does catch on and spread), we can divide both sides by ${\textstyle u}$ to obtain $$2(1-u)-a=0$$ $$2(1-u)=a$$ $$1-u={\dfrac {a}{2}}$$ $$u=1-{\dfrac {a}{2}}.$$ Answer: ${\textstyle {\color {blue}C}}$.