Science:Math Exam Resources/Courses/MATH101/April 2007/Question 06 (a)

MATH101 April 2007

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Question 06 (a)

Full-Solution Problems. Justify your answers and show all your work. Simplification of answers is not required. The population of fish in a lake is m million, where $m=m(t)$ varies with time t measured in years. The number of fish is currently 2 million. Suppose m satisfies the logistic-growth differential equation ${\frac {dm}{dt}}=16m\left(1-{\frac {m}{4}}\right)$ When will the number of fish equal 3 million? You may use the fact that the general solution to the logistic-growth differential equation $y'=ky(1-(y/K))$ is $y=K/(1+Ae^{-kt})$ where A is constant.

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
First solve for the constant A using the initial condition $m(0)=2$

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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. With $k=16$ and $K=4$, we know from the problem that $m(t)={\frac {4}{1+Ae^{-16t}}}$ Solving for A, we use the fact that $m(0)=2$ and see that $2=m(0)={\frac {4}{1+Ae^{-16(0)}}}={\frac {4}{1+A}}$ Cross multiplying, we see that $\displaystyle 2+2A=4$ and solving gives $A=1$. Thus, we have $m(t)={\frac {4}{1+e^{-16t}}}$. We want to solve for t when $m(t)=3$. Plugging this in yields $3={\frac {4}{1+e^{-16t}}}$ Cross multiplying, we see that $\displaystyle 3+3e^{-16t}=4$ Bringing the 3 over and dividing by 3 yields $\displaystyle e^{-16t}={\frac {1}{3}}$ Taking logarithms and then dividing by -16 gives $\displaystyle t={\frac {-\ln({\frac {1}{3}})}{16}}={\frac {\ln(3)}{16}}$ and this completes the problem.

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