Science:Math Exam Resources/Courses/MATH110/December 2017/Question 08 (b) (ii)

MATH110 December 2017

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Question 08 (b) (ii)
A second bacteria culture (containing a slow growing bacterium) has size ${\textstyle \displaystyle B(t)={\frac {4t+3}{1+t}}}$ bacteria at time ${\textstyle t}$ (in days). (ii) At ${\textstyle t=1}$, is the culture’s growth rate increasing or decreasing?

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
The growth rate is given by $B'(t)$. To see it is increasing or decreasing, we need to find the rate of change of the growth rate, which is the second derivative.

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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. Following the Hint, we need to see whether ${\textstyle B''(1)>0}$ or ${\textstyle B''(1)<0}$. We have seen in part (b)(i) that $$B'(t)={\dfrac {1}{(1+t)^{2}}}=(1+t)^{-2}.$$ Differentiating again using the power rule, we have $$B''(t)=-2(1+t)^{-3}=-{\dfrac {2}{(1+t)^{3}}}.$$ At time ${\textstyle t=1}$, we have $$B''(1)=-{\dfrac {2}{(1+1)^{3}}}=-{\dfrac {2}{2^{3}}}=-{\dfrac {1}{4}}<0.$$ Since ${\textstyle B''(1)<0}$, ${\textstyle B'(t)}$ is decreasing at ${\textstyle t=1}$. Answer: The culture’s growth rate at ${\textstyle t=1}$ is ${\textstyle {\color {blue}{\mbox{decreasing}}}}$.