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

MATH110 December 2017

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Question 08 (b) (i)
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). (i) How fast is the culture growing (in bacteria per day) 1 day after the culture was first established?

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 rate of growth is the derivative of the size. So we need to find ${\textstyle B'(1)}$.

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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 find ${\textstyle B'(1)}$. First we differentiate ${\textstyle B(t)}$. Instead of using the quotient rule, we can actually rewrite $$B(t)={\dfrac {4t+3}{1+t}}={\dfrac {(4t+4)-1}{1+t}}=4-{\dfrac {1}{1+t}}.$$ It is easier to use this form than to differentiate directly using the quotient rule. Indeed, using the power rule (note that ${\textstyle {\dfrac {1}{1+t}}=(1+t)^{-1}}$), we have $${\begin{aligned}B'(t)&=0-\left((1+t)^{-1}\right)'\\&=-(-1)(1+t)^{-2}\\&={\dfrac {1}{(1+t)^{2}}}.\end{aligned}}$$ When ${\textstyle t=1}$, $$B'(1)={\dfrac {1}{(1+1)^{2}}}={\dfrac {1}{2^{2}}}={\dfrac {1}{4}}.$$ So the rate of growth is ${\textstyle {\dfrac {1}{4}}}$. Answer: the rate of growth of the culture is ${\textstyle {\color {blue}{\dfrac {1}{4}}}}$.