Science:Math Exam Resources/Courses/MATH103/April 2011/Question 08

MATH103 April 2011

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Question 08

In a lab a bacterial colony is grown in a petri dish. The colony is circular and the area $A$ covered by the colony increases at a rate proportional to its circumference $C$ (because of the high nutrient concentration in the surrounding area): ${\frac {dA}{dt}}=k\cdot C.$ One day ($t=0$) the colony was observed to cover an area of $1[mm^{2}]$. The next day the colony had grown to cover $2[mm^{2}]$. What area does the colony cover after $T$ days? (Assume that the petri dish is large enough such that the colony never reaches the boundary.)

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
How are the area and circumference of a circle related? Can you relate both to the radius r(t)?

Hint 2
Take the derivative of $A(t)=\pi (r(t))^{2}$ and relate it to the given equation for $dA/dt$. Can you solve for r(t)?

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. We define $r(t)$ to be the radius (in [mm]) of the area $A(t)$ (in [mm2]) at time $t$. Then the circumference of $A$ is $C=2\pi r$ and we know that ${\frac {dA}{dt}}=k\cdot C=k\cdot 2\pi r(t).$ The area $A$ expressed in terms of $r$ is $A=r^{2}\pi$. We take the derivative on both sides and get ${\frac {d}{dt}}A={\frac {d}{dt}}(r^{2}\pi )=2r(t){\frac {dr}{dt}}\pi .$ We combine the two equations $dA/dt=2k\pi r(t)$ and $dA/dt=2r(t)dr/dt\pi$ above and cancel $\ 2\pi r(t)$ to get $\displaystyle {\frac {dr}{dt}}=k.$ Integrating leads to $r(t)\ =kt+const.$ To find $A(t)$ again we get back to the formula $A(t)\ =r(t)^{2}\pi =(kt+const)^{2}\pi .$ Now all that's left to do is find the constants $k$ and $const$. For that we use the information given: $A(t)=1$. So, we get $A(0)=(k\cdot 0+const)^{2}\pi =const^{2}\pi =1,$ which leads to $\displaystyle const={\frac {1}{\sqrt {\pi }}}.$ The unit of constant is [mm]. We can take the positive solution of the root, since negative numbers don't make sense in this context. Now we use $A(1)=\left(1\cdot k+{\frac {1}{\sqrt {\pi }}}\right)^{2}\pi =2,$ which gives $k={\frac {({\sqrt {2}}-1)}{\sqrt {\pi }}}.$ The unit of k is [mm/day]. Hence, the final equation for $A(t)$ is given by $A(t)=\left({\frac {({\sqrt {2}}-1)}{\sqrt {\pi }}}t+{\frac {1}{\sqrt {\pi }}}\right)^{2}\pi ,$ which you can simplify to $A(t)=\left(({\sqrt {2}}-1)t+1\right)^{2}.$

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