Science:Math Exam Resources/Courses/MATH103/April 2017/Question 05

MATH103 April 2017

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Question 05
In a petri dish with a radius of $r=3{\text{cm}}$ scientists placed a bacteria killing agent along a diameter. They found the density of bacteria, $\rho (x)$, increased with distance $x$ cm from the diameter according to $\rho (x)=kx$ ${\text{bacteria}}/{\text{cm}}^{2}$ for some constant $k$. Determine the total population size of bacteria, $N$, in the petri dish.

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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
As the density of bacteria is given for any point on petri dish, estimate the population of bacteria on the grey strip illustrated in the question. Use this to set up the integral that determines the entire population (the quantity $N$), but do this with care as some functions are only antiderivatives on certain intervals.

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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. To estimate the number of bacteria in the shaded area, recall that the length of the grey strip in the petri dish that is a distance of $x$ from the diameter is, by Pythagoras' theorem, $2{\sqrt {r^{2}-x^{2}}}$. Since the width of the shaded area is $dx$, and because $r=3$, the number of bacteria in the grey strip is the product $\rho (x){\text{ }}2{\sqrt {3^{2}-x^{2}}}{\text{ }}dx$ Moreover, as $\rho (x)=\rho (-x)$, the number of bacteria to the right of the diameter equals to the number of bacteria to the left of the diameter. So as $N$ is the collection of the bacteria from all such grey strips, the number of bacteria to the right of the diameter is $N/2=\int _{0}^{3}\rho (x){\text{ }}2{\sqrt {3^{2}-x^{2}}}\,dx=\int _{0}^{3}2{\sqrt {3^{2}-x^{2}}}kx\,dx$ Use integration by substitution with the substitution $u=x^{2}$ to obtain $\int 2{\sqrt {3^{2}-x^{2}}}kx\,dx=k\int 2x{\sqrt {3^{2}-x^{2}}}\,dx=k\int {\sqrt {3^{2}-u}}\,du=-{\frac {2k}{3}}(3^{2}-u)^{3/2}=-{\frac {2k}{3}}(3^{2}-x^{2})^{3/2}.$ So, as $-{\frac {2k}{3}}(3^{2}-x^{2})^{3/2}$ is an antiderivative of $2{\sqrt {3^{2}-x^{2}}}kx$ on $[0,3]$, the Fundamental Theorem of Calculus gives $N/2=-{\frac {2k}{3}}(3^{2}-x^{2})^{3/2}{\bigg |}_{x=0}^{3}=-{\frac {2k}{3}}((3^{2}-3^{2})^{3/2}-(3^{2}-0^{2})^{3/2})=18k$ And it follows that $N=2N/2=36k.$ Hence, the total population of the bacteria is $\color {blue}36k$