Science:Math Exam Resources/Courses/MATH103/April 2015/Question 04

MATH103 April 2015

  •  Q1 (a) (i)  •  Q1 (a) (ii)  •  Q1 (a) (iii)  •  Q1 (b) (i)  •  Q1 (b) (ii)  •  Q1 (b) (iii)  •  Q1 (c) (i)  •  Q1 (c) (ii)  •  Q1 (c) (iii)  •  Q1 (d) (i)  •  Q1 (d) (ii)  •  Q1 (e) (i)  •  Q1 (e) (ii)  •  Q1 (e) (iii)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q2 (d)  •  Q2 (e)  •  Q3 (a)  •  Q3 (b)  •  Q3 (c)  •  Q4  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q6 (a)  •  Q6 (b)  •  Q6 (c)  •  Q7 (a)  •  Q7 (b)  •  Q7 (c)  •  Q7 (d)  •  Q8  •  Q9  •  Q10 (a)  •  Q10 (b)  •  

   Other MATH103 Exams

  •  April 2006  •  April 2005  •  April 2011  •  April 2010  •  April 2009  •  April 2012  •  April 2013  •  April 2014  •  April 2015  •  April 2016  •  April 2017  •  

Question 04
Consider a swarm of ants distributed over a circular region. At distance $r$ from the centre of the region, the density of the ant population is observed to be $b(r)={\frac {1}{r^{2}+1}}$, measured in units of one thousand per square meter. What is the total number of ants within a radius of 2m? (Work must be shown for full marks.)

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
Find the integral expression for the population of ants within a radius of $2$ using its density function $b$.

Checking a solution serves two purposes: helping you if, after having used the hint, 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 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The most difficult part of this problem is setting up the integral. Consider a region formed between a circle of radius $2$ and a circle of radius $2-dr$ where $dr$ is infinitesimally thin. The population of ants inside this small region is given by $b(r)\times Area=b(r)2\pi rdr$ We can get the total population by adding up the contributions of an infinite number of these small regions in our circle of radius $2$, or in other words, we can get the total population by integrating the above equation from $0$ to $2$. Population = $\int _{0}^{2}{\frac {2\pi rdr}{r^{2}+1}}=2\pi \int _{0}^{2}{\frac {rdr}{r^{2}+1}}$ Use substitution: $u=r^{2}+1$ $du=2rdr$ ${\frac {du}{2}}=rdr$ $2\pi \int _{0}^{2}{\frac {rdr}{r^{2}+1}}={\frac {2}{2}}\pi \int _{0^{2}+1}^{2^{2}+1}{\frac {du}{u}}=\pi \int _{1}^{5}{\frac {du}{u}}$ $=\pi (ln(5)-ln(1))=\pi ln(5)$ Conclusion: There are $\color {blue}1000\times \pi ln(5)$ ants

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Let $P(t)$ be the population of ants within a radius of $t$ with units of one thousand. Then, it has a relation with the density as follows: $P(t)=\int _{B_{t}}bdA.$ where $B_{t}$ represents an open ball centered at the origin with the radius $t$ and $dA$ is an infinitesimal area. Writing the integral in the polar coordinate, we have $\int _{B_{t}}bdA=\int _{-\pi }^{\pi }\int _{0}^{t}b(r)rdrd\theta =2\pi \int _{0}^{t}b(r)rdr.$ Plugging given $b$ into the integral and using the change of variable $s=r^{2}+1$, we have $ds=2rdr$ and $\int _{0}^{t}b(r)rdr=\int _{0}^{t}{\frac {r}{r^{2}+1}}dr={\frac {1}{2}}\int _{1}^{t^{2}+1}{\frac {1}{s}}ds={\frac {1}{2}}\ln(1+t^{2}).$ Therefore, the population of ants within a radius of $2$ is $P(2)=2\pi \cdot {\frac {1}{2}}\ln(1+2^{2})=\color {blue}\pi \ln 5(\times 1000ants)$.