Science:Math Exam Resources/Courses/MATH104/December 2015/Question 03 (a)

MATH104 December 2015

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q2 (a)  •  Q2 (b)  •  Q3 (a)  •  Q3 (b)  •  Q3 (c)  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q6 (a)  •  Q6 (b)  •  Q7  •  Q8 (a)  •  Q8 (b)  •  Q8 (c)  •  Q9  •  Q10 (a)  •  Q10 (b)  •  Q10 (c)  •  Q10 (d)  •  Q11  •  

   Other MATH104 Exams

  •  December 2014  •  December 2011  •  December 2010  •  December 2012  •  December 2013  •  December 2015  •  December 2016  •  

Question 03 (a)
A (spherical) balloon is being deflated at the rate of 8 cm3/s. How fast is its radius changing when the radius is 2 cm? Note: The volume $V$ of a sphere of radius $r$ is ${\tfrac {4\pi }{3}}r^{3}.$

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
If the volume and radius of the balloon are considered as functions of time ($V(t)$ and $r(t),$ respectively), then the rate of change in the balloon's volume can be expressed as ${\tfrac {dV}{dt}}.$ Likewise, the rate of change in the balloon's radius is ${\tfrac {dr}{dt}}.$

Hint 2
We are also given the balloon's volume as a function of its radius: namely, $V(r)={\tfrac {4\pi }{3}}r^{3}.$ How might we use this to relate ${\tfrac {dV}{dt}}$ (which is given) to ${\tfrac {dr}{dt}}$ (which is what must be found)?

Hint 3
Consider using the chain rule.

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. Let $V(t)$ be the volume of the balloon (in cm3) at time $t$ (in s); let $r(t)$ be its radius (in cm) at time $t$ (in s). We know that $V(r)={\color {Orange}{\frac {4\pi }{3}}r^{3}}.$ To find ${\frac {dr}{dt}},$ the rate of change in the radius of the balloon, we observe that by the chain rule ${\begin{aligned}{\frac {dV}{dt}}&={\frac {dV}{dr}}\cdot {\frac {dr}{dt}}\\{\frac {dV}{dt}}&={\frac {d}{dr}}\left({\color {Orange}{\frac {4\pi }{3}}r^{3}}\right)\cdot {\frac {dr}{dt}}\\{\frac {dV}{dt}}&=4\pi r^{2}\cdot {\frac {dr}{dt}}\,.\\\end{aligned}}$ We are given that ${\frac {dV}{dt}}={\color {OliveGreen}-8}$ (the negative sign indicates that the volume is decreasing as the balloon is being deflated) and $r={\color {OrangeRed}2}$ at the moment in question. Thus ${\begin{aligned}{\color {OliveGreen}-8}&=4\pi ({\color {OrangeRed}2}^{2})\cdot {\frac {dr}{dt}}\\{\frac {dr}{dt}}&=-{\frac {8}{16\pi }}=-{\frac {1}{2\pi }}\,.\end{aligned}}$ Hence when the radius is 2 cm, it is decreasing at a rate of ${\color {blue}{\frac {1}{2\pi }}\,{\frac {\text{cm}}{\text{s}}}}\,.$