Science:Math Exam Resources/Courses/MATH110/April 2011/Question 06

MATH110 April 2011

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Question 06

Stalagmites are cone-shaped mineral deposits which rise from the floors of limestone caves. They form over millenia as dripping water deposits calcium carbonate onto the cave floor. Consider a cone-shaped stalagmite with a length always equal to five times its radius. Suppose the stalagmite's height increases at a rate of 0.13 millimetres per year. Write down two expressions for the rate of change of its volume: (i) with respect to its height h, and (ii) with respect to its radius r. Hint: the volume V of a cone with radius r and height h is equal to ${\frac {1}{3}}\pi r^{2}h$.

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
What fact about the proportions of the stalagmite can you use to relate the variables r and h? Once you have done this, you can write the volume formula either in terms of h (for part i) or r (for part ii).

Hint 2
Try differentiating the equation relating h and r. What does this tell you about the relationship between $dh/dt$ and $dr/dt$?

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 are given the volume of a cone. $V={\frac {1}{3}}\pi r^{2}h$ If we differentiate this expression as it is, we will end up with an expression relating $dV/dt$ to both variables of r and h. To avoid this, we need to write the volume formula in terms of just r or h, by writing one variable in terms of another. The question states that the length of the stalagmite is always five times its radius, so we can write $h=5r$, or, alternatively, $r=h/5$. (i) Finding an expression in terms of h: Using $r=h/5$, we plug it into our volume formula to get ${\begin{aligned}V&={\frac {1}{3}}\pi \left({\frac {h}{5}}\right)^{2}h\\&={\frac {1}{3}}\pi {\frac {h^{3}}{25}}\end{aligned}}$ Differentiating yields ${\begin{aligned}{\frac {dV}{dt}}&={\frac {1}{3}}\pi {\frac {3h^{2}}{25}}{\frac {dh}{dt}}\\&={\frac {\pi h^{2}}{25}}{\frac {dh}{dt}}\end{aligned}}$ And because we know that $dh/dt=0.13$ mm/year, we finally reach an expression relating the rate of change in volume to the height h ${\frac {dV}{dt}}={\frac {\pi h^{2}}{25}}0.13$ (ii) Finding an expression in terms of r: To find an expression in terms of r we simply use the relationship h = 5r to get ${\begin{aligned}V&={\frac {1}{3}}\pi r^{2}(5r)\\&={\frac {5}{3}}\pi r^{3}\end{aligned}}$ Differentiating yields ${\begin{aligned}{\frac {dV}{dt}}&={\frac {5}{3}}\pi 3r^{2}{\frac {dr}{dt}}\\&=5\pi r^{2}{\frac {dr}{dt}}\end{aligned}}$ To find an expression for $dV/dt$ solely in terms of $r$, we need to find $dr/dt$. We can do this by differentiating the equation $h=5r$ to get ${\frac {dh}{dt}}=5{\frac {dr}{dt}}$ ${\frac {0.13}{5}}={\frac {dr}{dt}}$ Plugging this into our formula for $dV/dt$ gives ${\begin{aligned}{\frac {dV}{dt}}&=5\pi r^{2}{\frac {0.13}{5}}\\&=\pi 0.13r^{2}\end{aligned}}$ in terms of r, as desired.

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