Science:Math Exam Resources/Courses/MATH110/April 2013/Question 07

MATH110 April 2013

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Question 07
The Bubble Nebula is an expanding sphere of gas and stellar ejecta in the constellation Cassiopeia. The radius of the Bubble Nebula is 3(1013) km, and it is expanding at a rate of 7(106) km/h. Determine how quickly the volume of the nebula is increasing. Hint: the volume of a sphere of radius r is (4/3)πr3

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
We are told the rate at which the radius of a sphere is changing and asked to find the rate at which the volume is changing. How can we use the volume equation provided to relate the rates?

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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. Since we are interested in the rate of change of the volume of the sphere while being given information about the rate of change of its radius we are facing a typical related rates problem. There are two variables here at stake: the radius of the sphere, $r$, and its volume, $V$. Now there isn't ONE sphere, since the sphere is expanding with time. This means that actually, both the radius and the volume are actually functions of time that we can write $r(t)$ and $V(t)$. We know the relationship that exists between the volume and the radius and a sphere, which at any time $t$ is expressed by the equation $\displaystyle V(t)={\frac {4}{3}}\pi (r(t))^{3}$ Since we are interested in the relationship that links the rates of change with respect of time for both the radius and the volume of the sphere, it makes sense to differentiate this equation on both sides with respect to time. We obtain: ${\begin{aligned}V'(t)&={\frac {4}{3}}\pi \cdot 3(r(t))^{2}\cdot r'(t)\\&=4\pi (r(t))^{2}\cdot r'(t)\end{aligned}}$ Now the problem tells us that there is a specific time, let's call it $t_{0}$, at which we know both the radius of the sphere and its instantaneous rate of change. In other words, if we use the data given in the statement of the problem, we have: $r(t_{0})=3\cdot 10^{13}$ $r'(t_{0})=7\cdot 10^{6}$ And we are interested in obtaining the rate of change of volume at that moment, that is $V'(t_{0})$. So we can use the equation we obtained above at the time $t_{0}$, substitute the available information and obtain that: ${\begin{aligned}V'(t_{0})&=4\pi (r(t_{0}))^{2}\cdot r'(t_{0})\\&=4\pi (3\cdot 10^{13})^{2}\cdot 7\cdot 10^{6}\\&=4\pi \cdot 9\cdot 10^{26}\cdot 7\cdot 10^{6}\\&=4\cdot 7\cdot 9\cdot \pi \cdot 10^{32}\\&=252\pi \cdot 10^{32}\end{aligned}}$ Since $252\pi$ is roughly equal to $800$, we can say that the Bubble Nebula is expanding at a rate of approximatively 8(1034) km3/h. Fun Fact: For info, the volume of the Earth is roughly 1012 km3 and the Sun's volume is roughly 1018 km3.

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