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datanewb
 4 years ago
Related rates 2 E4: Sand is pouring on a conical pile at 12m^3/min. The diameter of the base is always 3/2 the height. Find the rate at which the height is increasing when the pile is 2 m tall.
datanewb
 4 years ago
Related rates 2 E4: Sand is pouring on a conical pile at 12m^3/min. The diameter of the base is always 3/2 the height. Find the rate at which the height is increasing when the pile is 2 m tall.

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datanewb
 4 years ago
Best ResponseYou've already chosen the best response.0The question is located here: http://ocw.mit.edu/courses/mathematics/1801scsinglevariablecalculusfall2010/partcmeanvaluetheoremantiderivativesanddifferentialequations/problemset5/MIT18_01SC_pset2prb.pdf And the answer is here: http://ocw.mit.edu/courses/mathematics/1801scsinglevariablecalculusfall2010/partcmeanvaluetheoremantiderivativesanddifferentialequations/problemset5/MIT18_01SC_pset2sol.pdf

datanewb
 4 years ago
Best ResponseYou've already chosen the best response.0I get an answer that is different from the solution provided in the answer key. I'm going to show my work, and perhaps someone can tell me where I went wrong, or let me know that there is an error in the solutions.

datanewb
 4 years ago
Best ResponseYou've already chosen the best response.0\[r = \frac{3h}{4} \\ V = \frac{\pi r^2h}{3}\\ V = \frac{3\pi h^3}{16} \\ V' = \frac{9\pi h^2h'}{16}\\ V' = 12 ,h=2\\12 = \frac{9\pi 2^2h'}{16} \\h' = \frac{16}{3\pi}\]

datanewb
 4 years ago
Best ResponseYou've already chosen the best response.0Since this is a related rates question, I realized Leibniz notation would be more apt. \[V = \frac{3\pi h^2}{16} \\ \frac{dV}{dh} = \frac{9\pi h^2}{16} \\\frac{dV}{dt} = \frac{dh}{dt} * \frac{dV}{dh} \\ 12 = \frac{dh}{dt} * \frac{9\pi h^2}{16} \\ \frac{dh}{dt} = \frac{12 * 16}{9\pi h^2} \\ \frac{dh}{dt} = \frac{16}{3\pi}\] Still not sure why the solution gave a different answer, but I'm moving on. I can't see how my answer could be wrong.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I am not sure either, but I think that there is some mistake in the solution. They have set dr/dt= 12, whereas the problem states that dV/dt = 12 m^3/min. Also, the solution solves for dV/dt, but the problem asks you to find dh/dt.

datanewb
 4 years ago
Best ResponseYou've already chosen the best response.0Thank you for taking the time to look at this problem even though it was a closed question. I think you are right. Are you working through the OCW course as well?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Not really, I was trying to work through the ODE course but haven't for a while. I like to look through peoples' questions about Calculus because it is a good review and there are many parts of Calculus I didn't really understand when I took it. Plus, it is fun! Also, what is a closed question? I saw this label, but didn't know where to find an explanation for it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I guess this means the question is old and/or solved?

datanewb
 4 years ago
Best ResponseYou've already chosen the best response.0Well, on OpenStudy you are only able to to have one open question at a time. The closed questions are really an archive of all question asked. Some of the closed questions have been solved, some haven't, but it is assumed when the question is closed that the original poster has moved on to another question. However, I find it fun to look through the closed questions sometimes too, and solve one that never was solved.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you! I also had stumped on this error while trying to do all exercices! I also found a mistake with problem 2B1b. So yeah it's reassuring to know our answers are correct! :)
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