Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

Related rates 2 E-4: 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.

OCW Scholar - Single Variable Calculus
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
The question is located here:http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-c-mean-value-theorem-antiderivatives-and-differential-equations/problem-set-5/MIT18_01SC_pset2prb.pdf And the answer is here: http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-c-mean-value-theorem-antiderivatives-and-differential-equations/problem-set-5/MIT18_01SC_pset2sol.pdf
I 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.
|dw:1347937101924:dw|

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

\[r = \frac{3h}{4} \\ V = \frac{\pi r^2h}{3}\\ V = \frac{3\pi h^3}{16} \\ V' = \frac{9\pi h^2h'}{16}\\ V' = 12 ,h=2\\12 = \frac{9\pi 2^2h'}{16} \\h' = \frac{16}{3\pi}\]
Since 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.
I 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.
Thank 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?
Not 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.
I guess this means the question is old and/or solved?
Well, 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.
Thank you! I also had stumped on this error while trying to do all exercices! I also found a mistake with problem 2B-1b. So yeah it's reassuring to know our answers are correct! :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question