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rates of change

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Heinsbergrelatz

Posted: Wed Mar 03, 2010 8:24 am Post subject: rates of change

Forum Masters Degree

Joined: 01 Aug 2009
Posts: 598
Location: Singapore

as im continuously studying calculus alone at home, i seem to come up with alot of questions i tumble upon.. i was wondering if you guys could help me despite my continuous posts in this particular sub forum. Embarassed

one more thing i realized, the more i do maths, the more distant my understanding becomes. So regularly i do 5 hours of maths everyday, and every time i do it, it seems to puzzle me more and more. Confused

it really bugs me...

anyway here is the question; this is actually a ral simple question, where you apply the chain rule, and just find the derivative, but i just cant get around to the answer.

a trough of length 6m has a uniform cross-section which is an equilateral triangle with sides of 1m. Water leaks from the bottom of the trough at a constant rate of $0.1m^{3}/min$ . Find the tate of change in which the water level is falling at the instant when it is 20cm deep.

formula of a trough (triangular prism)- $\frac{1}{2} bhl$

thank you in advance
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$G_\mu_\nu + \Lambda g_\mu_\nu=\frac{8\pi G}{c^{4}} T_\mu_\nu$

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Arcane_Mathematician

Posted: Wed Mar 03, 2010 3:21 pm Post subject:

Forum Cosmic Wizard

Joined: 21 Feb 2009
Posts: 2478
Location: San Jose, California

Related Rates.

What you need to do for these kinds of problems is define the rate of change in the dimension you are looking for in terms of the total volume.

volume of your trough: $\frac{1}{2}lhb\ l=600\ b=\frac{2\sqrt{3}}{3}h$

$\frac{dV}{dt}=.1\frac{m^3}{min}$ and $\frac{dV}{dh}=\dfrac{600\frac{2\sqrt{3}}{3}h^2}{2}=200\sqrt{3}h^2$

$\frac{dV}{dt}=\frac{dV}{dh}\frac{dh}{dt}=200\sqrt{3}h^2\frac{dh}{dt}$ which means:

$200\sqrt{3}h^2\frac{dh}{dt}=.1\frac{m^3}{min}$

$\frac{dh}{dt}=\dfrac{1}{2000\sqrt{3}h^2}\frac{m^3}{min}$ and continue to plug and chug.
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Heinsbergrelatz

Posted: Sun Mar 14, 2010 8:04 am Post subject:

Forum Masters Degree

Joined: 01 Aug 2009
Posts: 598
Location: Singapore

thank you
_________________
------------------

$G_\mu_\nu + \Lambda g_\mu_\nu=\frac{8\pi G}{c^{4}} T_\mu_\nu$

-------------------

Fox

Posted: Tue Jul 27, 2010 6:40 pm Post subject:

New Member

Joined: 26 Jul 2010
Posts: 3

I'm new and have been browsing around the forums. I think I found an error in the response to this post. The formula I found for the height of an equilateral triangle is

$h=\frac{b \sqrt{3}} {2}$

solving for b

$b=\frac{2h} {\sqrt{3}}$

Last edited by Fox on Tue Jul 27, 2010 7:32 pm; edited 3 times in total

Harold14370

Posted: Tue Jul 27, 2010 7:02 pm Post subject:

Moderator

Joined: 13 Apr 2007
Posts: 3061
Location: Pennsylvania

Fox wrote:

I'm new and have been browsing around the forums. I think I found an error in the response to this post. The formula I found for the height of an equilateral triangle is

h={b3^1/2}/2

solving for b

b=2h/3^1/2

You say. $b=\frac{2h}\sqrt{3}$

Arcane says $b=\frac{2\sqrt{3}}{3}h$

If you multiply your equation by $\frac{\sqrt{3}}{\sqrt3}$ I think you will see they are the same.

Fox

Posted: Tue Jul 27, 2010 7:24 pm Post subject:

New Member

Joined: 26 Jul 2010
Posts: 3

oh, thanx

I also wondered why he converted length to centimeters, but left the volume change rate in cubic meters/ min.

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