How Fast Is Water Draining from the Tank?

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Homework Help Overview

The discussion revolves around a problem involving the rate of water draining from a tank, modeled by Toricelli's Law. The tank holds 3500 gallons and drains over a period of 50 minutes, with a mathematical expression provided for the volume of water remaining over time.

Discussion Character

  • Mathematical reasoning, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss different methods for finding the derivative of the volume function, with some suggesting the use of the chain rule instead of the quotient rule. There is also an exploration of the expected negative sign in the derivative, reflecting the decreasing volume of water.

Discussion Status

Participants are actively engaging with the problem, offering alternative approaches to differentiation and questioning the methods used by others. There is a recognition of the need for negative results in the context of the problem, indicating a productive exploration of the topic.

Contextual Notes

Some participants express confusion over the application of differentiation rules and the interpretation of the results, highlighting the complexity of the problem setup.

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Homework Statement



If a tank holds 3500 gallons of water, which drains from the bottom of the tank in 50 minutes, then Toricelli's Law gives the volume V of water remaining in the tank after t minutes as the following.

V = 3500 (1-t/50)^2 0 </=t<=50

Find the rate at which water is draining from the tank after the following amounts of time.

(a) 5 min

(b) 10 min

(c) 20 min


Homework Equations


Basic knowledge of derivatives. The quotient rule?


The Attempt at a Solution



I attempted to solve for V'(t).

V = 3500((1-(t/50)2)

V'(t) = 1(1-2t(2500)-t^2(1))/(2500)2

I then plugged in 5 for t = 5 and definitely got the wrong answer. The answer should be negative, since the water in the tank is decreasing, and it should be greatest at the smaller times since there is more water in the tank to leave it.
 
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I factored out the equation so that V=3500(1-t/50)² becomes V=3500(1-2t/50+t²/2500)
Then I derived that to get: V'(t)=3500(-2/50+2t/2500)
Much easier than quotient rule, I think.
Then sub in your values and get negative results as required.
 
There is no need to use the quotient rule- there is no variable in the deominator: just the chain rule: V'= (3500)(2)(1- t/50)(-1/50).

And, redargon, you didn't "factor out" anything, you multiplied what was already factored.
 
ah, indeed. :shy:
 

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