How Fast Is Water Draining from the Tank?

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SUMMARY

The discussion focuses on calculating the rate at which water drains from a tank using Toricelli's Law. The volume of water remaining in the tank after time t is given by the equation V = 3500(1 - t/50)² for 0 ≤ t ≤ 50. Participants explored the derivative V'(t) to find the rate of drainage at specific times: 5, 10, and 20 minutes. The consensus is that using the chain rule simplifies the differentiation process compared to the quotient rule, yielding negative results as expected since the water volume decreases over time.

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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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