Gas Tank Leak Rate Problem: Modeling and Calculating with V(t)=90(1-t/18)^2

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SUMMARY

The gas leak rate from a parked pickup truck's tank can be modeled using the function V(t) = 90(1 - t/18)^2, where V represents the volume of gas in liters after t hours, with t ranging from 0 to 18. To determine the leak rate at t = 12 hours, one must compute the first derivative, dV/dt, and evaluate it at t = 12. The correct leak rate is 3.33 liters per hour, confirming that the book's answer is incorrect.

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xdeanna
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The gas tank of a parked pickup truck develops a leak. The amount, V, in liters, of gas remaining in the tank after t hours can be modeled by the function
V(t)= 90(1-t/18)^2 , t is between 0 and 18.

b) how fast is the gas leaking from the tank at t=12h

I tried V(12) and V'(12) but still didn't get the answer at the back of the book. Shouldn't it be the first derivative at t=12?
 
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xdeanna said:
The gas tank of a parked pickup truck develops a leak. The amount, V, in liters, of gas remaining in the tank after t hours can be modeled by the function
V(t)= 90(1-t/18)^2 , t is between 0 and 18.

b) how fast is the gas leaking from the tank at t=12h

I tried V(12) and V'(12) but still didn't get the answer at the back of the book. Shouldn't it be the first derivative at t=12?

Yes, you should compute
[tex]\frac{dV}{dt}[/tex]
and evaluating for t =12

You should get 3.33 l/h
 
Thanks :) I got that too.. the book is wrong
 

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