Exponential decay

A tank contains 55 liters of water. Water is flowing out at the rate of 7% per minute. How long does it take to drain the tank?

Correct answer:

t = min INF
t2 =  14.2857 min

Step-by-step explanation:

p1=7%=1007=0.07 q=1p1=10.07=10093=0.93 V1=55 l V2=V1 q=55 0.93=201023=51.15 l V3=V2 q=51.15 0.93=47.5695 l V4=V3 q=47.5695 0.9344.2396 l V5=V4 q=44.2396 0.9341.1429 l  V(n) = V(n1)  q V(n) = V1   e t/τ q = e t/τ  τ=1/lnq=1/ln0.9313.7797 min  Vn = 0  t=
t2=1/p1=1/0.07=14.2857 min

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Showing 3 comments:
t = infinity time to drain at rate 7% per minute of remaining volume. (exponential decay curve never touch  or cross zero - line y=0).  Time constant τ = 13.77 min

t2 = 14.286 min if we took 7% as linear volume decay, but this is not correct in this case.

see more on https://en.wikipedia.org/wiki/Exponential_decay

Math Student
Thank you so much for your assistance.
I tried to solve using geometric sequence.
At t=~13.77 mins, there's ~20.2 liters of water left in the tank.

I do realize that the exponential function will never be zero. However, logically there must be a finite time when the tank will be emptied.
Still can't resolve the situation.

However, thank you again for your time and effort.

Yes, the right answer is then: the tank is empty in infinity time... T=13.77 mins is time constant, all decays can be normalized in time to have the same curve, but technically emptied at a time approx. 3-5 time constants. After time constant 13.77 mins the tank is on 63%... etc



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