residual = (initial amount)(1 -k)^t . . . . . . assuming k is a positive number
Where T is the half-life, this formula can also be expressed as ...
residual = (initial amount)(1/2)^(t/T)
Then the relationship between k and T is ...
(1 -k)^t = (1/2)^(t/T)
or ...
1 -k = (1/2)^(1/T)
This lets us write k in terms of T as ...
k = 1 -(1/2)^(1/T)
and it lets us write T in terms of k as ...
log(1-k) = (1/T)log(1/2)
T = log(1/2)/log(1-k)
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The attached spreadsheet table implements these formulas to compute T from k and vice versa. Formatting is in % and to four decimal places as required by the problem statement.
Let denote the amount of salt in the tank at time . We're told that .
For , the salt flows in at a rate of (1/5 lb/gal)*(5 gal/min) = 1 lb/min. When the regulating mechanism fails, 20 lbs of salt is dumped and no more salt flows for . We can capture this in terms of the unit step function and Dirac delta function as
(in lb/min)
The salt from the mixed solution flows out at a rate of
Then the amount of salt in the tank at time changes according to
Let denote the Laplace transform of , . Take the transform of both sides to get