Let y(t) represent the level of water in inches at time t in hours. Then we are given ...
y'(t) = k√(y(t)) . . . . for some proportionality constant k
y(0) = 30
y(1) = 29

We observe that a function of the form
y(t) = a(t - b)²
will have a derivative that is proportional to y:
y'(t) = 2a(t -b)

We can find the constants "a" and "b" from the given boundary conditions.
At t=0
30 = a(0 -b)²
a = 30/b²
At t=1
29 = a(1 - b)² . . . . . . . . . substitute for t
29 = 30(1 - b)²/b² . . . . . substitute for a
29/30 = (1/b -1)² . . . . . . divide by 30
1 -√(29/30) = 1/b . . . . . . square root, then add 1 (positive root yields extraneous solution)
b = 30 +√870 . . . . . . . . simplify

The value of b is the time it takes for the height of water in the tank to become 0. It is 30+√870 hours ≈ 59 hours 29 minutes 45 seconds

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3 years ago

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Pls help will mark brainliest