A1000 gallon tank, initially full of water, develops a leak at the bottom. given that 200 gallons of water leak out in the first 10 minutes, find the amount of water, a(t), left in the tank t minutes after the leak develops if the water drains off at a rate proportional to the amount of water present.

Let the amount of water remaining be a, then: da/dt = -ka, since the rate of loss of water is proportional to the amount remaining in the tank, with k as proportionality constant.
Solving the equation gives: da/a = -kdt
Integrating: ln(a0 / a) = kt
Substituting the given a0 = 1000 gallons and a = 800 gallons (since 200 has leaked out) at the given t = 10 minutes:
ln(1000 / 800) = 10k
k = 0.0223 (rate constant)
Then we substitute back into the equation to get:
ln(1000 / a) = 0.0223t
1000 / a = e^(0.0223t)
a = 1000e^(-0.0223t)