The amount of rabbits on an island doubles every 3 years. Suppose that the growt

Question

The amount of rabbits on an island doubles every 3 years. Suppose that the growth is continuous Find the number of rabbits present after x years if they start with 100 rabbits. Determine how many rabbits are present after 7 years if they start with 100 rabbits. When will the population hit 6, 000 if they start with 100? (Chapter 7 Review #20) A tank contains 100L of pure water. Brine that contains 0.1kg of salt per liter enters the tank at a rate of 10 L per minute. The solution is kept thoroughly mixed and drains from the tank at the same rate. How much salt is in the tank after 6 minutes?

Explanation / Answer

15.
Let y be the number of rabbits at time t
we have dy/dt = (2/3)
a) so at time t = 0 , y(0) = 100
so, y = x(0) + (2/3) * t = 100 + ((2/3) * x) is the answer
b) for x = 7,
y = 100 + ((2/3)*7) = 105
c) for x = 6000 ,
6000 = 100 + ((2/3) *t)
t = 8850 years is the answer
16.
Let V be the volume of water in tank,
Initial Volume of water = V0 = 100 L
Conc of entering stream = Cin = 0.1 Kg/L
Rate of inflow = Fin = 10 L/min
Rate of outflow = F out = 10 L/mon
we have dV/dt = Fin - Fout =0
so by mass balance of salt in tank
d(V* C)/dt = (Fin * Cin) - (Fout *Cout)
100* dC/dt = (10 * 0.1) - (10*C)
so C = C1* e^(-0.1 *t) + 0.1
so initially C =0
so 0 = c1 + 0.1
C1 = -0.1
so C = -0.1 * e^(-0.1 *t) + 0.1
at time t = 6 minutes,
C = (-0.1 * e^(-0.1 *6)) + 0.1
C = 0.0451 Kg/L
So amount of slat in tank = v * C = 100 * 0.0451 = 4.512 Kg of salt is in the tank after 6 minutes

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