A 200gallon tnak currently holds 30 pounds of salt dissolved
(a)Set up and solve a differential equation for the amount of salt S(t) in solution at time t.
(b)How many gallons of solution are in the tank when there are 40 pounds of salt in solution?
(c) How much salt is in the tank when it overflows?
Solution
(a) let S(t) is the amount of salt present in the tank at time t
At t=0 ,
S(0) = 30 pounds
dS(t)/dt = 4 pounds/minute = 4 /60 pounds / sec = 1/15 pounds/sec
dS(t) = 1/15 dt (differential equation )
integrating on both sides
15S(t) = t + constant
at t=0 S(t) = 30
consatnt = 15 *30 = 450
a) S(t) = (t+450)/15
b) Given S(t) = 40
40 = (t + 450) /15
t= 150 s
total solution inthe tank T(t)
dT(t)/dt = 2/60 gallons/sec
integrating on both sides
30T(t) = t + consatnt
T(0) = 50
constant = 30*50 = 1500
T(t) = (t+1500)/30
GIven to find T(150)
T(150) = (150+1500)/30 = 55 gallons
b) Solution =55 gallons
c)to find t when T(t) = 200 gallons as the tank overflows when the tank is full and capavity of the tank is 200 gallons
200 = (t+1500)/30
t= 4500 s
S(4500) = (4500+45)/15
= 303 pounds
c) 303 pounds of salt
a) S(t) = (t+450)/15
b) Solution =55 gallons when 40 pounds of salt is present
c) 303 pounds of salt when overflows

