Answer:
a. x'(t) = kx(5000-x)
b. R_in = 7 lbs/min
R_out = 5c(t) lbs/min
c(t) = A(t)/(800 -3t) lbs/gal
A'(t) = R_in - R_out; A(0) = 200 lbs
Step-by-step explanation:
A. When proportionality is joint between x and y, the expression describing that is kxy, where k is the constant of proportionality. Here, the rate of change is jointly proportional to number who have adopted (x) and number of the community of 5000 who have not (5000-x). The corresponding equation is ...
x'(t) = kx·(5000-x)
We know the initial number of adopters must be greater than 0, or the equation's only solution would be x=0.
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B. The rate of influx of salt is ...
R_in = (3.5 lbs/gal)×(2 gal/min) = 7 lbs/min
The number of gallons in the tank at time t is ...
v = 800 gal + (2 gal/min - 5 gal/min)(t min) = (800 -3t) gal
For amount of salt A(t) pounds, the concentration c(t) will be ...
c(t) = A(t)/v = A(t)/(800 -3t) . . . . lbs/gal
The outflow rate will be the product of outflow volume and concentration:
R_out = (c(t) lbs/gal)(5 gal/min) = 5c(t) lbs/min
And the differential equation for A(t) is ...
A'(t) = R_in -R_out . . . with initial condition A(0) = 200 lbs.
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When you go to solve the equation for A'(t), it will need to be cast in terms of A(t):
A'(t) = 7 -5A(t)/(800 -3t)
