Question

Write a differential equation to represent each situation below. Do NOT solve them. a. A new technology is introduced into a community of 5000 people. If the rate at which the technology is adopted in the community is jointly proportional to the number of people who have adopted the technology and the number of people who have not adopted it, write a differential equation to represent the number of people, x(t), who have adopted the technology by time t. b. A tank with a capacity of 1000 gal originally contains 800 gal of water with 200 lbs of salt in the solution. Water containing 3.5 lbs of salt per gal is entering at a rate of 2 gal/min, and the mixture is allowed to flow out of the tank at a rate of 5 gal/min. Write a differential equation reflecting the information above, clearly stating the requested intermediate results below. Let A(t) represent the amount of salt (in pounds) in the tank after t minutes. Do NOT solve the DE! Show units with each factor in the three setup steps below, and simplify each expression, showing the resulting units as well. *Be sure to include the initial conditions for this DE in your final equation. R_in = Concentration of salt in the tank: c(t) = R_out = Differential equation: A(0) =

Answer 1

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)