(C) Logistic growth is modeled by equations of the form ,
where L is the upper limit. The graph shows L = 1000, so the differential
equation must be . Only equation C is of this
form (k = 0.003).
(D) We start with x = 3 and y = 100. At x = 3, ,
moving us to x = 3 + 2 = 5 and y = 100 + 5 = 105. From there
, so when x = 5 + 2 = 7 we estimate y = 105 + (−1) =
104.
(C) We separate the variables in the given d.e., then solve:
Since y(0) = 180, ln 112 = c. Then
When t = 10, y = 68 + 112e−1.1 105°F.
(E) The solution of the d.e. in Question 48, where y is the temperature of
the coffee at time t, is
y = 68 + 112e−0.11t.
We find t when y = 75°F:
