Question

Ques Soln a) It is mentioned in the problem that the rate y' is proportional to y and 1-y by some proportional factor k. Hence the equation is simply as follows dy/dt = ky(1-y) though direction comparison from the logistic equation shown below dy/dt = ky(1 - y/A) we can already infer that A. Now we solve for B by using given data that 104 student out of 1000 student hear rumor at 8:00 AM means 104/1000 x 100 = 10.4% student hear at t=0 = 10%. so (y,t) = (0.10, 0). Hence from logistic equation formula y = A / (1 - e^-kt/B) we can solve for the unknown value B since -kt=0 when t=0 y = A / (1 - e^-kt/B) (y,t) = (0.1, 0) 0.1 = 1 / (1 - e^-k(0)/B) 0.1 = 1 / (1 - 1/B) B / (B-1) = 0.1 B = -1/9 Now using our second data set which is (y,t) = (0.5, 4) means 50% student hear at 12:00 Noon. y = A / (1 - e^-kt/B) 0.5 = 1 / (1 + e^-4k/9) 0.5 (1 + 9e^-4k) = 1 e^-4k = 1/9 k = 0.55 So final equation is y = 1 / (1 + 9e^-0.55t) For 90% to hear at t=? (y,t) = (0.9, t) 0.9 = 1 / (1 + 9e^-0.55t) t = 7.98 hours Final Ans

          Ques Soln
a) It is mentioned in the problem that the rate y' is proportional to y and 1-y by some proportional factor k. Hence the equation is simply as follows
dy/dt = ky(1-y)
though direction comparison from the logistic equation shown below
dy/dt = ky(1 - y/A)
we can already infer that A. Now we solve for B by using given data that 104 student out of 1000 student hear rumor at 8:00 AM means 104/1000 x 100 = 10.4% student hear at t=0 = 10%.
so (y,t) = (0.10, 0). Hence from logistic equation formula
y = A / (1 - e^-kt/B)
we can solve for the unknown value B since -kt=0 when t=0
y = A / (1 - e^-kt/B)
(y,t) = (0.1, 0)
0.1 = 1 / (1 - e^-k(0)/B)
0.1 = 1 / (1 - 1/B)
B / (B-1) = 0.1
B = -1/9
Now using our second data set which is (y,t) = (0.5, 4) means 50% student hear at 12:00 Noon.
y = A / (1 - e^-kt/B)
0.5 = 1 / (1 + e^-4k/9)
0.5 (1 + 9e^-4k) = 1
e^-4k = 1/9
k = 0.55
So final equation is
y = 1 / (1 + 9e^-0.55t)
For 90% to hear at t=?
(y,t) = (0.9, t)
0.9 = 1 / (1 + 9e^-0.55t)
t = 7.98 hours Final Ans

Ques Soln
a) It is mentioned in the problem that the rate y' is proportional to y and 1-y by some proportional factor k. Hence the equation is simply as follows
dy/dt = ky(1-y)
though direction comparison from the logistic equation shown below
dy/dt = ky(1 - y/A)
we can already infer that A. Now we solve for B by using given data that 104 student out of 1000 student hear rumor at 8:00 AM means 104/1000 x 100 = 10.4% student hear at t=0 = 10%.
so (y,t) = (0.10, 0). Hence from logistic equation formula
y = A / (1 - e^-kt/B)
we can solve for the unknown value B since -kt=0 when t=0
y = A / (1 - e^-kt/B)
(y,t) = (0.1, 0)
0.1 = 1 / (1 - e^-k(0)/B)
0.1 = 1 / (1 - 1/B)
B / (B-1) = 0.1
B = -1/9
Now using our second data set which is (y,t) = (0.5, 4) means 50% student hear at 12:00 Noon.
y = A / (1 - e^-kt/B)
0.5 = 1 / (1 + e^-4k/9)
0.5 (1 + 9e^-4k) = 1
e^-4k = 1/9
k = 0.55
So final equation is
y = 1 / (1 + 9e^-0.55t)
For 90% to hear at t=?
(y,t) = (0.9, t)
0.9 = 1 / (1 + 9e^-0.55t)
t = 7.98 hours Final Ans

Added by Julio R.

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