The number of students that hear a rumor on a small college campus t days after the rumor starts is modeled by the logistic function R(t)= 3,000 1+Bekt. If 5 students initially heard the rumor and 149 students heard the rumor after 1 day, then how long will it take 2,700 students to hear the rumor?
To solve this you need to find the unknown constants B and k.
The information given:
R(0) = 5
R(1) = 149
will allow us to do this.
First plug in initial point R(0) = 5 and solve for B (e^0 = 1)
\(\frac{3000}{1+ B} = 5\)
\(1+B = \frac{3000}{5} = 600\)
\(B = 599\)
Next plug in R(1) = 149 and solve for k (actually only solve for e^k)
\(\frac{3000}{1+599e^k} = 149\)
\(e^k = \frac{3000-149}{(599)(149)}\)
Finally we can solve for t when R(t) = 2700.
Plug in the fraction for e^k since e^(kt) = (e^k)^t
\(\frac{3000}{1 +599( \frac{3000-149}{(599)(149)})^t } = 2700\)
\(599 (\frac{3000-149}{(599)(149)})^t = \frac{3000}{2700} - 1 = \frac{1}{9}\)
\(t = \frac{ln ( \frac{1}{(599)(9)}) }{ln(\frac{3000-149}{(599)(149)})} = 2.495\)
The logistic function is \(S\) shaped curve which use to determine the statistical distribution over the energy states of a system.
The time will take for 2,700 student to hear the rumor is 2.495 days.
Given:
The given logistic function is,
\(R(t)=\frac{3000}{1+Be^{kt}}\)
As per the question, at initial stage 5 students heard the rumor. At initial stage \(t=0\).
Calculate the constant \(B\) at initial stage.
\(R(t)=\frac{3000}{1+Be^{kt}}\\R(0)=\frac{3000}{1+Be^{k\times 0}}\\R(0)=\frac{3000}{1+B}\)
Substitute \(R(0)=5\).
\(\begin{aligned}5=\frac{3000}{1+B}\\1+B= \frac{3000}{5}\\B=599\\\end\)
As per the question, after 1 day 149 students heard the rumor. After 1 day stage \(t=1\).
\(R(t)=\frac{3000}{1+Be^{kt}}\\R(1)=\frac{3000}{1+Be^{k\times 1}}\\R(1)=\frac{3000}{1+B^k}\)
Substitute \(R(1)=149\) and \(B=599\).
\(\begin{aligned}149=\frac{3000}{1+599e^k}\\e^k=\frac{(3000-1149)}{(599)(149)} \\\end{aligned}\)
Calculate the time take to hear rumor by 2,7000 students.
\(\begin {aligned}2700=\frac{3000}{1+599(\frac{3000-149}{599\times 149})^t }\\\frac{3000}{2700}-1 =599(\frac{3000-149}{599\times 149})^t }\\t=\frac{ln(\frac{1}{(599)(9)})}{ln(\frac{3000-149}{599\times 149}) }\\t=2.495\:\rm days\\\end{aligned}\)
Thus, the time will take for 2,700 student to hear the rumor is 2.495 days.
Learn more about volume and pressure here:
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