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Magazine Chapter 5: Q4E (page 281)
Suppose for a certain disease described by the SIR model it is determined that a= 0.003 and b= 0.5.
- In the SI-phase plane, sketch the trajectory corresponding to the initial condition that one person is infected and 700 persons are susceptible.
- From your graph in part (a), estimate the peak number of infected persons. Compare this with the theoretical prediction\({\bf{S = }}\frac{{\bf{k}}}{{\bf{a}}}\)=167 persons when the epidemic is at its peak.
Short Answer
The equation is\({\bf{I = - S + }}\frac{{{\bf{500}}}}{{\bf{3}}}{\bf{lnS - 390}}{\bf{.8467}}\)
The peak of infected persons is 295.
Step by step solution
Use SIR Model
While using the SIR model\({\bf{i = - s + }}\frac{{\bf{k}}}{{\bf{a}}}{\bf{lns + K}}\).
Since \({\bf{i = }}\frac{{\bf{I}}}{{\bf{N}}}{\bf{,s = }}\frac{{\bf{S}}}{{\bf{N}}}\)then;
\({\bf{I = - S + }}\frac{{\bf{k}}}{{\bf{a}}}{\bf{lnS + K}}\)
Where K is some constant.
Put a= 0.003 and b= 0.5 then
\(\begin{aligned}{c}{\bf{I = - S + }}\frac{{{\bf{0}}{\bf{.5}}}}{{{\bf{0}}{\bf{.003}}}}{\bf{lnS + K}}\\{\bf{I = - S + }}\frac{{{\bf{500}}}}{{\bf{3}}}{\bf{lnS + K}}\end{aligned}\)
Now, put I=1, S=700 then;
\(\begin{aligned}{c}{\bf{1 = - 700 + }}\frac{{{\bf{500}}}}{{\bf{3}}}{\bf{ln700 + K}}\\{\bf{K = - 390}}{\bf{.8467}}\end{aligned}\)
The equation is\({\bf{I = - S + }}\frac{{{\bf{500}}}}{{\bf{3}}}{\bf{lnS - 390}}{\bf{.8467}}\).
Trajectory
Find the result
While seeing the graph it shows that the peak of infected persons is 295. This is kind of far off from the theoretical prediction.
This is the required result.
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