Porisma

Worked solution

Solve logistic growth and sketch its changing curvature

A fish population satisfies dP/dt=0.02P(100−P)dP/dt=0.02P(100-P), with tt in weeks.

(i) Find the range of PP for strictly increasing population. Initially P=20P=20.

(ii) Show that P=100/(1+4e−2t)P=100/(1+4e^{-2t}).

(iii) Describe the long-run population and sketch PP against tt.

[2013/CJC/II/4 (modified)]

CJC · Discussion · Question 10

Answer

(i) 0<P<1000<P<100. (ii) P=100/(1+4e−2t)P=100/(1+4e^{-2t}). (iii) P→100P\to100 from below, with inflection (ln⁡2,50)(\ln2,50).

(i) Determine the increasing range

0.02P(100−P)>00.02P(100-P)>0 precisely when 0<P<1000<P<100. The equilibrium values P=0P=0 and P=100P=100 are not strictly increasing.

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(ii) Integrate using partial fractions

∫dP/[P(100−P)]=∫0.02 dt\int dP/[P(100-P)]=\int0.02\,dt. Since 1/[P(100−P)]=1100[1/P+1/(100−P)]1/[P(100-P)]=\tfrac1{100}[1/P+1/(100-P)], integration gives ln⁡∣P∣−ln⁡∣100−P∣=2t+C′\ln|P|-\ln|100-P|=2t+C^\prime. Equivalently (100−P)/P=Ae−2t(100-P)/P=Ae^{-2t}, so P=100/(1+Ae−2t)P=100/(1+Ae^{-2t}). At t=0,P=20t=0,P=20, A=4A=4, proving the requested formula.

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(iii) Sketch the initial value, inflection and long-run limit

The population starts at (0,20)(0,20) and increases towards 100100 from below as t→∞t\to\infty. For the curvature, P′′=0.02P′(100−2P)P^{\prime\prime}=0.02P^\prime(100-2P): it is positive while P<50P<50, zero at P=50P=50, and negative afterwards. Thus the graph is initially convex, has an inflection at (ln⁡2,50)(\ln2,50) and then becomes concave down towards the asymptote P=100P=100.

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Logistic fish population P=100/(1+4 exp(−2t)) for t at least zero: starts at (0,20), initially bends upward, passes through inflection (ln2,50), then bends downward towards dashed asymptote P=100.
Logistic fish population P=100/(1+4 exp(−2t)) for t at least zero: starts at (0,20), initially bends upward, passes through inflection (ln2,50), then bends downward towards dashed asymptote P=100.
Logistic fish population P=100/(1+4 exp(−2t)) for t at least zero: starts at (0,20), initially bends upward, passes through inflection (ln2,50), then bends downward towards dashed asymptote P=100.

Common mistakes

Starting below half the limiting population means the curve initially bends upwards.

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