Question

A model for the spread of a rumor is given by the equation

\(p\left( t \right) = \frac{{\bf{1}}}{{{\bf{1}} + a{e^{ - kt}}}}\)

Where \(p\left( t \right)\) is the proportion of the population that knows the rumor at time t and a and k are positive constants.

a) When will half the population have heard the rumor?

b) When is the rate of spread of the rumor greatest?

c) Sketch the graph of p.

Short answer

a) After \(t = \frac{{\ln a}}{k}\), the rumor will reach half of the population.

b) The rate of spread of rumor is greatest at \(t = \frac{{\ln a}}{k}\) , i.e., by the time when half of the population have heard the rumor.

c) The graph of p is shown below:

Step-by-step solution

Find the time when half the population heard rumor

Substitute\(\frac{1}{2}\)for\(p\left( t \right)\)in the equation\(p\left( t \right) = \frac{1}{{1 + a{e^{ - kt}}}}\).

\(\begin{array}{c}\frac{1}{2} = \frac{1}{{1 + a{e^{ - kt}}}}\\1 + a{e^{ - kt}} = 2\\a{e^{ - kt}} = 1\\{e^{ - kt}} = \frac{1}{a}\\ - kt = - \ln a\\t = \frac{{\ln a}}{k}\end{array}\)

Thus, after time \(t = \frac{{\ln a}}{k}\), the rumor will reach half of the population.

Find the time when the rate of spread of rumor is greatest

Differentiate the equation\(p\left( t \right) = \frac{1}{{1 + a{e^{ - kt}}}}\).

\(\begin{array}{c}p'\left( t \right) = \frac{{\rm{d}}}{{{\rm{d}}t}}\left( {\frac{1}{{1 + a{e^{ - kt}}}}} \right)\\ = \frac{{ak{e^{ - kt}}}}{{{{\left( {1 + a{e^{ - kt}}} \right)}^2}}}\end{array}\)

Differentiate the equation\(p'\left( t \right) = \frac{{ak{e^{ - kt}}}}{{{{\left( {1 + a{e^{ - kt}}} \right)}^2}}}\).

\(\begin{array}{c}p''\left( t \right) = \frac{{\rm{d}}}{{{\rm{d}}t}}\left( {\frac{{ak{e^{ - kt}}}}{{{{\left( {1 + a{e^{ - kt}}} \right)}^2}}}} \right)\\ = \frac{{{{\left( {1 + a{e^{ - kt}}} \right)}^2}\left( { - a{k^2}{e^{ - kt}}} \right) - ak{e^{ - kt}} \cdot 2\left( {1 + a{e^{ - kt}}} \right)\left( { - ak{e^{ - kt}}} \right)}}{{{{\left( {1 + a{e^{ - kt}}} \right)}^4}}}\\ = \frac{{\left( {1 + a{e^{ - kt}}} \right)\left( { - ak{e^{ - kt}}} \right)\left( {k\left( {1 + a{e^{ - kt}}} \right) - 2ak{e^{ - kt}}} \right)}}{{{{\left( {1 + a{e^{ - kt}}} \right)}^4}}}\\ = \frac{{ - a{k^2}{e^{ - kt}}\left( {a{e^{ - kt}} - 1} \right)}}{{{{\left( {1 + a{e^{ - kt}}} \right)}^3}}}\end{array}\)

If\(p''\left( t \right) > 0\), then;

\(\begin{array}{c}a{e^{ - kt}} > 1\\ - kt > \ln \frac{1}{a}\\t < \frac{{\ln a}}{k}\end{array}\)

So,\(p'\left( t \right)\)is increasing for\(t < \frac{{\ln a}}{k}\)and\(p'\left( t \right)\)is decreasing for\(t > \frac{{\ln a}}{k}\).

Thus, the rate of spread of rumor is greatest at \(t = \frac{{\ln a}}{k}\) , i.e., by the time when half of the population has heard the rumor.

Sketch the graph of \(p\left( t \right)\)

ubstitute 0 for \(t\) in the equation \(p\left( t \right) = \frac{1}{{1 + a{e^{ - kt}}}}\).

\(\begin{array}{c}p\left( 0 \right) = \frac{1}{{1 + a{e^{ - 0}}}}\\ = \frac{1}{{1 + a}}\end{array}\)

And,\(\mathop {\lim }\limits_{t \to \infty } p\left( t \right) = 1\).

So\(p = 1\)is the horizontal asymptote.

The figure below represents the graph of \(p\left( t \right)\):