Step 1

Given: \(\displaystyle{N}{\left({t}\right)}\) denotes the diversification function, which counts the number of taxa and r denotes the intrinsic rate of diversification.

Step 2

Calculation:

a) The exponential model is described as:

\(\displaystyle{\frac{{{d}{N}}}{{{\left.{d}{t}\right.}}}}={r}_{{{e}}}{N}\)

\(\displaystyle\Rightarrow\ {\frac{{{d}{N}}}{{{N}}}}={r}_{{{e}}}{\left.{d}{t}\right.}\)

Integrate both sides

\(\displaystyle\Rightarrow\ {\int_{{{N}{\left({0}\right)}}}^{{{N}{\left({t}\right)}}}}\ {\frac{{{d}{N}}}{{{N}}}}={r}_{{{e}}}\ {\int_{{{0}}}^{{{t}}}}\ {\left.{d}{t}\right.}\)

\(\Rightarrow\ \ln\ |N|_{N(0)}^{N(t)}=r_{e}t\mid_{0}^{t}\)

\(\displaystyle\Rightarrow\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{N}{\left({0}\right)}}}}\right]}={r}_{{{e}}}{t}\)

\(\Rightarrow\ \begin{array}{|c|}\hline r_{e}=\frac{1}{t}\ \ln\ \left[\frac{N(t)}{N(0)}\right]\\ \hline \end{array}\)

Step 3

b) The logistic model is described as:

\(\displaystyle{\frac{{{d}{N}}}{{{\left.{d}{t}\right.}}}}={r}_{{{l}}}{N}\ {\left({1}\ -\ {\frac{{{N}}}{{{K}}}}\right)}\)

\(\displaystyle\Rightarrow\ {\frac{{{d}{N}}}{{{\left.{d}{t}\right.}}}}={\frac{{{r}_{{{l}}}}}{{{K}}}}\ {\left({N}\right)}{\left({K}\ -\ {N}\right)}\)

\(\displaystyle\Rightarrow\ {\left[{\frac{{{K}}}{{{N}{\left({K}\ -\ {N}\right)}}}}\right]}{d}{N}={r}_{{{l}}}{\left.{d}{t}\right.}\)

\(\displaystyle\Rightarrow\ {d}{N}\ {\left[{\frac{{{1}}}{{{N}}}}\ +\ {\frac{{{1}}}{{{K}\ -\ {N}}}}\right]}={r}_{{{l}}}{\left.{d}{t}\right.}\)

\(\displaystyle\Rightarrow\ {\frac{{{d}{N}}}{{{N}}}}\ +\ {\frac{{{d}{N}}}{{{K}\ -\ {N}}}}={r}_{{{l}}}{\left.{d}{t}\right.}\)

Integrate both sides

\(\displaystyle{\int_{{{N}{\left({0}\right)}}}^{{{N}{\left({t}\right)}}}}\ {\frac{{{d}{N}}}{{{N}}}}\ +\ {\int_{{{N}{\left({0}\right)}}}^{{{N}{\left({t}\right)}}}}\ {\frac{{{d}{N}}}{{{k}\ -\ {N}}}}={r}_{{{l}}}\ {\int_{{{0}}}^{{{t}}}}\ {\left.{d}{t}\right.}\)

\(\Rightarrow\ln|N|\mid_{N(0)}^{N(t)}\ -\ \ln\ |K\ -\ N|\mid_{N(0)}^{N(t)}=r_{l}\mid_{0}^{t}\ dt\)

\(\displaystyle\Rightarrow\ {\ln{\ }}{\left({\frac{{{N}{\left({t}\right)}}}{{{N}{\left({0}\right)}}}}\right)}\ -\ {\ln{\ }}{\left({\frac{{{K}\ -\ {N}{\left({t}\right)}}}{{{K}\ -\ {N}{\left({0}\right)}}}}\right)}={r}_{{{l}}}{t}\)

\(\displaystyle\Rightarrow\ {\left[{\ln{{\left({N}{\left({t}\right)}\right)}}}\ -\ {\ln{{\left({K}\ -\ {N}{\left({t}\right)}\right)}}}\right]}\ +\ {\left[{\ln{{\left({N}{\left({0}\right)}\right)}}}\ +\ {\ln{{\left({K}\ -\ {N}{\left({0}\right)}\right)}}}\right]}={r}_{{{l}}}{t}\)

\(\displaystyle\Rightarrow\ {\ln{{\left[{\frac{{{N}{\left({t}\right)}}}{{{K}\ -\ {N}{\left({t}\right)}}}}\right]}}}\ +\ {\ln{\ }}{\left[{\frac{{{K}\ -\ {N}{\left({0}\right)}}}{{{N}{\left({0}\right)}}}}\right]}={r}_{{{l}}}{t}\)

\(\Rightarrow\ \begin{array}{|c|}\hline r_{l}=\frac{1}{t}\ \ln\ \left[\frac{N(t)}{K\ -\ N(t)}\right]\ +\ \frac{1}{t}\ \ln\ \left[\frac{K\ -\ N(0)}{N(0)}\right] \\ \hline \end{array}\)

Step 4

c) For exponential model:

\(\displaystyle{r}_{{{e}}}={\frac{{{1}}}{{{T}}}}\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{N}{\left({0}\right)}}}}\right]}\)

As \(\displaystyle{N}{\left({10}\right)}={1000}\) and \(\displaystyle{N}{\left({0}\right)}={1}:\)

\(\displaystyle\Rightarrow\ {r}_{{{e}}}={\frac{{{1}}}{{{10}}}}\ {\ln{\ }}{\left[{\frac{{{N}{\left({10}\right)}}}{{{N}{\left({0}\right)}}}}\right]}\)

\(\displaystyle\Rightarrow\ {r}_{{{e}}}=\ {\frac{{{1}}}{{{10}}}}\ {\ln{\ }}{\left[{\frac{{{1000}}}{{{1}}}}\right]}\)

\(\Rightarrow\ \begin{array}{|c|}\hline r_{e}=0.69 \\ \hline \end{array}\)

For logistic model:

\(\displaystyle{r}_{{{l}}}={\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{K}\ -\ {N}{\left({0}\right)}}}{{{N}{\left({0}\right)}}}}\right]}\ +\ {\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{K}\ -\ {N}{\left({t}\right)}}}}\right]}\)

As \(\displaystyle{N}{\left({10}\right)}={1000},\ {N}{\left({0}\right)}={1}\) and \(\displaystyle{K}={1001}:\)

\(\displaystyle\Rightarrow\ {r}_{{{l}}}={\frac{{{1}}}{{{10}}}}\ {\left[{\ln{\ }}{\left({\frac{{{1001}\ -\ }}{{{1}}}}\right)}\ +\ {\ln{\ }}{\left({\frac{{{1000}}}{{{1001}\ -\ {1000}}}}\right)}\right]}\)

\(\displaystyle\Rightarrow\ {r}_{{{l}}}={\frac{{{1}}}{{{10}}}}\ {\left[{\ln{\ }}{\left({1000}\right)}\ +\ {\ln{{\left({1000}\right)}}}\right]}\)

\(\displaystyle\Rightarrow\ {r}_{{{l}}}={\frac{{{2}}}{{{10}}}}\ {\ln{{\left({1000}\right)}}}\)

\(\Rightarrow\ \begin{array}{|c|}\hline r_{l}=1.381 \\ \hline \end{array}\)

In case 2, \(\displaystyle{N}{\left({10}\right)}={1000},\ {N}{\left({0}\right)}={1}\) and \(\displaystyle{K}={10},{000}:\)

\(\displaystyle\Rightarrow\ {r}_{{{l}}}={\frac{{{1}}}{{{10}}}}{\left[{\ln{\ }}{\left({\frac{{{10000}\ -\ {1}}}{{{1}}}}\right)}\ +\ {\ln{\ }}{\left({\frac{{{1000}}}{{{10000}\ -\ {100}}}}\right)}\right]}\)

\(\displaystyle\Rightarrow\ {r}_{{{l}}}={\frac{{{1}}}{{{10}}}}{\left[{\ln{{\left({9999}\right)}}}\ +\ {\ln{\ }}{\left({\frac{{{1000}}}{{{9000}}}}\right)}\right]}\)

\(\displaystyle\Rightarrow\ {r}_{{{l}}}={\frac{{{1}}}{{{10}}}}{\left[{\ln{\ }}{\left({9999}\right)}\ +\ {\ln{\ }}{\left({\frac{{{1}}}{{{9}}}}\right)}\right]}\)

\(\displaystyle\Rightarrow\ {r}_{{{l}}}={\frac{{{1}}}{{{10}}}}{\left[{\ln{\ }}{\left({9999}\ \cdot\ {\frac{{{1}}}{{{9}}}}\right)}\right]}\)

\(\displaystyle\Rightarrow\ {r}_{{{l}}}={\frac{{{1}}}{{{10}}}}{\ln{{\left({1111}\right)}}}\)

\(\Rightarrow\ \begin{array}{|c|}\hline r_{l}=0.7 \\ \hline \end{array}\)

Step 5

d) From part (c), it is observed that the calculated rate r in exponential model is less than the estimated rate in logistic model for both values of \(\displaystyle{K}{\left({K}={10001}\right.}\) and \(\displaystyle{K}={10000}{)}.\)

The above observation justifies the following quote:

"There must be a general tendency for calculated values of \(\displaystyle{\left[{r}\right]}\) to represent underestimates for exponential rates because some radiation will have followed distincly sigmoid paths during the interval evaluated."

e) From part (b):

\(\displaystyle{r}_{{{e}}}={\frac{{{1}}}{{{t}}}}{\left[{\ln{\ }}{\left[{\frac{{{K}\ -\ {N}{\left({0}\right)}}}{{{N}{\left({0}\right)}}}}\right]}\ +\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{K}\ -\ {N}{\left({t}\right)}}}}\right]}\right]}\)

\(\displaystyle\Rightarrow\ {r}_{{{e}}}={\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{N}{\left({0}\right)}}}}\ \cdot\ {\frac{{{K}\ -\ {N}{\left({0}\right)}}}{{{K}\ -\ {N}{\left({t}\right)}}}}\right]}\)

\(\displaystyle\Rightarrow\ {r}_{{{e}}}={\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{N}{\left({0}\right)}}}}\ \cdot\ {\left({\frac{{{1}\ -\ {\frac{{{N}{\left({0}\right)}}}{{{K}}}}}}{{{1}\ -\ {\frac{{{N}{\left({t}\right)}}}{{{K}}}}}}}\right)}\right]}\)

As \(\displaystyle{\frac{{{N}}}{{{K}}}}\ {\left\langle\ {1}\right.}\)

\(\displaystyle\Rightarrow\ {r}_{{{e}}}={\frac{{{1}}}{{{t}}}}{\left[{\ln{\ }}{\left({\frac{{{N}{\left({t}\right)}}}{{{N}{\left({0}\right)}}}}\ \cdot\ {1}\right)}\right]}\)

\(\Rightarrow\ \begin{array}{|c|}\hline r_{e}=\frac{1}{t}\ \ln\ \left[\frac{N(t)}{N(0)}\right]\\ \hline \end{array}\)

Hence, the exponential model is a good approximation to logistic model when

\(\displaystyle{\frac{{{N}}}{{{K}}}}\ {\left\langle\ {1}.\right.}\)