Question

Suppose the rate at which a population of $100$weasels catches weasel-mumps is $r\left(t\right)=8-2\ln (t+1)$weasels per day, where $t>0$measures the number of days after the initial outbreak. Find a formula that represents the number $W\left(t\right)$of weasels that have contracted weasel-mumps $t$days after the initial outbreak. Approximately how long will it until all but one of the weasels are infected?

Short answer

The solution is$1000-202\ln \left(101\right)$.

Step-by-step solution

Step 1. Given information.

The rate of weasel-mump infection is$r\left(t\right)=8-2\ln (t+1)$.

Step 2. Find a formula that represents the number W(t).

$\begin{array}{rcl}\int r\left(t\right)& =& \int \left\{8-2\ln (t+1)\right\}dt\\ & =& 8\int dt-2\int \ln (t+1)dt\\ & =& 8t-2\left[\ln (t+1)\int dt-\int \left\{\frac{d}{dt}\left(\ln \right(t+1\left)\right)\int dt\right\}dt\right]\\ & =& 8t-2\left[\ln (t+1)·t-\int \frac{1}{t+1}·tdt\right]\\ & =& 8t-2t·\ln (t+1)+2\int \left\{1-\frac{1}{t+1}\right\}dt\\ & =& 8t-2t·\ln (t+1)+2\int dt+2\int \frac{1}{t+1}dt\\ & =& 8t-2t·\ln (t+1)+2t-2\left[\int \frac{1}{u}du\right]\dots \dots \dots \dots \dots \left\{u=t+1,du=dt\right\}\\ & =& 8t-2t·\ln (t+1)-2\ln \left|u\right|\\ & =& 10t-2t·\ln (t+1)-2\ln \left|t+1\right|\end{array}$

Step 3. Now calculate the time.

$\begin{array}{rcl}W\left(t\right)& =& 10t-2t·\ln (t+1)-2\ln \left|t+1\right|\\ W\left(100\right)& =& 10\left(100\right)-2\left(100\right)·\ln (100+1)-2\ln \left|100+1\right|\\ & =& 1000-200\ln \left(101\right)-2\ln \left(101\right)\\ & =& 1000-202\ln \left(101\right)\end{array}$

Step 4. Simplified answer.

Hence, the required value is$1000-202\ln \left(101\right)$.