(i) Separate variables and integrate:
\(\int \frac{1}{x} \, dx = -k \int \frac{1}{\sqrt{t}} \, dt\)
\(\ln x = -2k\sqrt{t} + C\)
Use initial condition \(x = 100\) when \(t = 0\):
\(\ln 100 = C\)
Thus, \(\ln x = -2k\sqrt{t} + \ln 100\)
(ii) Given \(t = 25\) when \(x = 80\):
\(\ln 80 = -2k\sqrt{25} + \ln 100\)
\(\ln 80 = -10k + \ln 100\)
\(10k = \ln 100 - \ln 80\)
\(k = \frac{\ln 100 - \ln 80}{10}\)
Find \(t\) when \(x = 40\):
\(\ln 40 = -2k\sqrt{t} + \ln 100\)
\(2k\sqrt{t} = \ln 100 - \ln 40\)
\(\sqrt{t} = \frac{\ln 100 - \ln 40}{2k}\)
Substitute \(k\):
\(\sqrt{t} = \frac{\ln 100 - \ln 40}{2 \times \frac{\ln 100 - \ln 80}{10}}\)
\(\sqrt{t} = \frac{10(\ln 100 - \ln 40)}{2(\ln 100 - \ln 80)}\)
\(t = \left( \frac{10(\ln 100 - \ln 40)}{2(\ln 100 - \ln 80)} \right)^2\)
\(t \approx 64.1\)