1. **State the problem:** We have a leaking water tank with volume $V$ at time $t$ hours, and the rate of change of volume is proportional to the volume itself: $$\frac{dV}{dt} = -kV$$ where $k$ is a positive constant. 2. **Show that $V = V_0 e^{-kt}$ is a solution:** Start with the proposed solution: $$V = V_0 e^{-kt}$$ Differentiate with respect to $t$: $$\frac{dV}{dt} = V_0 \cdot (-k) e^{-kt} = -k V_0 e^{-kt} = -k V$$ This matches the differential equation, so $V = V_0 e^{-kt}$ is indeed a solution. 3. **Find $k$ given $V_0 = 3000$ litres and $V(3) = 1900$ litres:** Use the formula: $$V = V_0 e^{-kt}$$ Substitute values: $$1900 = 3000 e^{-3k}$$ Divide both sides by 3000: $$\frac{1900}{3000} = e^{-3k}$$ $$\frac{\cancel{1900}}{\cancel{3000}} = e^{-3k}$$ (no common factors to cancel, so just write as fraction) Take natural logarithm: $$\ln\left(\frac{1900}{3000}\right) = -3k$$ Solve for $k$: $$k = -\frac{1}{3} \ln\left(\frac{1900}{3000}\right)$$ Calculate: $$\frac{1900}{3000} = 0.6333\ldots$$ $$\ln(0.6333) = -0.4572$$ $$k = -\frac{1}{3} \times (-0.4572) = 0.1524$$ Rounded to 4 decimal places: $$k = 0.1523$$ 4. **Find the time when volume is 250 litres:** Use the formula again: $$250 = 3000 e^{-0.1523 t}$$ Divide both sides by 3000: $$\frac{250}{3000} = e^{-0.1523 t}$$ Take natural logarithm: $$\ln\left(\frac{250}{3000}\right) = -0.1523 t$$ Calculate: $$\frac{250}{3000} = 0.08333$$ $$\ln(0.08333) = -2.4849$$ Solve for $t$: $$t = -\frac{1}{0.1523} \times (-2.4849) = \frac{2.4849}{0.1523} = 16.32 \text{ hours}$$ 5. **Convert time to day and clock time:** The leak started at 7pm Wednesday (time $t=0$). Add 16.32 hours: 16 hours and 0.32 hours = 16 hours and $0.32 \times 60 = 19.2$ minutes. 7pm + 16 hours = 11am next day (Thursday). Add 19 minutes: 11:19am Thursday. **Final answer:** Mr Huxley discovered the leak at approximately 11:19am on Thursday.