(a)
\( \frac{dx}{dt} = 5 \, \text{mh}^{-1} \)

Water height rate: \( \frac{dy}{dt} = 0.2 \, \text{m/h} \)
Beach equation: \( y = 0.02x^2 \)
Derivative: \( \frac{dy}{dx} = 0.04x \)
At \( x = 1 \): \( \frac{dy}{dx} = 0.04 \)
Chain rule: \( \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \)
Substitute: \( 0.2 = 0.04 \times \frac{dx}{dt} \)
Solve: \( \frac{dx}{dt} = \frac{0.2}{0.04} = 5 \)

Result: \( \frac{dx}{dt} = 5 \, \text{mh}^{-1} \) [3]

(b)
(i) Time ≈ 9.26 hours
Snail position: \( (X, Y) \), \( Y = 0.02X^2 \)
Speed: \( \sqrt{\left( \frac{dX}{dt} \right)^2 + \left( \frac{dY}{dt} \right)^2} = 1 \)
Since \( \frac{dY}{dt} = 0.04X \cdot \frac{dX}{dt} \):
\( \left( \frac{dX}{dt} \right)^2 (1 + (0.04X)^2) = 1 \)
\( \frac{dX}{dt} = \frac{1}{\sqrt{1 + 0.0016X^2}} \)
Time: \( t = \int_1^{10} \sqrt{1 + 0.0016X^2} \, dX \)
Substitute: \( u = 0.04X \), \( dX = 25 \, du \), limits \( u = 0.04 \) to \( u = 0.4 \)
\( t = 25 \int_{0.04}^{0.4} \sqrt{1 + u^2} \, du \)
Integral: \( \frac{u}{2} \sqrt{1 + u^2} + \frac{1}{2} \ln \left( u + \sqrt{1 + u^2} \right) \)
Evaluate: \( t \approx 25 \times (0.41015 – 0.040016) \approx 9.26 \)

Result: Time ≈ 9.26 hours [3]

(ii) Snail reaches (10, 2) before water
Water height: \( y = 0.2t \)
At \( y = 2 \): \( t = \frac{2}{0.2} = 10 \, \text{hours} \)
Snail time: \( t \approx 9.26 \, \text{hours} \)
Since \( 9.26 < 10 \), snail arrives first
Alternative: At \( t = 9.26 \), water height = \( 0.2 \times 9.26 = 1.852 < 2 \)

Result: Snail reaches (10, 2) before water [2]