Find an expression for the volume of water $V{\text{ }}({{\text{m}}^3})$ in the trough in terms of $\theta$.

Calculate $\frac{{{\text{d}}\theta }}{{{\text{d}}t}}$ when $\theta = \frac{\pi }{3}$.

area of segment $= \frac{1}{2} \times {0.5^2} \times (\theta - \sin \theta )$     M1A1

$V = {\text{area of segment}} \times 10$

$V = \frac{5}{4}(\theta - \sin \theta )$     A1

[3 marks]

METHOD 1

$\frac{{{\text{d}}V}}{{{\text{d}}t}} = \frac{5}{4}(1 - \cos \theta )\frac{{{\text{d}}\theta }}{{{\text{d}}t}}$     M1A1

$0.0008 = \frac{5}{4}\left( {1 - \cos \frac{\pi }{3}} \right)\frac{{{\text{d}}\theta }}{{{\text{d}}t}}$     (M1)

$\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = 0.00128{\text{ }}({\text{rad}}\,{s^{ - 1}})$     A1

METHOD 2

$\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = \frac{{{\text{d}}\theta }}{{{\text{d}}V}} \times \frac{{{\text{d}}V}}{{{\text{d}}t}}$     (M1)

$\frac{{{\text{d}}V}}{{{\text{d}}\theta }} = \frac{5}{4}(1 - \cos \theta )$     A1

$\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = \frac{{4 \times 0.0008}}{{5\left( {1 - \cos \frac{\pi }{3}} \right)}}$     (M1)

$\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = 0.00128\left( {\frac{4}{{3125}}} \right)({\text{rad }}{s^{ - 1}})$     A1

[4 marks]