IB Mathematics AHL 3.11 Relationships in trigonometric functions AA HL Paper 3
Question
The diagram below shows the boundary of the cross-section of a water channel.
The equation that represents this boundary is y=16sec(\(\frac{\pi x}{36}\))-32 where x and y are both measured in cm.
The top of the channel is level with the ground and has a width of 24cm. The maximum depth of the channel is 16 cm.
Find the width of the water surface in the channel when the water depth is 10 cm. Give your answer in the form a arccosb where a,b ∈\(\mathbb{R}\).
▶️Answer/Explanation
Ans
10 cm water depth corresponds to 16sec(\(\frac{\pi x}{36}\)0=k or equivalent i.e. making a trigonometrical function the subject of the equation.
\(cos(\frac{\pi x}{36})=\frac{8}{13}\)
\(\frac{\pi x}{36}=\pm arccos\frac{8}{13}\)
Width of water surface is \(\frac{72}{\pi}\) arccos \(\frac{8}{13}\) (cm)