The cross-section of a designer architectural arch is modelled by the curve shown in a coordinate plane, with the equation
x=16sin2y+3,0⩽y⩽π2 x = 16\sin^2 y + 3, \quad 0 \leqslant y \leqslant \frac{\pi}{2} x=16sin2y+3,0⩽y⩽2πwhere x x\,x and y y\,y are spatial coordinates measured in decimetres. The point P(k,π6)P\left(k, \frac{\pi}{6}\right)P(k,6π) lies on the curve.
Verify that k=7k = 7k=7.
(i) Find dxdy\frac{dx}{dy}dydx in terms of yyy.
(ii) Hence show that dydx=12(x−3)(19−x)\frac{dy}{dx} = \frac{1}{2\sqrt{(x-3)(19-x)}}dxdy=2(x−3)(19−x)1.
The normal to the curve at PPP intersects the xxx-axis at the point NNN.
Determine the exact area of triangle OPNOPNOPN, where OOO is the origin. Give your answer in the form aπ+bπ2a\pi + b\pi^2aπ+bπ2 where aaa and bbb are constants to be found.