Find \( \int \cos^2(3x) \, dx \) - HSC - SSCE Mathematics Extension 1 - Question 12 - 2018 - Paper 1

Using the geometry of the ferris wheel, the height ( h ) of the carriage can be expressed as:
[ h = 20 \sin(\theta) ]
To find ( \frac{dh}{d\theta} ), we differentiate with respect to ( \theta ):
[ \frac{dh}{d\theta} = 20 \cos(\theta) ]
Thus, we have shown that ( \frac{dh}{d\theta} = 20 \cos(\theta) ).

When the height ( h ) is 15 metres, we can set up the equation:
[ h = 20 - 15 = 5 ]
Using ( h = 20 \sin(\theta) ), we find:
[ 20 \sin(\theta) = 15 \Rightarrow \sin(\theta) = \frac{15}{20} = 0.75 ]
Thus, ( \theta = \sin^{-1}(0.75) ).
Next, we calculate the speed of the height increase. Given that ( \frac{dh}{dt} = \frac{dh}{d\theta} \cdot \frac{d\theta}{dt} ):
[ \frac{dh}{dt} = 20 \cos(\theta) \cdot \frac{d\theta}{dt} ]
Where ( \frac{d\theta}{dt} = 1.5 ) rad/min. So, substituting values, we find the speed when ( \theta ) is approximately ( 0.644 ,\text{rad} ):
[ \frac{dh}{dt} = 20 \cos(0.644) \cdot 1.5 \approx 19.8 , \text{m/min} ]
Thus, the speed at which the top of the carriage is rising is approximately 19.8 m/min.

To find the derivative of ( f(x) = \sin^{-1} x + \cos^{-1} x ), we differentiate both components:
[ f'(x) = \frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-x^2}} = 0 ]
Hence, we have shown that ( f'(x) = 0 ).

Since the derivative of ( f(x) ) is zero, ( f(x) ) must be constant. To find the constant, we evaluate ( f(0) = \sin^{-1}(0) + \cos^{-1}(0) = 0 + \frac{\pi}{2} = \frac{\pi}{2} ). Thus,
we have ( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} ).

The function ( f(x) = \sin^{-1} x + \cos^{-1} x ) is constant and equals ( \frac{\pi}{2} ). The sketch should show a horizontal line at the value of ( \frac{\pi}{2} ) across the domain:
( -1 \leq x \leq 1 ).
This indicates that the function maintains a constant value of ( \frac{\pi}{2} ) for all ( x ) within this interval.