(a) Simplify : (frac{1}{1 – cos theta} + frac{1}{1 + cos theta}) and leave your answer in terms of (sin theta).
(b) Find the equation of the line joining the stationary points of (y = x^{2} (x – 3)) and the distance between them.
Explanation
(a) (frac{1}{1 – cos theta} + frac{1}{1 + cos theta})
= (frac{(1 + cos theta) + (1 – cos theta)}{1 – cos^{2} theta})
= (frac{2}{1 – cos^{2} theta})
= (frac{2}{sin^{2} theta})
(b) (y = x^{2} (x – 3))
(frac{mathrm d y}{mathrm d x} = x^{2} + 2x(x – 3))
= (x^{2} + 2x^{2} – 6x = 3x^{2} – 6x)
At stationary point, (frac{mathrm d y}{mathrm d x} = 0)
(3x^{2} – 6x = 0)
(3x(x – 2) = 0 implies x = text{0 or 2})
If x = 0, (y = 0^{2}(0 – 3) = 0)
If x = 2, (y = 2^{2}(2 – 3) = -4)
The points are (0, 0) and (2, -4).
Gradient = (frac{-4 – 0}{2 – 0} = frac{-4}{2} = -2)
Equation : (frac{y – (-4)}{x – 2} = -2)
(frac{y + 4}{x – 2} = -2 implies y + 4 = -2(x – 2))
(y + 4 = 4 – 2x implies y + 2x = 0)
(y = -2x)
Distance : (sqrt{(2 – 0)^{2} + (-4 – 0)^{2}})
= (sqrt{4 + 16} = sqrt{20})
= (2sqrt{5} units)