Arc length of parametric curve

\includegraphics{figure_1} The parametric equations of the curve \(C\) shown in Figure 1 are $$x = a(t - \sin t), \quad y = a(1 - \cos t), \quad 0 \leq t \leq 2\pi$$ Find, by using integration, the length of \(C\). (Total 6 marks)

A curve is defined parametrically by $$x = t^3 + 5, \quad y = 6t^2 - 1$$ The arc length between the points where \(t = 0\) and \(t = 3\) on the curve is \(s\).

1. Show that \(s = \int_{0}^{3} 3t\sqrt{t^2 + A} \, \text{d}t\), stating the value of the constant \(A\). [4 marks]
2. Hence show that \(s = 61\). [4 marks]

The arc of the curve with equation \(y = 4 - \ln(1-x^2)\) from \(x = 0\) to \(x = \frac{3}{4}\) has length \(s\).

1. Show that \(s = \int_0^{\frac{3}{4}} \frac{\sqrt{1+x^2}}{1-x^2} \, dx\). [4 marks]
2. Find the value of \(s\), giving your answer in the form \(p + \ln N\), where \(p\) is a rational number and \(N\) is an integer. [6 marks]

A curve is defined parametrically by the equations $$x = \frac{3}{2}t^3 + 5$$ $$y = t^{\frac{3}{2}} \quad (t \geq 0)$$ Show that the arc length of the curve from \(t = 0\) to \(t = 2\) is equal to 26 units. [5 marks]