A2 June 2024 Paper 1 Q15
15 A curve is defined parametrically by the equations
\[x = \frac{3}{2}t^3 + 5\]\[y = t^{\frac{9}{2}} \qquad (t \geqslant 0)\]Show that the arc length of the curve from \(t = 0\) to \(t = 2\) is equal to 26 units. [5 marks]
| Scheme | Marks | AO |
|---|---|---|
| Obtains \(\dot{x} = \dfrac{9}{2}t^2\) and \(\dot{y} = \dfrac{9}{2}t^{\frac{7}{2}}\) | B1 | 1.1b |
| Obtains their \(\sqrt{\dot{x}^2 + \dot{y}^2}\) | M1 | 1.1a |
| Writes integrand in the form \(kt^2\sqrt{(1 + t^3)}\) | M1 | 3.1a |
| Obtains \(\lambda(1 + t^3)^{\frac{3}{2}}\) | A1 | 2.2a |
| Completes a reasoned argument to obtain 26 AG | R1 | 2.1 |
| (5 marks) |
Typical solution
\[\dot{x} = \frac{9}{2}t^2 \text{ and } \dot{y} = \frac{9}{2}t^{\frac{7}{2}}\]\[\dot{x}^2 + \dot{y}^2 = \frac{81}{4}(t^4 + t^7)\]Arc length
\[\begin{aligned}s &= \int_0^2 \sqrt{\frac{81}{4}(t^4 + t^7)}\,\mathrm{d}t \\ &= \frac{9}{2}\int_0^2 t^2\sqrt{(1 + t^3)}\,\mathrm{d}t \\ &= \left[(1 + t^3)^{\frac{3}{2}}\right]_0^2 \\ &= 9^{\frac{3}{2}} - 1^{\frac{3}{2}} \\ &= 26\end{aligned}\]