# Question 4:

## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $U$ is the curtate cycloid and $V$ is the sine wave | B1 | Must identify both correctly |
| Because the sine wave has a sinusoidal shape / the curtate cycloid has flatter troughs and sharper peaks, or equivalent reasoning | | Accept any valid geometric justification e.g. sine wave is symmetric about its midline, curtate cycloid has wider troughs |

## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| For the curtate cycloid: when $y = 0$, $b\cos\theta = 0$, so $\theta = \pi/2$ | M1 | Setting $y=0$ and solving, or finding $x$ values when $y=0$ |
| At $\theta = \pi/2$: $x = a\cdot\frac{\pi}{2} - b\sin\frac{\pi}{2} = \frac{a\pi}{2} - b$ | | |
| At $\theta = -\pi/2$: $x = -\frac{a\pi}{2} + b$ | | |
| At $\theta = 0$: $x = 0$, $y = b$ (maximum) | | |
| $d$ is the horizontal distance from the axis of symmetry of one arch to where $y=0$ | | |
| $d = \frac{a\pi}{2} - (\frac{a\pi}{2} - b) = b$ | A1 | Shown clearly that $d = b$ |

## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| The wavelength of the curtate cycloid is $2\pi a$ and the sine curve with the same wavelength and height has equation $y = b\cos\left(\frac{x}{a}\right)$ | M1 | Comparing the two curves with same wavelength ($2\pi a$) and same height ($b$) |
| Both curves pass through the same points (same height, same wavelength, same value of $d = b$), so they are very similar in shape | A1 | Concluding that since $d=b$ for both curves they share this geometric property, justifying similarity |

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