Find y''_xx for the cycloid x = a (t- sin t), y = a (1

Find $y''_{xx}$ for the cycloid $x = a\bigl(t-\sin t\bigr),\ y = a\bigl(1-\cos t\bigr)$.

Step 1: Derivatives with respect to $t$

\[ \dot x = a\bigl(1-\cos t\bigr), \qquad \dot y = a\sin t. \]

Step 2: First derivative, simplified by half angles

Using $\sin t = 2\sin\tfrac{t}{2}\cos\tfrac{t}{2}$ and $1-\cos t = 2\sin^{2}\tfrac{t}{2}$: \[y'_{x} = \frac{a\sin t}{a\bigl(1-\cos t\bigr)} = \frac{2\sin\tfrac{t}{2}\cos\tfrac{t}{2}}{2\sin^{2}\tfrac{t}{2}} = \cot\frac{t}{2}.\]

Step 3: Differentiate with respect to $t$

The inner derivative $\tfrac12$ appears as a factor: \[\bigl(y'_{x}\bigr)'_{t} = -\frac{1}{\sin^{2}\tfrac{t}{2}}\cdot\frac12 = -\frac{1}{2\sin^{2}\tfrac{t}{2}}.\]

Step 4: Divide by $\dot x$

Writing $\dot x = 2a\sin^{2}\tfrac{t}{2}$ again, the two squared sines combine into a fourth power: \[y''_{xx} = \frac{-\dfrac{1}{2\sin^{2}\tfrac{t}{2}}}{2a\sin^{2}\tfrac{t}{2}} = -\frac{1}{4a\sin^{4}\tfrac{t}{2}}.\]

Step 5: A neater form

The ordinate itself is $y = a\bigl(1-\cos t\bigr) = 2a\sin^{2}\tfrac{t}{2}$, so $\sin^{4}\tfrac{t}{2} = \dfrac{y^{2}}{4a^{2}}$ and

\[ y''_{xx} = -\frac{1}{4a}\cdot\frac{4a^{2}}{y^{2}} = -\frac{a}{y^{2}}. \]

The curvature of the cycloid is thus governed entirely by its height above the base line.

Step 6: Verification

\[t = \pi \text{ (top of the arch): } \sin\frac{\pi}{2} = 1 \;\Longrightarrow\; y''_{xx} = -\frac{1}{4a}\]

\[\text{From the neat form: } y = 2a \;\Longrightarrow\; -\frac{a}{4a^{2}} = -\frac{1}{4a} \;\;\checkmark\]

\[y''_{xx}\lt 0 \text{ for every } t\neq2k\pi: \text{ each arch is concave down throughout} \;\;\checkmark\]