Find dy/dx given x = a(t- sin t), y = a(1- cos t) —
Find $\dfrac{dy}{dx}$ given $x = a(t-\sin t)$, $y = a(1-\cos t)$ — the cycloid.
Explanation of Method
The cycloid, with a cusp at each point where it meets the ground.
Step 1: Differentiate with respect to $t$
\[ x'_{t} = a\bigl(1-\cos t\bigr), \qquad y'_{t} = a\sin t. \]
Step 2: Form the ratio
The constant $a$ cancels — scaling the wheel does not change the shape of the arch: $\frac{dy}{dx} = \frac{a\sin t}{a\bigl(1-\cos t\bigr)} = \frac{\sin t}{1-\cos t}$.
Step 3: Simplify with half-angle identities
Using $\sin t = 2\sin\dfrac{t}{2}\cos\dfrac{t}{2}$ and $1-\cos t = 2\sin^{2}\dfrac{t}{2}$, one factor of $\sin\dfrac{t}{2}$ cancels: \[\frac{dy}{dx} = \frac{2\sin\frac{t}{2}\cos\frac{t}{2}}{2\sin^{2}\frac{t}{2}} = \cot\frac{t}{2}.\]
Step 4: Verification
\[t = \pi:\quad (x,y) = \bigl(a\pi,\,2a\bigr),\quad \frac{dy}{dx} = \cot\frac{\pi}{2} = 0 \text{ — horizontal tangent at the top} \;\;\checkmark\]
\[t \to 0^{+}:\quad \cot\frac{t}{2} \to +\infty \text{ — vertical tangent at the cusp} \;\;\checkmark\]
\[t = \frac{\pi}{2}:\quad \frac{dy}{dx} = \cot\frac{\pi}{4} = 1 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\frac{dy}{dx} = \dfrac{\sin t}{1-\cos t} = \cot\dfrac{t}{2}, \qquad t \neq 2k\pi\,}\]
Remark
Remark. At $t = 2k\pi$ both $x'_{t}$ and $y'_{t}$ vanish, so the parametrisation is singular; the curve genuinely has a cusp there, where the rolling wheel's marked point is momentarily at rest. Compare Problem 193, whose $-\cot\tfrac{t}{2}$ is this answer reflected — the same arch traced the other way.