On the cycloid x=a(t- sin t), y=a(1- cos t), find

On the cycloid $x=a(t-\sin t),\ y=a(1-\cos t)$, find the point that divides the first arch in the ratio $1:3$.

We use the cycloid's known arc-length formula and solve for the dividing parameter.

Step 1: Total arch length

One arch ($t\in[0,2\pi]$) has total length $L_{\text{total}}=8a$.

Step 2: Required partial length

Dividing in ratio $1:3$ means the first piece is $\frac14$ of the total:

\[L_1=\frac14\cdot8a=2a\]

Step 3: Arc length from the cusp

The cycloid's speed is $\sqrt{(dx/dt)^2+(dy/dt)^2}=2a\sin\frac t2$, so

\[s(t)=\int_0^t2a\sin\frac\tau2\,d\tau = 4a\left[-\cos\frac\tau2\right]_0^t\Big/2 \cdot 2 = 4a\left(1-\cos\frac t2\right)\]

Step 4: Solve for $t$

\[4a\left(1-\cos\frac t2\right)=2a \implies \cos\frac t2=\frac12 \implies \frac t2=\frac\pi3 \implies t=\frac{2\pi}3\]

Step 5: Coordinates of the dividing point

\[x=a\left(\frac{2\pi}3-\sin\frac{2\pi}3\right)=a\left(\frac{2\pi}3-\frac{\sqrt3}2\right)=\frac{a(4\pi-3\sqrt3)}{6}\]

\[y=a\left(1-\cos\frac{2\pi}3\right)=a\left(1+\frac12\right)=\frac{3a}2\]

Step 6: Verify the ratio

The remaining piece: $s(2\pi)-s\left(\frac{2\pi}3\right)=4a(1-\cos\pi)-4a\left(1-\cos\frac\pi3\right)=8a-2a=6a$, and indeed $2a:6a=1:3$. ✓

\[\boxed{\left(\frac{a(4\pi-3\sqrt3)}{6},\,\frac{3a}2\right)\ \text{at}\ t=\frac{2\pi}3}\]