With Parametric equations, the arc length formula is written with equation \eqref{1}.
$$ArcLength=\int_{a}^{b}\sqrt{1+\left(\frac{dy/dt}{dx/dt}\right)^{2}}\,dx \tag{1} \label{1}$$
and for convenience might be expressed as
$$f(t)=\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}dt$$
Example:
Let $x(t)=t\cdot cos(t)$ and $y(t)=t\cdot sin(t).$ This curve will be a really nice spiral between $0$ and $20\pi.$
Example 1: 1.20.6. Calculate the length of a cycloid curve for a wheel of radius 8 inches.
The cycloid parametric equations are $x(t)=r(t-sin(t))$ and $y(t)=r(1-cos(t)).$
Answer: The only thing to do here is plug into the equation for parametric arc length.
$$f(t)=\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}dt=\sqrt{r^{2}(t-sin(t))^{2}+r^{2}(1-cos(t))^{2}}dt$$
$$=r\cdot\sqrt{2-2cos(t)}dt$$
Normally, I would stop at this point and integrate, but this can be beneficially simplified using a trig identity. The identity is $\cos\theta=1-2\sin^2\left(\frac{\theta}{2}\right).$ Factoring the radical gets $\sqrt{2(1-cos(t))}.$ Applying the identity gets $\sqrt{2(1-(1-2sin^{2}(t/2))}=\sqrt{4sin^{2}(t/2)}=2sin(\frac{t}{2}).$
$$f(t)=r\cdot2sin(\frac{t}{2})dt \tag{2} \label{2}$$
Equation \eqref{2} is a great deal easier to integrate.
$$\int_{0}^{2\pi}r\cdot2sin(\frac{t}{2})dt=-r\cdot4cos(\frac{t}{2})\Big|_{0}^{2\pi}=8r=64$$