Consider a curve given by \[ x=f(t),\; y=g(t),\; a\leq t\leq b, \] where the curve is traversed exactly once for \(t\) varying from \(a\) to \(b\). If \(f'\) and \(g'\) are continuous on \([a,b]\), then the arc length \(L\) of the curve is \[ L=\int_a^b \sqrt{ [f'(t)]^2+[g'(t)]^2}\;dt=\int_a^b \sqrt{ \left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2}\;dt. \]

Note that if the curve is \(y=f(x)\) on \([a,b]\), then its natural parametrization is \[x=t,\; y=f(t),\; a\leq t\leq b,\] and consequently the above arc length formula becomes that in section Arc Length: \[ L=\int_a^b \sqrt{ \left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2}\;dt =\int_a^b \sqrt{ 1+\left( \frac{dy}{dx} \right)^2}\;dx. \]

Example. Find the circumference of a circle of radius \(r\).

Solution. Consider the following circle of radius \(r\): \[ x=r\cos t,\; y=r\sin t,\; 0\leq t\leq 2\pi. \] Here \(\displaystyle\frac{dx}{dt}=-r\sin t\) and \(\displaystyle\frac{dy}{dt}=r\cos t\). Then the circumference \(L\) is \[\begin{align*} L &=\int_0^{2\pi} \sqrt{ \left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2}\;dt\\ &=\int_0^{2\pi} \sqrt{ (-r\sin t)^2+(r\cos t)^2}\;dt\\ &=\int_0^{2\pi} \sqrt{ r^2(\sin^2 t+\cos^2 t)}\;dt\\ &=\int_0^{2\pi} r\;dt\\ &=\left. rt \right\vert_0^{2\pi}\\ &=2\pi r. \end{align*}\] Note that if we consider the following parametrization of the circle \[ x=r\cos t,\; y=r\sin t,\; 0\leq t\leq 4\pi, \] where the circle is traversed twice, then the arc length formula gives twice the circumference: \[\int_0^{4\pi} \sqrt{ \left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2}\;dt\\ =\int_0^{4\pi} \sqrt{ (-r\sin t)^2+(r\cos t)^2}\;dt =4\pi r. \] Therefore, the arc length formula gives us the arc length of a curve from its parametrization for which the curve is traversed exactly once.

Example. Find the arc length of one arch of the cycloid: \[ x=r(t-\sin t),\; y=r(1-\cos t), \] which is traced by a fixed point on a circle of radius \(r\) rolling on the \(x\)-axis.

Solution. Here \(\displaystyle\frac{dx}{dt}=r(1-\cos t)\) and \(\displaystyle\frac{dy}{dt}=r\sin t\). Since one arch of the cycloid is traced by varying \(t\) from \(0\) to \(2\pi\), its arc length \(L\) is \[\begin{align*} L &=\int_0^{2\pi} \sqrt{ \left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2}\;dt\\ &=\int_0^{2\pi} \sqrt{ (r(1-\cos t))^2+(r\sin t)^2}\;dt\\ &=\int_0^{2\pi} \sqrt{ r^2(1-2\cos t+\cos^2 t+\sin^2 t)}\;dt\\ &=\int_0^{2\pi} r \sqrt{2(1-\cos t)}\;dt\\ &=\int_0^{2\pi} r \sqrt{4\sin^2\left(\frac{t}{2} \right)}\;dt\\ &=\int_0^{2\pi} 2r \sin\left(\frac{t}{2} \right)\;dt \;\;\left(\text{since } \sin\left(\frac{t}{2} \right)\geq 0 \text{ for } 0\leq t \leq 2\pi \right)\\ &=\left. -4r \cos\left(\frac{t}{2} \right) \right\vert_0^{2\pi}\\ &=-4r\cos(\pi)+4r\cos(0) \\ &=8r.\\ \end{align*}\]

Suppose that a particle is moving in the \(xy\)-plane along a parametric curve \(c(t)=(x(t),y(t))\) where time \(t\) starts from \(t=a\). Then the distance \(s(t)\) traveled over the time interval \([a,t]\) is \[ s(t)=\int_a^t \sqrt{ [x'(t)]^2+[y'(t)]^2}\;dt. \] Since the speed is the rate of change of distance traveled with respect to time, the speed at time \(t=c\) is \[ s'(c)=\left. \left(\frac{d}{dt} \int_a^t \sqrt{ [x'(t)]^2+[y'(t)]^2}\;dt \right) \right\vert_{t=c} =\left.\sqrt{ [x'(t)]^2+[y'(t)]^2} \right\vert_{t=c}.\]

Example. Suppose that a particle is moving along the curve \[x=e^t+e^{-t},\; y=2t,\; 0\leq t <\infty.\]

1. Find the speed (in m/s) at \(t=2\) sec.
   
   
2. Find the distance traveled during the time interval \([0,3]\).
   
   

Solution. Here \(x'(t)=e^t-e^{-t}\) and \(y'(t)=2\). \[\sqrt{ [x'(t)]^2+[y'(t)]^2} =\sqrt{(e^t-e^{-t})^2+2^2} =\sqrt{e^{2t}+e^{-2t}+2} =\sqrt{(e^t+e^{-t})^2} =e^t+e^{-t}\] (a) The speed at \(t=2\) sec is \[ s'(2)=\left.\sqrt{ [x'(t)]^2+[y'(t)]^2} \right\vert_{t=2} =\left. e^t+e^{-t} \right\vert_{t=2} =e^2+e^{-2} \text{ m/s}.\] (b) The distance traveled during the time interval \([0,3]\) is \[\begin{align*} s(3) &=\int_0^3 \sqrt{ [x'(t)]^2+[y'(t)]^2}\;dt\\ &=\int_0^3 (e^t+e^{-t})\;dt\\ &= \left. (e^t-e^{-t}) \right\vert_0^3\\ &= e^3-e^{-3} \text{ m}. \end{align*}\]

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