Note

Because of the square root, the arc length integral cannot be evaluated explicitly except in special cases, but we can always approximate it numerically.

THEOREM 1 Arc Length

Let \(c(t)=(x(t),y(t))\), where \(x'(t)\) and \(y'(t)\) exist and are continuous. Then the arc length \(s\) of \(c(t)\) for \(a\le t\le b\) is equal to \begin{equation} \label{12.length.formula} s = \int_a^b \sqrt{x'(t)^2 + y'(t)^2}\, dt\tag{3} \end{equation}

EXAMPLE 1

Use Eq. 3 to calculate the arc length of a circle of radius \(R\).

Solution With the parametrization \(x=R\cos\theta\), \(y=R\sin\theta\), \begin{align*} x'(\theta)^2+y'(\theta)^2 &= (-R\sin\theta)^2 + (R\cos\theta)^2 = R^2(\sin^2\theta+\cos^2\theta) = R^2 \end{align*}

We obtain the expected result: \begin{align*} s &= \int_{0}^{2\pi} \sqrt{x'(\theta)^2+y'(\theta)^2}\, d\theta = \int_{0}^{2\pi} R\, d\theta = 2\pi R \end{align*}

EXAMPLE 2 Length of the Cycloid

Calculate the length \(s\) of one arch of the cycloid generated by a circle of radius \(R=2\) (Figure 11.15).

Figure 11.15: One arch of the cycloid generated by a circle of radius 2.

Solution We use the parametrization of the cycloid in Eq. (6) of Section 1: \begin{align*} x(t) &= 2(t - \sin t),& y(t) &= 2(1 - \cos t)\\ x'(t) &= 2(1 -\cos t), & y'(t)&=2 \sin t \end{align*} Thus, \begin{align*} x'(t)^2+y'(t)^2 &= 2^2(1-\cos t)^2+2^2\sin^2t\\ & = 4 - 8\cos t + 4\cos^2t+4\sin^2t \\ &= 8 -8\cos t\\ &=16\sin^2\frac{t}2\qquad\textrm{(Use the identity recalled in the margin.)} \end{align*}

One arch of the cycloid is traced as \(t\) varies from 0 to \(2\pi\), and thus \[ s = \int_{0}^{2\pi} \sqrt{x'(t)^2+y'(t)^2}\, dt = \int_{0}^{2\pi} 4 \sin \frac{t}2 \, dt = -8\cos\frac{t}2\bigg|_0^{2\pi}= -8(-1)+8 = 16 \]

Note that because \(\sin \tfrac{t}2\ge 0\) for \(0\le t\le 2\pi\), we did not need an absolute value when taking the square root of \(16\sin^2\tfrac{t}2\).

Note

In Chapter 13, we will discuss not just the speed but also the velocity of a particle moving along a curved path. Velocity is “speed plus direction” and is represented by a vector.

THEOREM 2 Speed Along a Parametrized Path

The speed of \(c(t)=(x(t), y(t))\) is \[ \boxed{\bbox[#fef7e5,5pt]{ \mathrm{Speed} = \frac{ds}{dt} = \sqrt{x'(t)^2 + y'(t)^2} }}\]

EXAMPLE 3

A particle travels along the path \(c(t) = (2t,1+t^{3/2})\). Find:

• (a) The particle’s speed at \(t=1\) (assume units of meters and minutes).
• (b) The distance traveled \(s\) and displacement \(d\) during the interval \(0\le t\le 4\).

Figure 11.17: The path \(c(t) = (2t,1+t^{3/2})\).

Solution We have \begin{align*} x'(t) &= 2,\qquad y'(t) = \frac32t^{1/2} \end{align*}

The speed at time \(t\) is \begin{align*} s'(t) &= \sqrt{x'(t)^2+y'(t)^2} = \sqrt{4 + \frac94t} \quad\text{m/min} \end{align*}

• (a) The particle’s speed at \(t=1\) is \(s'(1)= \sqrt{4+ \frac94}= 2.5 \text{m/min}\).
• (b) The distance traveled in the first 4 min is \[ s = \int_0^4 \sqrt{4+ \frac94t}\, dt = \frac8{27}\left(4+ \frac94t\right)^{3/2}\bigg|_0^4 = \frac{8}{27}\bigl(13^{3/2} - 8\bigr) \approx 11.52 \mathrm{m} \]

The displacement \(d\) is the distance from the initial point \(c(0)=(0,1)\) to the endpoint \(c(4)=(8,1+4^{3/2})=(8,9)\) (see Figure 4): \[ d = \sqrt{(8-0)^2+(9-1)^2} =8\sqrt 2 \approx 11.31 \mathrm{m} \]

EXAMPLE 4 Angular Velocity

Calculate the speed of the circular path of radius \(R\) and angular velocity \(\omega\). What is the speed if \(R=3\) m and \(\omega = 4\) rad/s?

Figure 11.18: A particle moving on a circle of radius \(R\) with angular velocity \(\omega\) has speed \(|\omega R|\).

Solution We have \(x=R\cos\omega t\) and \(y=R\sin\omega t\), and \[ x'(t)=-\omega R\sin \omega t,\qquad y'(t)= \omega R\cos \omega t \]

The particle’s speed is \begin{align*} \frac{ds}{dt} &= \sqrt{x'(t)^2+y'(t)^2} = \sqrt{(-\omega R\sin \omega t)^2+(\omega R\cos \omega t)^2}\\ & = \sqrt{\omega^2 R^2(\sin^2\omega t+\cos^2\omega t)} = |\omega| R \end{align*}

Thus, the speed is constant with value \(|\omega|R\). If \(R=3\) m and \(\omega = 4\) rad/s, then the speed is \(|\omega| R = 3(4) = 12 \text{m/s}\).

THEOREM 3 Surface Area

Let \(c(t)=(x(t),y(t))\), where \(y(t)\ge 0\), \(x(t)\) is increasing, and \(x'(t)\) and \(y'(t)\) are continuous. Then the surface obtained by rotating \(c(t)\) about the \(x\)-axis for \(a\le t \le b\) has surface area \begin{equation} \label{surfareaformula} S = 2\pi \int_a^b y(t)\sqrt{x'(t)^2+y'(t)^2}\,dt\tag{4} \end{equation}

EXAMPLE 5

Calculate the surface area of the surface obtained by rotating the tractrix \(c(t)=(t-\tanh t, \textrm{sech } t)\) about the \(x\)-axis for \(0 ≤ t < \infty\).

Figure 11.19: Surface generated by revolving the tractrix about the \(x\)-axis.

Solution Note that the surface extends infinitely to the right (Figure 11.19). We have \[ x'(t) = \frac{d}{dt}(t-\tanh t) = 1-\textrm{sech }^2t,\qquad y'(t)=\frac{d}{dt}\textrm{sech } t = -\textrm{sech } t\tanh t \]

Using the identities \(1 - \textrm{sech }^2 t = \tanh^2 t\) and \(\textrm{sech }^2t=1-\tanh^2t\), we obtain \begin{align*} x'(t)^2+y'(t)^2 &= (1-\textrm{sech }^2t)^2 +(-\textrm{sech } t\tanh t)^2 \\ &= (\tanh^2t)^2 +(1-\tanh^2t)\tanh^2t=\tanh^2t \end{align*}

The surface area is given by an improper integral, which we evaluate using the integral formula recalled in the margin: \begin{align*} S &= 2\pi\int_0^{\infty} \textrm{sech } t\sqrt{\tanh^2t} \,dt=2\pi\int_0^{\infty} \textrm{sech } t \tanh t \,dt=2\pi\lim_{R\to\infty} \int_0^R \textrm{sech } t \tanh t \,dt\\ &=2\pi\lim_{R\to\infty} (-\textrm{sech } t)\Big|_0^R = 2\pi \lim_{R\to\infty}(\textrm{sech } 0 - \textrm{sech } R)=2\pi\textrm{sech } 0 = 2\pi \end{align*}

Here we use that \(\textrm{sech } R = \frac1{e^R+e^{-R}}\) tends to zero (because \(e^R\to\infty\) while \(e^{-R}\to 0\)).