8.3: Arc Length and Curvature

Just like the area of a plane region can be found using calculus, so too can the length of a plane curve. Along the way the mystery mentioned in a footnote in Chapter 1 will finally be solved: what is the length of the hypotenuse of a right triangle with infinitesimal sides?

For a function \(y=f(x)\) denote by \(s\) the length of the piece of that curve over an interval \(\ival{a}{b}\), as in Figure [fig:arclength] (a). Call \(s\) the curve’s arc length over \(\ival{a}{b}\).

By the Microstraightness Property, for \(a \le x < b\) the curve is a straight line of length \(\ds\) over the infinitesimal interval \(\ival{x}{x+\dx}\), as in Figure [fig:arclength] (b), where \(\ds>0\) is the infinitesimal change in \(s\) over that interval. Notice that you cannot simply apply the Pythagorean Theorem here, since that would make \(\ds = \sqrt{(\dx)^2+(\dy)^2} = \sqrt{0+0} = 0\), which is false. The trick is to divide all sides of this infinitesimal right triangle by \(\dx\), which yields the similar—and noninfinitesimal—right triangle shown in Figure [fig:arclength](c). The Pythagorean Theorem can then be applied to that triangle:

\[\frac{\ds}{\dx} ~=~ \sqrt{1^2 + \left(\dydx\right)^2} \quad\Rightarrow\quad \ds ~=~ \sqrt{1 + \left(\dydx\right)^2}\;\dx \nonumber \]

Summing up those infinitesimal lengths \(\ds\) then yields the arc length \(s\):

That such a formula exists is of course good news, but as you have probably guessed, the integral cannot be evaluated in a closed form except for a few functions.¹ In most cases numerical integration methods will be required.

Example \(\PageIndex{1}\): arclength1

Find the arc length of the curve \(y=\cosh\,x\) over \(\ival{0}{1}\).

Solution

Since \(\dydx = \sinh\,x\), then the arc length \(s\) is:

\[\begin{aligned} s ~&=~ \int_0^1 \sqrt{1 + \left(\dydx\right)^2}~\dx ~=~ \int_0^1 \sqrt{1 + \sinh^2 x}~\dx\\ &=~ \int_0^1 \cosh\,x~\dx ~=~ \sinh\,x~\Biggr|_0^1 ~=~ \sinh\,1 ~-~ \sinh\,0 ~=~ \sinh\,1 ~\approx~ 1.1752\end{aligned} \nonumber \]

Example \(\PageIndex{1}\): Catenary

A catenary—a hanging uniform cable whose ends are fastened at the same height \(h\) a distance \(L\) apart—has its lowest point—the apex—a distance \(a>0\) above the ground. It can be shown² that with the apex at \((0,a)\), the equation of the catenary is \(y = a\,\cosh\,\tfrac{x}{a}\). Find the arc length of the catenary.

Solution

The figure on the right shows the catenary. By symmetry the total arc length \(s\) is twice the arc length over \(\ival{0}{L/2}\):

\[\begin{aligned} s ~&=~ 2\,\int_0^{L/2} \sqrt{1 + \left(\dydx\right)^2}~\dx ~=~ 2\,\int_0^{L/2} \sqrt{1 + \sinh^2 \tfrac{x}{a}}~\dx\

\[4pt] &=~ 2\,\int_0^{L/2} \cosh\,\tfrac{x}{a}~\dx ~=~ 2a\,\sinh\,\tfrac{x}{a}~\Biggr|_0^{L/2} ~=~ 2a\,\sinh\,\tfrac{L}{2a}\end{aligned} \nonumber \]

Suppose you tried to find the circumference \(s\) of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), with \(a>b>0\), which has eccentricity \(e=\frac{c}{a}\), where \(c=\sqrt{a^2-b^2}\). By symmetry, \(s\) is quadruple the arc length of the upper hemisphere \(y=b\sqrt{1-\frac{x^2}{a^2}}\) over \(\ival{0}{a}\):

\[E(k,\phi) ~=~ \int_0^{\phi} \sqrt{1 - k^2\,\sin^2 \theta}~\dtheta ~, \nonumber \]

with \(k=e\) and \(\phi = \frac{\pi}{2}\). This special case is denoted by \(E(k) = E(k,\frac{\pi}{2})\). Thus, the circumference \(s\) of the ellipse is:

\[s ~=~ 4a\,E(e) ~=~ 4a\,E(e,\tfrac{\pi}{2}) \nonumber \]

The integral \(E(k)\) for \(0<k<1\) cannot be evaluated in a closed form. There are tables³ for certain values of \(k\) between 0 and 1, but a number of scientific computing applications have built-in functions to evaluate elliptic integrals.

For example, suppose you want to find the circumference \(s\) of the ellipse \(\frac{x^2}{25}+\frac{y^2}{9}=1\). Then \(a=5\), \(b=3\), \(c=\sqrt{a^2-b^2}=4\), and \(e=\frac{c}{a}=0.8\), so \(s=4a\,E(e)=20\,E(0.8)\). In the Python-based open-source mathematical software system Sage⁴, the elliptic integral \(E(k,\phi)\) is provided by the function elliptice(\phi,k^2). Use \(k=e=0.8\) and \(\phi = \frac{\pi}{2}\):

In [1]: 20*elliptic_e(pi/2,0.8^2)
Out[1]: 25.5269988633981

The circumference is thus approximately 25.5269988633981. In Octave/MATLAB the function ellipke(e^2) evaluates the elliptic integral \(E(e)\), with one extra step:

MATLAB>> [K,E] = ellipke(0.8^2);
MATLAB>> 20*E
ans = 25.526998863398131

The parametric formula for arc length can be derived by dividing all sides of the infinitesimal right triangle in Figure [fig:arclength](b) by \(\dt\), then applying the Pythagorean Theorem to the resulting noninfinitesimal right triangle:

\[\dsdt ~=~ \sqrt{\left(\dxdt\right)^2 + \left(\dydt\right)^2} \quad\Rightarrow\quad \ds ~=~ \sqrt{\left(\dxdt\right)^2 + \left(\dydt\right)^2}~\dt \nonumber \]

Sum up those infinitesimal lengths \(\ds\) to obtain the arc length \(s\):

Example \(\PageIndex{1}\): arclencos3sin3

Find the arc length of the parametric curve \(x=\cos^3 t\), \(y=\sin^3 t\), \(0 \le t \le \pi/2\).

Solution

Since \(\dxdt=-3\cos^2 t\,\sin t\) and \(\dydt=3\sin^2 t\,\cos t\), then the arc length \(s\) is:

\[\begin{aligned} s ~&=~ \int_0^{\pi/2} \sqrt{\left(\dxdt\right)^2 + \left(\dydt\right)^2}~\dt ~=~ \int_0^{\pi/2} \sqrt{(-3\,\cos^2 t\;\sin\,t)^2 + (3\,\sin^2 t\;\cos\,t)^2}~\dt\\ &=~ \int_0^{\pi/2} \sqrt{9\,\cos^4 t\;\sin^2 t ~+~ 9\,\sin^4 t\;\cos^2 t}~\dt ~=~ \int_0^{\pi/2} 3\,\sqrt{\sin^2 t\;\cos^2 t\;(\cos^2 t ~+~ \sin^2 t)}~\dt\\ &=~ \int_0^{\pi/2} 3\,\sin\,t\;\cos\,t~\dt \quad\text{(since $\cos\,t \ge 0$ and $\sin\,t \ge 0$ for $0 \le t \le \pi/2$)}\\ &=~ \frac{3}{2}\,\sin^2 t~\Biggr|_0^{\pi/2} ~=~ \frac{3}{2}\end{aligned} \nonumber \]

The polar formula for arc length can be considered a special case of the parametric formula. A polar curve \(r=r(\theta)\) for \(\alpha\le\theta\le\beta\) has Cartesian coordinates \(x=r(\theta)\,\cos\,\theta\) and \(y=r(\theta)\,\sin\,\theta\), so that

\[\frac{\dx}{\dtheta} ~=~ \frac{\dr}{\dtheta}\,\cos\,\theta ~-~ r\,\sin\,\theta \quad\text{and}\quad \frac{\dy}{\dtheta} ~=~ \frac{\dr}{\dtheta}\,\sin\,\theta ~+~ r\,\cos\,\theta ~. \nonumber \]

It is left as an exercise to show that putting these derivatives into formula ([eqn:arclengthparam])—using the parameter \(\theta\) instead of \(t\)—yields the polar arc length formula:

Prove that the circumference of a circle of radius \(R\) is \(2\pi R\).

Solution: Use the polar curve \(r=R\) for \(0\le\theta\le 2\pi\). Then \(\frac{\dr}{\dtheta}=0\), so:

\[s ~=~ \int_0^{2\pi} \sqrt{r^2 + \left(\frac{\dr}{\dtheta}\right)^2}~\dtheta ~=~ \int_0^{2\pi} \sqrt{R^2 + 0^2}~\dtheta ~=~ \int_0^{2\pi} R~\dtheta ~=~ R\,\theta~\Biggr|_0^{2\pi} ~=~ 2\pi R \quad\checkmark \nonumber \]

In Chapter 4 you saw one simple measure of curvature: the second derivative. From Figure [fig:x2curvature] it is clear that the parabola \(y=x^2\) is less curved at the point \((1,1)\) than at the origin, yet \(\frac{d^2y}{\dx^2}=2\) at each point. So the second derivative—the rate of change of the slopes \(\dydx\) of the tangent lines—does not fully capture the curvature of a curve; more information is needed. It turns out that the rate of change of the angles of the tangent lines is the key to curvature.

First consider a curve with arc length \(s\) between two points \(A\) and \(B\) on the curve. Let \(\alpha\) be the angle between the tangent lines to the curve at \(A\) and \(B\), as in Figure [fig:avgcurvature](a).

For the same arc length \(s\) but larger angle \(\alpha\) as in Figure [fig:avgcurvature](b), the curvature appears greater. This suggests that curvature should measure \(\alpha\) relative to \(s\):

Similar to how the instantaneous rate of change of a function at a point is the average rate of change over an infinitesimal interval, the curvature of a curve at a point can be defined as the average curvature over an infinitesimal length of the curve:

The idea is that moving an infinitesimal length \(\ds\) along the curve induces an infinitesimal difference \(\dphi\) in the angles of the tangent lines. Roughly, as \(B\) moves toward \(A\):

\[\lim_{B \to A}~\bar{\kappa}_{AB} ~=~ \lim_{B \to A}~ \frac{\alpha}{s} ~=~ \frac{\dphi}{\ds} ~=~ \kappa \nonumber \]

Curvature can be viewed as the instantaneous rate of change of direction of a curve, but with respect to arc length (i.e. distance traveled) instead of time.

For a curve \(y=f(x)\), recall from formula ([eqn:tangentangle]) in Section 3.1 that \(\phi = \phi(x) = \tan^{-1} f'(x)\) at each point \((x,f(x))\) on the curve. Thus, by the Chain Rule:

\[\kappa ~=~ \frac{\dphi}{\ds} ~=~ \frac{d\,(\tan^{-1} f'(x))}{\ds} ~=~ \frac{\dfrac{f''(x)~\dx}{1 + (f'(x))^2}}{\ds} \nonumber \]

So since \(\ds=\sqrt{1 + (f'(x))^2}\,\dx\) by formula ([eqn:arclength]), then:

The above formula makes \(\kappa\) a function of \(x\). Note also that \(\kappa=0\) for a straight line, and that the curve \(y=f(x)\) is concave up if \(\kappa > 0\) and concave down if \(\kappa < 0\).

Find the curvature of the curve \(y=x^2\) for \(x=0\) and \(x=1\).

Solution: For \(f(x)=x^2\), \(f'(x)=2x\) and \(f''(x)=2\), so for any \(x\) the curvature \(\kappa=\kappa(x)\) is:

\[\kappa ~=~ \frac{f''(x)}{(1 + (f'(x))^2)^{3/2}} ~=~ \frac{2}{(1+4x^2)^{3/2}} \nonumber \]

In particular, \(\kappa(0)=2\) and \(\kappa(1)=\frac{2}{5^{3/2}} \approx 0.1789\). So \(y=x^2\) has more curvature at the origin than at \((1,1)\).

For a parametric curve \(x=x(t)\), \(y=y(t)\), the curvature \(\kappa\) will be a function of the parameter \(t\). Since \(\dydx = \frac{y'(t)}{x'(t)}\) by formula ([eqn:paramderiv1]) in Section 7.6, then by formula ([eqn:paramderiv2]):

\[\frac{d^2y}{\dx^2} ~=~ \frac{\Ddt\,\left(\Dydx\right)}{\Dxdt} ~=~ \frac{\Ddt\,\left(\dfrac{y'(t)}{x'(t)}\right)}{x'(t)} ~=~ \frac{x'(t)\,y''(t) ~-~ y'(t)\,x''(t)}{(x'(t))^3} \nonumber \]

So by formula ([eqn:curvaturefcn]):

\[\kappa ~=~ \frac{\dfrac{d^2y}{\dx^2}}{\left(1 + \left(\Dydx\right)^2\right)^{3/2}} ~=~ \frac{\dfrac{x'(t)\,y''(t) ~-~ y'(t)\,x''(t)}{(x'(t))^3}}{\left(1 \;+\; \left(\dfrac{y'(t)}{x'(t)}\right)^2\right)^{3/2}} ~=~ \frac{x'(t)\,y''(t) ~-~ y'(t)\,x''(t)}{\left((x'(t))^2\right)^{3/2} \,\left(1 \;+\; \left(\dfrac{y'(t)}{x'(t)}\right)^2\right)^{3/2}} \nonumber \]

Simplify the denominator to obtain the parametric curvature formula:

The derivation of the curvature formula in polar coordinates is left as an exercise:

Find the curvature of a circle of radius \(R\).

Solution: Use the polar curve \(r=r(\theta)=R\), so that \(r'(\theta) = 0 = r''(\theta)\):

\[\kappa ~=~ \frac{(r(\theta))^2 \;+\; 2\,(r'(\theta))^2 \;-\; r(\theta)\,r''(\theta)}{\left((r(\theta))^2 \;+\; (r'(\theta))^2\right)^{3/2}} ~=~ \frac{R^2 \;+\; 2 \cdot 0^2 \;-\; R \cdot 0}{(R^2 \;+\; 0^2)^{3/2}} ~=~ \frac{R^2}{R^3} ~=~ \frac{1}{R} \nonumber \]

A circle thus has constant curvature, as you would expect by symmetry. It turns out that any planar curve with constant curvature is either a line or part of a circle.⁵

[sec8dot3]

For Exercises 1-10, find the arc length of the given curve over the given interval.

3

\(y = x^{3/2}\) ; \(1 \le x \le 4\)

\(y = x^2\) ; \(0 \le x \le 1\)

\(y = x^{2/3}\) ; \(1 \le x \le 8\)

3

\(y = \dfrac{x^2}{4} - \dfrac{\ln\,x}{2}\) ; \(1 \le x \le 2\)

\(y = \dfrac{x^4}{4} + \dfrac{1}{8x^2}\) ; \(1 \le x \le 2\)

\(y = \ln\,\dfrac{e^x + 1}{e^x - 1}\) ; \(1 \le x \le 2\vphantom{\dfrac{x^4}{4}}\)

2

\(x = e^t\,\cos\,t\), \(y = e^t\,\sin\,t\) ; \(0 \le \theta \le \pi\)

\(x = \cos\,t \;+\; t\,\sin\,t\), \(y = \sin\,t \;-\; t\,\cos\,t\) ; \(0 \le t \le \pi\)

2

polar curve \(r = 1 + \cos\,\theta\) ; \(0 \le \theta \le 2\pi\)

polar curve \(r = e^{\theta}\) ; \(0 \le \theta \le 2\)

Find the arc length of the curve in Example \(\PageIndex{1}\) for \(0 \le t \le \pi\).

Use formula ([eqn:arclengthparam]) to find the circumference of the unit circle using two different parametrizations:

2

1. \(x=\cos\,t\), \(y=\sin\,t\), \(0 \le t \le 2\pi\vphantom{\dfrac{1-t^2}{1+t^2}}\)
2. \(x=\dfrac{1-t^2}{1+t^2}\), \(y=\dfrac{2t}{1+t^2}\), \(-\infty < t < \infty\)

For Exercises 13-18 find the curvature of the given curve at the indicated points. [[1.]]

3

\(y = \sin\,x\) at \(x=0\) and \(x=\frac{\pi}{2}\)

\(y = \ln\,x\) at \(x=1\)

\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) at \((a,0)\) and \((0,b)\)

3

\(y = e^x\) at \(x=0\)

\(x^2 - y^2 = 1\) at \((1,0)\)

\(r=1+\cos\,\theta\) at \(\theta=0\)

[[1.]]

2

Prove formula ([eqn:arclengthpolar]).

Prove formula ([eqn:curvaturepolar]).

Let \(\alpha\) and \(\beta\) be nonzero constants. Show that the arc length \(s\) of \(y=\beta\,\sin\,\frac{x}{\alpha}\) over the interval \(\ival{0}{x_0}\) can be put in terms of the elliptic integral \(E(k,\phi)\):

\[s ~=~ \sqrt{\alpha^2 + \beta^2}\,\cdot\,E\left(\sqrt{\frac{\beta^2}{\alpha^2 + \beta^2}},\; \frac{x_0}{\alpha}\right) \nonumber \]

For \(-1<k<1\) and \(-1\le x \le 1\), define \(u(x)\) and \(K\) by

\[u(x) ~=~ \int_0^x \frac{\dt}{\sqrt{1-t^2}\;\sqrt{1-k^2\,t^2}} \quad\text{and}\quad K ~=~ u(1) ~=~ \int_0^1 \frac{\dt}{\sqrt{1-t^2}\;\sqrt{1-k^2\,t^2}} \nonumber \]

so that all the square roots are defined and positive.

1. Show that \(u\) is an increasing function of \(x\) and thus has an inverse function, call it \(x = \sn\,u\), with domain \(\ival{-K}{K}\) and range \(\ival{-1}{1}\).
2. Define \(\;\cn\,u = \sqrt{1 - \sn^2 u}\;\) and \(\;\dn\,u = \sqrt{1 - k^2 \sn^2 u}\). Show that:

   \[\ddu\,(\sn\,u) ~=~ \cn\,u\;\dn\,u \quad,\quad \ddu\,(\cn\,u) ~=~ -\sn\,u\;\dn\,u \quad\text{, and}\quad \ddu\,(\dn\,u) ~=~ -k^2\,\sn\,u\;\cn\,u \nonumber \]

   The functions \(\sn\,u\), \(\cn\,u\), and \(\dn\,u\) are called the Jacobian elliptic functions.

3. Suppose that \(\sin\,\phi = \sn\,u\). Show that \(E(k,\phi) = \displaystyle\int_0^u \dn^2 v~\dv\).

The ends of a 50 ft long catenary are fastened 40 ft apart. Use a numerical method to find how much the apex dips below the ends. (Hint: Solve for \(a\) in Example

Example \(\PageIndex{1}\): arclengthcatenary

Add text here.

Solution

, then use symmetry.)