Arc Length

If the curve is a polygon, we can easily find its length; we just add the lengths of the line segments that form the polygon. (We can use the distance formula to find the distance between the endpoints of each segment.)

However, in general it can be very diffcult to find length of some curve. We use the same approach as with areas and volumes. We are going to define the length of a general curve by first approximating it by a polygon and then taking a limit as the number of segments of the polygon is increased.

Suppose that a curve C is described by the parametric equations x=f(t), y=g(t), `a<=t<=b`.

Let's assume that C is smooth in the sense that the derivatives f'(t) and g'(t) are continuous and not simultaneously zero for a<t<b. (This ensures that C has no sudden change in direction.) We divide the parameter interval into n subintervals of equal width `Delta t`. If `t_0`, `t_1`,`t_2`, . . . , `t_n` are the endpoints of these subintervals, then `x_i=f(t_i)` and `y_i=g(t_i)` are the coordinates of points `P_i=(x_i,y_i)` that lie on C and the polygon with vertices `P_0`,`P_1`, . . . , `P_n` approximates C. The length L of C is approximately the length of this polygon and the approximation gets better as we let n increase. Therefore, we define the length of curve C to be the limit of the lengths of these inscribed polygons: `L=lim_(n->oo)sum_(i=1)^n|P_(i-1)P_i|`.

We need a more convenient expression for L. Let `Deltax_i=x_i-x_(i-1)` and `Delta y_i=y_i-y_(i-1)` then the length of the i-th line segment of the polygon is `|P_(i-1)P_i|=sqrt((Delta x_i)^2+(Delta y_i)^2)`.

But by the definition of derivative `f'(t_i)~~(Delta x_i)/(Delta t)` and `g'(t_i)~~(Delta y_i)/(Delta t)` if `Delta t` is small. Therefore, `Delta x_i~~f'(t_i)Delta t` and `Delta y_i~~g'(t_i)Delta t`.

Thus, we have that `|P_(i-1)P_i|=sqrt((f'(t_i)Delta t)^2+(g'(t_i)Delta t)^2)=sqrt((f'(t_i))^2+(g'(t_i))^2)Delta t`.

So, `L=lim_(n->oo)sqrt((f'(t_i))^2+(g'(t_i))^2)Delta t`.

We recognize in this expression limit of Riemann sum, therefore we have

Arc Length Formula. If a smooth curve with parametric equations x=f(t), y=g(t), `a<=t<=b`, is traversed exactly once as t increases from a to b, then its length is `L=int_a^bsqrt(f'(t)+g'(t))dt`.

Note, that this formula is correct, provided that we rule out situations where a portion of the curve is traced out more than once.

In case when function is given as y=f(x), `a<=x<=b` then we can regard x as parameter, therefore parametric equations are x=x and y=f(x), so `L=int_a^bsqrt(1+(f'(x))^2)dx`.

In case when function is given as x=f(y), `a<=y<=b` then we can regard y as parameter, therefore parametric equations are x=f(y) and y=y, so `L=int_a^bsqrt(1+(f'(y))^2)dy`.

Because of the presence of the root sign, the calculation of an arc length often leads to an integral that is very difficult or even impossible to evaluate explicitly. In this case we need to use numerical integration.

Example 1. Find length of curve `x=e^tcos(t)`, `y=e^tsin(t)` where `0<=t<=pi`.

Using Product Rule we have that `(dx)/(dt)=(e^t)'cos(t)+e^t(cos(t))'=e^t(cos(t)-sin(t))` and `(dy)/(dt)=(e^t)'sin(t)+e^t(sin(t))'=e^t(cos(t)+sin(t))`.

Therefore, `L=int_0^pi sqrt((e^t(cos(t)-sin(t)))^2+(e^t(cos(t)+sin(t)))^2)dt=`

`=int_0^pi e^tsqrt(cos^2(t)+sin^2(t)-2sin(t)cos(t)+sin^2(t)+cos^2(t)+2sin(t)cos(t))dt=`

`=sqrt(2)int_0^pi e^tdt=sqrt(2)e^t|_0^pi=sqrt(2)(e^pi-1)`.

Example 2. Find length of curve y=sin(x), `0<=x<=pi`.

Since `y'=cos(x)` then `L=int_0^pi sqrt(1+cos^2(x))dx`. We can find exact value of this integral, thus, we need to use numeric integration. Let's use Simpson's Rule with n=10.

`Delta x=(pi-0)/10=pi/10` and `f(x)=sqrt(1+cos^2(x))`.

`L~~(Delta x)/3(f(0)+4f(pi/10)+2f((2pi)/10)+4f((3pi)/10)+2f((4pi)/10)+4f((5pi)/10)+2f((6pi)/10))+`

`+(Delta x)/3(4f((7pi)/10)+2f((8pi)/10)+4f((9pi)/10)+f(pi))~~3.8202`.

Example 3. Find length of curve `x=y^(3/2)`, `0<=y<=1`.

Since `x'=3/2sqrt(y)` then `L=int_0^1sqrt(1+(3/2sqrt(y))^2)dy=int_0^1 sqrt(1+9/4 y)dy=1/2 int_0^1 sqrt(4+9y)dy=`

`=1/2 *2/27 (9y+4)^(3/2)|_0^1=1/27(13^(3/2)-8)`.

Example 4. Find the length of one arch of cycloid, `x=r(theta-sin(theta))`, `y=r(1-cos(theta))`.

One arch of cycloid is described by the parameter interval `0<=theta<=2pi`.

Since `(dx)/(dtheta)=r(1-cos(theta))` and `(dy)/(dtheta)=rsin(theta)` then

`L=int_0^(2pi) sqrt((r(1-cos(theta)))^2+(rsin(theta))^2)dtheta=r int_0^(2pi)sqrt(1-2cos(theta)+cos^2(theta)+sin^2(theta))dtheta=`

`=sqrt(2)r int_0^(2pi) sqrt(1-cos(theta))dtheta`.

Now we need to use double angle formula `1-cos(theta)=2sin^2(theta/2)`.

Now integral can be rewritten as `2r int_0^(2pi)|sin(theta/2)|dtheta`.

Since `sin(theta/2)` is positive on `[0,2pi]` then we can drop absolute value bars and now we have following integral:



This result says that the length of one arch of a cycloid is eight times the radius of the generating circle.