Home  18.013A  Chapter 23 


Example 1
We begin by computing arclength of our helix from ${s}_{0}=0$ to ${s}_{1}=2\pi $ .
The parametric equations for $P$ are $x=\mathrm{cos}s,y=\mathrm{sin}s$ , and $z=s$ . We therefore obtain
To compute arclength we want to sum the lengths $dl$ of each of our infinitesimal pieces of $P$ .
Since we are integrating $d\stackrel{\u27f6}{l}$ which is $\widehat{T}dl$ where $\widehat{T}$ is the unit vector in the direction tangent to $P$ , which is the direction of $\frac{d\stackrel{\u27f6}{\rho}}{ds}$ , we can find $dl$ from this by taking its dot product with the unit vector $\widehat{T}$ , which obeys $\widehat{T}=\frac{d\stackrel{\u27f6}{\rho}/ds}{\leftd\stackrel{\u27f6}{\rho}/ds\right}$ .
We find then that arc length on $P$ is given by
The integrand here is then the absolute value of $\frac{d\stackrel{\u27f6}{\rho}}{ds}$ which is ${\left({\left(\frac{dx}{ds}\right)}^{2}+{\left(\frac{dy}{ds}\right)}^{2}+{\left(\frac{dz}{ds}\right)}^{2}\right)}^{1/2}$ .
Here it is ${\left({(\mathrm{sin}s)}^{2}+{(\mathrm{cos}s)}^{2}+1\right)}^{1/2}$ which is ${2}^{1/2}$ .
The value of the integral and the arclength of this helix is therefore ${2}^{1/2}({s}_{1}{s}_{0})$ or ${2}^{3/2}\pi $ .
Example 2
Compute the work done by the force of gravity when the naughty child throws a stone of mass $M$ out the window and it lands on the ground 30 feet below the window. The work done here is given by the line integral over the path of the stone of $\stackrel{\u27f6}{F}\xb7d\stackrel{\u27f6}{l}$ . Since $\stackrel{\u27f6}{F}=Mg\widehat{k}$ , the integrand in the final ordinary integral $ds$ becomes $Mg\frac{dz}{ds}$ and the work done by gravity is
in appropriate units no matter what the path was, since the force is a gradient and the integral therefore path independent.
