]> 23.2 Examples

## 23.2 Examples

Example 1

We begin by computing arc-length of our helix from $s 0 = 0$ to $s 1 = 2 π$ .

The parametric equations for $P$ are $x = cos ⁡ s , y = sin ⁡ s$ , and $z = s$ . We therefore obtain

$d x d s = − sin ⁡ s , d y d s = cos ⁡ s , d z d s = 1$

To compute arc-length we want to sum the lengths $d l$ of each of our infinitesimal pieces of $P$ .

Since we are integrating $d l ⟶$ which is $T ^ d l$ where $T ^$ is the unit vector in the direction tangent to $P$ , which is the direction of $d ρ ⟶ d s$ , we can find $d l$ from this by taking its dot product with the unit vector $T ^$ , which obeys $T ^ = d ρ ⟶ / d s | d ρ ⟶ / d s |$ .

We find then that arc length on $P$ is given by

$∫ P T ^ · d l ⟶ = ∫ s = s 0 s 1 d ρ ⟶ d s · d ρ ⟶ d s | d ρ ⟶ d s | d s = ∫ s 0 s 1 | d ρ ⟶ d s | d s$

The integrand here is then the absolute value of $d ρ ⟶ d s$ which is $( ( d x d s ) 2 + ( d y d s ) 2 + ( d z d s ) 2 ) 1 / 2$ .

Here it is $( ( sin ⁡ s ) 2 + ( cos ⁡ s ) 2 + 1 ) 1 / 2$ which is $2 1 / 2$ .

The value of the integral and the arc-length of this helix is therefore $2 1 / 2 ( s 1 − s 0 )$ or $2 3 / 2 π$ .

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 $F ⟶ · d l ⟶$ . Since $F ⟶ = − M g k ^$ , the integrand in the final ordinary integral $d s$ becomes $− M g d z d s$ and the work done by gravity is

$∫ s 0 s 1 ( − M g ) d z d s d s = − M g ∫ z 0 z 0 − 30 d z = 30 M g$

in appropriate units no matter what the path was, since the force is a gradient and the integral therefore path independent.