Length of path examples

# Multivariable calculus and vector analysis

## A set of on-line readings

### Non-Math ML version

### Length of path examples

#### Example 1

Write a parameterization for the straight-line path from the point (1,2,3) to the point (3,1,2). Find the path length.

Solution: The vector from (1,2,3) to (3,1,2) is d = (3, 1, 2) - (1, 2, 3) = (2,-1,-1). We can parametrized the line segment by

c(t) = (1, 2, 3) + t(2,-1,-1), 0 ≤ t ≤ 1 |  |

To find path length, we calculate

c'(t) | = (2,-1,-1) |  |
||c'(t)|| | = = |  |

Therefore, the length of the path is

∫ ab||c'(t)||dt = ∫ 01dt = |  |

Clearly, it was silly to calculate the length this way. We knew the length of the line segment must be ||d|| = . But, this simply illustrates the method of calculating path length.

#### Example 2

Another parameterization for the line segment of example 1 is

p(t) = (1, 2, 3) + (et - 1)(2,-1,-1), 0 ≤ t ≤ log 2. |  |

Find the length of the line segment using this parametrization.

This might seem tricky at first. You can just take it by faith, skip the rest of this paragraph, and proceed. Or, just notice that (et - 1) is zero when t = 0, and it is 1 when t = log 2. Indeed, a particle with position p(t) at time t does move along the straight line from (1,2,3) to (3,1,2) as t goes from 0 to log 2. It just doesn’t move at a constant speed. You can read about another example where particles move along the same curve but at different speeds.

Solution: We simply use the definition of path length to find the length of the line segment using using this parameterization. We calculate

p'(t) | = et(2,-1, 1) |  |
||p'(t)|| | = et||(2,-1,-1)|| = et, |  |

so the path length is

∫ ab||p'(t)||dt | = ∫ 0log 2etdt |  |
| = (elog 2 - e0) |  |
| = (2 - 1) = , |  |

which agrees with example 1.

Examples 1 and 2 illustrate an important principle. The length of a path does not depend on its parametrization. Of course, this makes sense, as the distance a particle travels along a particular route doesn’t depend on its speed.

#### Example 3

Find the length of the helix parametrized by c(t) = (cos t, sin t,t) for 0 ≤ t ≤ 6π. (This was the example used in the previous reading.)

Solution: We calculate

c'(t) | = (- sin t, cos t, 1) |  |
||c'(t)|| | = = . |  |

The length is

∫ 06πdt = = 6π ≈ 26.7. |  |

## Contents by lecture

### Part 1

### Part 2

### Part 3

### Part 4

### Part 5

### Part 6

### Part 7

### Part 8

### Part 9

### Part 10

### Part 11

### Part 12

- The length of a path*
- Quiz 12: Length of path quiz*
- Physical interpretation of a parametrization and its derivative
- Length of a path examples

### Part 13

### Part 14

### Part 15

### Part 16

### Part 17

### Part 18

### Part 19

### Part 20

### Part 21

### Part 22

### Part 23

### Part 24

### Part 25

### Part 26

### Part 27