# 12. Arc Length of Curve: Parametric, Polar Coordinates

by M. Bourne

## Arc Length of a Curve which is in Parametric Coordinates

We'll first look at an example then develop the formula for the general case.

### Example 1 - Race Track

In the Curvilinear Motion section, we had an example where a race car was travelling around a curve described in parametric equations as:

`x(t) = 20 + 0.2t^3`,`y(t) = 20t − 2t^2`

where

xandyare in meters andtis time in seconds.

What is the distance travelled by the car in the first 8 seconds?

### Solution

The graph of this case is given below.

It is based on plotting the *x-* and *y*-points at times between `t = 0` and `t = 8`.

So for example, at `t = 0`,

`x(0) = 20`, and `y(0) = 0`, so the car starts at `(20, 0)`.

At `t = 3`,

`x(3) = 25.4`, and `y(3) = 42`, so the car is at `(25.4, 42)`.

Finally, at `t = 8`, the car is at

`x(8) = 122.4`, and `y(8) = 32`, that is `(122.4, 32)`.

The parametric curve (*x*(*t*), *y*(*t*)), showing assorted points.

**Estimate: **An inspection of the graph shows our final answer should be around 150 m.

We extend the concept from Arc Length of a Curve to the parametric case.

We start with the expression that we met in the earlier section:

`"length"=r=int_a^bsqrt((dx)^2+(dy)^2`

Differentiating with respect to *t* and squaring gives:

`((dr)/(dt))^2=((dx)/(dt))^2+((dy)/(dt))^2`

Taking the positive square root of each side:

`(dr)/(dt)=sqrt(((dx)/(dt))^2+((dy)/(dt))^2`

Then, integrating with respect to *t* from *t = t*_{1} to *t* = *t*_{2} gives us the formula for the length of a curve in parametric equations form:

`"length"=r` `=int_(t_1)^(t_2)sqrt(((dx)/(dt))^2+((dy)/(dt))^2`

### Back to Example 1

Find the required length travelled by the race car using the given formula.

## Arc Length of a Curve in Polar Coordinates

Once again we start with an example to get a sense of what we are trying to find.

### Example 2 - Golden Spiral

A Golden Spiral has the characteristic such that for every quarter turn (`90^@` or `π/2` in radians), the distance from the center of the spiral increases by the golden ratio `Phi = 1.6180`.

See more on this interesting topic at Golden Spiral, and see background at Polar Coordinates.

The formula for a golden spiral is as follows:

r(θ) = 1.618013e^{0.30635θ}

### Solution

Find the length of the spiral from the center to the point where it has rotated two complete revolutions.

Following is the spiral whose length we need to find. It traces out the angle from `θ = 0` to `θ = 4π` (2 revolutions).

**Estimate: **It is actually quite difficult to estimate the length of this curve by inspection. But it is reasonable to imagine we can approximate it with a circle, radius 40 and this would give a length (circumference) of

`C = 2πr = 2π(40) = 80π ≈ 251`

Next we'll meet the equation for the length.

## General Form of the Length of a Curve in Polar Form

In general, the arc length of a curve *r*(θ) in polar coordinates is given by:

`L=int_a^bsqrt(r^2+((dr)/(d theta))^2)d theta`

where θ spans from θ =

ato θ =b.

### Back to Example 2

Use the above formula to find the length of the Golden Spiral, rotated 2 revolutions.

## Archimedean Spiral

We can see Archimedean Spirals in the spring mechanism of clocks.

Watch mechanism [Image source]

An Archimedean Spiral has general equation in **polar coordinates**:

`r = a + bθ`

where

ris the distance from the origin;

ais the start point of the spiral; and

baffects the distance between each arm. (The distance is actually given by `2pib`.)

### Example 3

Find the length of a flat clock spring which is in the shape of a spiral having 7.5 turns, where the inner radius is 5 mm and the outer end radius is 15.5 mm.

(You can see more background on this question at Length of Archimedean Spiral.)

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