But in derivation terms - with dx and dy also shown as 'legs' of a right triangle:

Where we treat ds as the differential of arc length.

ds

^{2}= dx

^{2}+ dy

^{2}

Or: ds =

**Ö**(dx

^{2}+ dy

^{2}) = dx

**Ö**[ 1 + (dy/dx)

^{2}]

This can then be integrated (between appropriate limits, say x1 and x2) to give the total length of a curve.. Thence the arc length is given by:

where the function f(x) defines the curve for which the arc length is evaluated between the points x1 and x2. Consider then finding the arc length between x1 = 0 and x2 = 8 for the parabola section below:

For which we have: f(x) = [(x)]

^{½}

Then evaluating the integral as shown, one obtains: s = 8.732

A more difficult problem involves a polar curve, i.e. a curve in polar coordinates such as:

This curve has the function: r(q) = q - sin (q)

**q = -p**and

**q = p**.

The relevant integral for the arc length for this example is:

Performing the integration one then obtains: L = 8.764

Which interested and energized readers are invited to check!

**Parametric Representations:**

In advanced treatments of curvature including
determining the curvature of a complex curve, parametric representations are the norm. This usually entails taking the tangent T to the curve at a point. From this the curvature k of the arc can also be computed. For example, one can give the parametric representation of a specific
type of curve using:

x
= 6 sin 2 t, y = 6 cos 2 t, z = 5t

For
which the tangent to the curve at a point is given by:

**T**= d

**R**/ ds =

**i**(dx/ds) +

**j**(dy/ds) +

**k**(dz/ds)

=

**i**(dx/dt) (dt/ds) +

**j**(dy/dt)(dt/ds) +

**k**(dz/dt) (dt/ds)

Then:

dx/dt= 12 cost 2t, dy/dt = -12 sin 2t, dz/dt = 5

Therefore, the tangent unit vector is:

**|**

**T**

**|**

**=**1 = (dt/ds)

^{2}[(12 cost 2t)

^{2}+ (-12 sin 2t)

^{2}+ 5

^{2}]

**|**

**T**

**|**

**=**1 = (dt/ds)

^{2}[(144 + 25)] = (dt/ds)

^{2}(169)

**So that:**(dt/ds)

^{ }= 1/ 13

And:

**T**= 1/ 13 [12 cost 2t)**i**+ (-12 sin 2t)**j**+ 5**k**] dt/ds
The

*curvature*can then be obtained from: k =**|****d****T/**ds**|**
Thus,
any curve can be given by a parametric representation: u1 = u1(t)
and u2= u2(t)

For
such a curve, consider now the distance between two infinitely near points on
the surface, i.e. the distance or interval ds between two specified points.

**Example problem**:
A
curve is given in spherical coordinates x

^{i}by:
x

^{1}= t, x^{2 }= arcsin 1/t, x^{3}= (t2 – 1)^{1/2}^{}

Compute
the length of the arc between t = 1 and t = 2

Solution:

(ds/ dt)

^{2}= (dx^{1}/ dt)^{ 2}+ (x^{1 })^{ 2}(dx^{2}/ dt)^{ 2}+ (x^{1 }sin x^{ 2})^{2}(dx^{3}/ dt)^{ 2}
And:

(dx

^{1}/ dt)^{ 2}= 1
(dx

^{2}/ dt)^{ 2}= [ -1/ t^{2 }/ Ö {1 – ( 1/ t^{2 })} = 1/ t^{2 }(t^{2 }- 1)
(dx

^{3}/ dt)^{ 2}= 2t/ Ö2(t^{2 }- 1) = t^{2 }/ (t^{2 }- 1)
Whence:

(ds/ dt)

^{2}=
1 + t

^{2 }· 1/ t^{2 }(t^{2 }- 1) + (t · 1/ t^{ }**)**^{ 2 }· t^{2 }/ (t^{2 }- 1)
= 2 t

^{2 }/ (t^{2 }- 1)
Then the length of the curve is:

L = ò

_{1 }^{2}^{ }Ö2 t/ (t^{2 }- 1)^{1/2}^{ }dt = Ö2(t^{2 }- 1) ]_{1}^{2 }**= Ö6**^{ }*Problem for Math Mavens*:

For the parametric example with:

**T**= 1/ 13 [12 cost 2t)

**i**+ (-12 sin 2t)

**j**+ 5

**k**] dt/ds,

Find the

__curvature__and the length of the curve from t = 0 to t = p.
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