To find the length of a curve, it is essential that the curve is described in parametric form (convert it, if need be:)
`ai+bj+ck`
As we (should) remember, this means that at any given input we will have a vector pointing at the current point of the graph.

`2ti+3t2j+t3k`
evaluated at t=4 will be (8,48,64). Taking the Origin (0,0,0) as our reference point, this is both the vector pointing to that point on the graph and the coordinates of said point in the extended cartesian plane.

What we still want, though, is the length of this infintesimally small segment of the graph. Therefore, we find the perpendicular (to the vector) component of the equation at this point by taking the derivative and normalizing it.

Again, picture the vector pointing to a point on the graph. Perpendicular to that vector, at its tip, lies the derivative vector tangent to the graph. By shortening this derivative vector to a unit length (normalizing, so we add the squared components and take their square root: (i2+j2+k2)1/2))we have eliminated any effect of the distance from the origin.

Since we have an infintesimal amount of differentiated points, it would only make sense that we integrate this mess over the parameter span we want to consider. So, to find the length of function r(t) from t=0 to t=2:

``` _2
( `
\  ||r'(t)||
,_)
0
```
The biggest obstacle in these problems is integrating the square root. If you're doing this for a class, hope that your professor is the type that designs his problems so a lot of things fall out and be SURE to know your trig identities. Other than that, you're pretty much dependent on your Integral tables.

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