Drawing part of a Bézier curve by reusing a basic Bézier-curve-function?

What you need is the De Casteljau algorithm. This will allow you to split your curve into whatever segments you'd like.

However, since you're dealing with just cubic curves, I'd like to suggest a slightly easier to use formulation that'll give you a segment from t0 to t1 where 0 <= t0 <= t1 <= 1. Here's some pseudocode:

u0 = 1.0 - t0
u1 = 1.0 - t1

qxa =  x1*u0*u0 + bx1*2*t0*u0 + bx2*t0*t0
qxb =  x1*u1*u1 + bx1*2*t1*u1 + bx2*t1*t1
qxc = bx1*u0*u0 + bx2*2*t0*u0 +  x2*t0*t0
qxd = bx1*u1*u1 + bx2*2*t1*u1 +  x2*t1*t1

qya =  y1*u0*u0 + by1*2*t0*u0 + by2*t0*t0
qyb =  y1*u1*u1 + by1*2*t1*u1 + by2*t1*t1
qyc = by1*u0*u0 + by2*2*t0*u0 +  y2*t0*t0
qyd = by1*u1*u1 + by2*2*t1*u1 +  y2*t1*t1

xa = qxa*u0 + qxc*t0
xb = qxa*u1 + qxc*t1
xc = qxb*u0 + qxd*t0
xd = qxb*u1 + qxd*t1

ya = qya*u0 + qyc*t0
yb = qya*u1 + qyc*t1
yc = qyb*u0 + qyd*t0
yd = qyb*u1 + qyd*t1

Then just draw the Bézier curve formed by (xa,ya), (xb,yb), (xc,yc) and (xd,yd).

Note that t0 and t1 are not exactly percentages of the curve distance but rather the curves parameter space. If you absolutely must have distance then things are much more difficult. Try this out and see if it does what you need.

Edit: It's worth noting that these equations simplify quite a bit if either t0 or t1 is 0 or 1 (i.e. you only want to trim from one side).

Also, the relationship 0 <= t0 <= t1 <= 1 isn't a strict requirement. For example t0 = 1 and t1 = 0 can be used to "flip" the curve backwards, or t0 = 0 and t1 = 1.5 could be used to extend the curve past the original end. However, the curve might look different than you expect if you try to extend it past the [0,1] range.

Edit2: More than 3 years after my original answer, MvG pointed out an error in my equations. I forgot the last step (an extra linear interpolation to get the final control points). The equations above have been corrected.

In an answer to another question I just included some formulas to compute control points for a section of a cubic curve. With u = 1 − t, a cubic bezier curve is described as

B(t) = u³P₁ + 3u²t P₂ + 3ut²P₃ + t³P₄

P₁ is the start point of the curve, P₄ its end point. P₂ and P₃ are the control points.

Given two parameters t₀ and t₁ (and with u₀ = (1 − t₀), u₁ = (1 − t₁)), the part of the curve in the interval [t₀, t₁] is described by the new control points

• Q₁ = u₀u₀u₀P₁ + (t₀u₀u₀ + u₀t₀u₀ + u₀u₀t₀) P₂ + (t₀t₀u₀ + u₀t₀t₀ + t₀u₀t₀) P₃ + t₀t₀t₀P₄
• Q₂ = u₀u₀u₁P₁ + (t₀u₀u₁ + u₀t₀u₁ + u₀u₀t₁) P₂ + (t₀t₀u₁ + u₀t₀t₁ + t₀u₀t₁) P₃ + t₀t₀t₁P₄
• Q₃ = u₀u₁u₁P₁ + (t₀u₁u₁ + u₀t₁u₁ + u₀u₁t₁) P₂ + (t₀t₁u₁ + u₀t₁t₁ + t₀u₁t₁) P₃ + t₀t₁t₁P₄
• Q₄ = u₁u₁u₁P₁ + (t₁u₁u₁ + u₁t₁u₁ + u₁u₁t₁) P₂ + (t₁t₁u₁ + u₁t₁t₁ + t₁u₁t₁) P₃ + t₁t₁t₁P₄

Note that in the parenthesized expressions, at least some of the terms are equal and can be combined. I did not do so as the formula as stated here will make the pattern clearer, I believe. You can simply execute those computations independently for the x and y directions to compute your new control points.

Note that a given percentage of the parameter range for t in general will not correspond to that same percentage of the length. So you'll most likely have to integrate over the curve to turn path lengths back into parameters. Or you use some approximation.