Hello,

 I am getting stuck on the following exercise. 

A curve p1 is Bezier form has coefficients:

    --  --    --  --    --  --
    | -2 |    | -1 |    | x0 |
    |  1 |  , |  0 |  , | y0 |
    --  --    --  --    --  --

and another curve p2 has coefficients

    --  --    --  --    --  --
    | -2 |    | -1 |    | x0 |
    |  1 |  , |  0 |  , | y0 |
    --  --    --  --    --  --

For what values x0, y0 do the two curves meet with equal first derivative?
For what values x2, y2 do the two curves meet with equal second derivative?

If you could, please explain your thought processes while solving
this, and also, explain what the problem is asking in different words.
 Thanks!

---

Request for Question Clarification by maniac-ga on 14 Oct 2006 09:39 PDT

Hello Garbonzo,

It appears the coefficients for x0, y0 were repeated and x2, y2 were omitted.
Can you clarify the question (otherwise the curves p1, p2 would be
equal at all points)?

You may also want to explain the use of the coefficients in your
question - there are more than one way to represent a bezier curve
(another example is the use of 4 control points), we would like to
make sure you get the correct answer.

  --Maniac

---

Clarification of Question by garbonzo-ga on 14 Oct 2006 09:56 PDT

oh poo,

i copied and pasted, and failed to edit.

here are the coefficients in the 2nd curve

    --  --    --  --    --  --
    | x0 |    |  1 |    | x2 |
    | y0 |  , |  0 |  , | y2 |
    --  --    --  --    --  --

sorry. This is the question as it is posed on a sample midterm, for
preperation for the midterm I have this coming Tuesday. Thanks again.

---

Request for Question Clarification by maniac-ga on 14 Oct 2006 13:21 PDT

Hello Garbonzo,

Hmm. An interesting question. Apparently the coefficients you provided
are the three data points of a pair of quadratic bezier curves as
described at
  http://en.wikipedia.org/wiki/Bézier_curve
I created a simple worksheet to play with the formulas and noted there
appears to be an infinite number of solutions to both equations. As a
side note, I've found this is a helpful first step in solving these
kinds of problems (allowing you to confirm results visually as well as
formally).

When Y0 is one (1), the first curve is horizontal. The second curve
starts horizontal and then moves towards the X2/Y2 coordinates. This
satisfies the first condition (equal first derivative).

In a similar manner, if Y2 is one (1), both the first and second
curves are horizontal. This satisfies the second condition (equal
second derivative).

Of course, this is not a formal proof - there are certainly other
researchers here (e.g., mathtalk) who could work through it in a much
more formal manner. It would probably be something like the following:

At the wikipedia site, the equation is represented like:
 B(t) = (1-t)^2*P0+2*t*(1-t)*P1+t^2*P2
where P0, P1, and P2 represent two element vectors with the [X,Y] values.

Note that at t=0, this simplifies to B(t) = P0, and at t=1 this
simplifies to B(t) = P2.

Reorder the terms in powers of t like:
  B(t) = t^2*(P0-P1+P2) + t*(2*P0+2P1) + (P0)

The first derivative of B(t) in terms of t is:
  B'(t) = 2*t*(P0-P1+P2) + (2*P0+2P1)
and second derivative is:
  B''(t) = 2*(P0 - P1 + P2)

You would then compute the formulas for T=1 (curve p1) and T=0 (curve
p2) equate, and solve to compare the slopes / accelerations at that
point.

If you are satisfied with the above as the answer, please let me know
and I'll repost as an answer. If you really need a formal solution,
another researcher (or commenter) may post one.

  --Maniac

---

Clarification of Question by garbonzo-ga on 16 Oct 2006 12:54 PDT

Hello maniac, (that sounds weird...)

  That answer is suitable. Please feel free to repost that as your
answer, and I will happily "fork up the dough". Thanks a bunch.

garbonzo

Hello Garbonzo,

Thanks for the kind words. Let me recap and expand (and correct!) a
little bit on the answer.

First - the values you provided are the X / Y coordinates for the
control points of a Quadratic Bezier Curve as described at
  http://en.wikipedia.org/wiki/Bezier_curve
[apparently GA doesn't reformat the URL with the é properly]

As illustrated at the same page (scroll down), a Quadratic Bezier
Curve has slope limiting as follows:
 at t=0, slope of P0 -> P1
 at t=1, slope of P1 -> P2
Hmm. It appears my spreadsheet is in error - let me fix it. [...]

OK. With the corrected worksheet, there is still an infinite number of
solutions where the first condition appears to be true.

The corrected solution is where Y0 = 0 and X0 is in the range (-1, 1).
This is a condition where

p1 (P0, P1, P2) has P1 -> P2 has slope zero.
p2 (P0, P1, P2) has P0 -> P1 has slope zero.

Correcting my previous message, making the second derivative equal,
the simplest solution is where

X0 = 0, Y0 = 0
and
X2 = 2, Y2 = 1

In this case, the resulting curves p1 and p2 are symmetric around the
origin, and thus the second derivative has to be equal on both sides
of Y=0. Note in this case that both (X0, Y0) and (X2, Y2) must be
specified for the condition to be true.

Reviewing the formal analysis I described before, I found at least one
more mistake. The corrected versions are:
  B(t) = t^2*(P0-2*P1+P2) + t*(-2*P0+2*P1) + (P0)
The first derivative of B(t) in terms of t is:
  B'(t) = 2*t*(P0-2*P1+P2) + (-2*P0+2*P1)
and second derivative is:
  B''(t) = 2*(P0 - 2*P1 + P2)
I suggest you confirm this with your own analysis. It looks odd to me
that the order of the three points
  P0, P1, P2
or
  P2, P1, P0
doesn't change the acceleration - so I may have made an additional
mistake [I'll think about it some more & post a clarification if I
find it]

However, based on the above and doing a little testing with the
worksheet, it appears that the condition of:
  Y0 = 0 and X0 is in the range (-1, 1)
is still OK when X2=2, Y2=-1. Taking a look at equating B''(t) for the
two splines, the point (X0, Y0) falls out during simplification and
the other points are symmetric - so the acceleration is still
"constant" and equal for both splines.

If any part of the answer is unclear or if you need additional
information on some part of the answer, please make a clarification
request. I will be glad to help further.

Good luck with your upcoming test.

  --Maniac

That helped. Thanks.