- Given three control points on the
*xy*-plane (-1,0), (0,1) and (2,0), do the following:- Write down its Bézier curve equation.
- Expand this equation to its equivalent conventional form.
- Since there are three control points, there are three Bézier coefficients. Write down their equations and sketch their graphs.
- Use your calculator to find enough number of points using the conventional parametric form and sketch the curve.
- Find points on the curve that correspond to
*u*= 0, 0.25, 0.5, 0.75 and 1 with the conventional form. - Use de Casteljau's algorithm to find points on the curve
corresponding to
*u*= 0, 0.25, 0.5, 0.75 and 1. - Subdivide the Bézier curve at
*u*= 0.4 and list the control points of the resulting curve segments. - Increase the degree of this curve to three and list the new set of control points. Then, increase the degree to four and list the new set of control points.

- In the variation diminishing property, what if you have a line or a plane that passes through a control point or contains a line segment of the control polyline? Suggest a proper counting of intersection points and verify your claim with examples.
- A Bézier curve of degree 2 defined by three control
points
**P**_{0},**P**_{1}and**P**_{2}is a portion of a conic section. What type of this conic section is it? Is it a portion of a parabola, a hyperbola or an ellipse? You can assume the given control points are in the*xy*-coordinate plane. - Suppose Bézier curve
**C**(*u*) (*resp.*,**D**(*u*)) of degree*n*is defined by control points**P**_{0},**P**_{1}, ...,**P**_{n}(*resp.*,**O**_{0},**Q**_{1}, ...,**Q**_{n}). If the curves are identical (*i.e.*,**C**(*u*) =**D**(*u*) for every*u*in [0,1]), then the corresponding control points are also identical (*i.e.*,**P**_{i}=**Q**_{i}for all 0 <=*i*<=*n*).

First show that if (1-*Hint:**u*)**A**+*u***B**is a zero vector for every*u*in [0,1], then**A**and**B**are both zero vectors. Then, work the de Casteljau's algorithm backward to show that**P**_{i}-**Q**_{i}is a zero vector for all 0 <=*i*<=*n*. - Suppose Bézier curve
**C**(*u*) of degree*n*is defined by control points**P**_{0},**P**_{1}, ...,**P**_{n}.- Prove the following:
- Show that curve
**C**(*u*) can be rewritten to the following*matrix*form:where entry

*m*_{ij}is defined as follows:

*u*^{0}= 1,*u*^{1},*u*^{2}, ....,*u*^{n}. This is the so-called*monomial form*and the basis functions are*u*^{0}= 1,*u*^{1},*u*^{2}, ....,*u*^{n}. However, the use of this monomial form is computationally unstable. - Prove the following:
- Show that the maximum of
*B*_{n,i}(*u*) occurs at*u*=*i/n*and that the maximum value is - Verify the following results with your calculus knowledge:
- the derivative of
*B*_{n,i}(*u*): - the derivative of Bézier curve
**p**(*u*):

- the derivative of
- The discussion of joining two Bézier curves with
*C*^{1}-continuity assumes the domain of the curves is [0,1]. Suppose the domain of the first curve is [0,*s*] and the domain of the second curve is [*s*,1]. Redo the calculation. What is your conclusion? Is there any modification required? - Prove the following:
where

**D**_{i}^{k}'s are the*k*-th difference points and*C*(*k*,*j*) is the binomial coefficient defined as follows:With this formula, we can express a higher derivative using the original control points rather than using finite difference points.

- After subdividing a Bézier curve of degree
*p*at*s*, we have two Bézier curves of degree*p*, one on interval [0,*s*] while the other on [*s*,1]. Show that these two curves are of*C*^{1}continuous at the joining point.

: Suppose the last two control points of the curve on [0,*Hint**s*] are**P**_{p-1}and**P**_{p}, and the first two control points of the curve on [*s*,1] are**Q**_{0}and**Q**_{1}. Then, we have**P**_{p-1},**P**_{p}=**Q**_{0}and**Q**_{1}are on the same line, and the ratio of the distance from**P**_{p-1}to**P**_{p}=**Q**_{0}and the distance from**P**_{p}=**Q**_{0}to**Q**_{1}is equal to*s*due to subdivision. Now, change the variables of both curves so that they have domain on [0,1]. A simple calculation will lead to the desired conclusion.