Q1. Write the equation of an implicit curve or a surface. (1 point)
A1. The equation should be in the form f(x,y,z)=0, so x^2+y^2+z^2-r^2=0, where x^2 means x to the power of 2, is an implicit equation of a surface (a sphere).
Q2. For two curve segments to join smoothly at a point, is it required that their first derivatives at the point be the same or their second derivative? (1 point)
A2. Their first derivatives.
Q3. What has to be satisfied among the control points of two quadratic B-spline curve segments in order for them to join smoothly? Give an example in a drawing. What about two cubic B-spline curve segments? (4 points)
A3. It is required that they share two control points. More specifically, the last two control points of one quadratic B-spine segment should be the same as the first two control points of another quadratic B-spline segment. If the curves are cubic, they should share three control points.
Q4. What is the difference between a Bezier curve and a rational Bezier curve? Are they the same? If not, which one is better and in what way? (3 points)
A4. Bezier and rational Bezier curves do not produce the same shape. The conrol points in a rational Bezier curve have weights associated with them, which may be varied to produce different curves even with the same control points. In a Bezier curve, once the control points are defined, the curve becomes fixed. Rational Bezier curves can generate exact conics, but Bezier curves can only approximate them. Since the basis functions of both Bezier and rational Bezier curves can be precomputed, drawing them would require the same amount of time, and since rational Bezier curves have the ability to design more shapes than Bezier curves, rational Bezier curves are preferred over Bezier curves in shape design.
.
Q5. It is required to draw a cylinder. Should we use a Bezier formulation, a Hermite bicubic formulation, or a surface of sweep formulation? Assuming the base of the cylinder is a circle of radius r in the xy plane centered at the origin and the height of the cylinder is h, find the equation of the cylinder. (6 points)
A5. A cylinder cannot be defined by a Hemite bicubic or a Bezier surface because these formulations cannot reproduce exact conics, and base of a cylinder is a conic. The formulation to use is the surface of sweep. To obtain a cylinder, we should sweep the circular base of the cylinder along its axis. Therefore,
x=rcos(2 pi u)
y=rsin(2 pi u)
z=vh
Q6. Find the equation of a Hermite cubic curve that will pass through points (0,0) and (1,1) and will be tangent to the x axis at (0,0) and have slope 1 at (1,1). (5 points)
A6. Since the curve has to pass through points (0,0), (1,1), it means P(0) = (x(0), y(0)) = (0, 0) and P(1) = (x(1), y(1)) = (1, 1). Also, since the curve should be tangent to x axis at the start and have a slpoe of 1 at the end, we should have P'(0) = (x'(0), y'(0)) = (1, 0) and P'(1) = (x'(1), y'(1) = (1, 1). Substituting these into the equation of the Hemite cubic curve, we obtain
x=u
y=-u^3+2*u^2
Note that this is only one solution. Other solutions would be obtained when increasing or decreasing the size of the tangent ventors at the start and the end of the curve.