Problems
- Compute the curvature, normal vector or binormal vector of the
following parabola:
f(u) = ( u, 1 + u2,
u + u2 )
- Consider the following two curve segments with the origin their
joining point:
f(u) = ( u, -u2, 0 )
g(v) = ( v, 0, v2 )
where u and v are in [-1,0] and [0,1], respectively,
Are they C1, C2 or
G2 continuous
at the origin? Are they curvature continuous?
- Consider the following two circular arcs joining at the origin:
f(u) = ( cos(u + PI/2),
-(1 + sin(u + PI/2)), 0 )
g(v) = ( -cos(v + PI/2), 0,
1 - sin(v + PI/2) )
where both u and v are in the range of 0 and PI.
Note that circular arcs f(u) and g(v)
lie on the xy- and xz-coordinate planes,
respectively. Analyze the continuity at the origin.
- Eliminate the parameter u from the following equations:
x = f(u) = au2
+ bu + c
y = g(u) = pu2
+ qu + r
What is the type of the curve? It must be a conic; but which one?
Some calculations is necessary.
(Hint: If p is not zero, then solving for u
from the second equation and plugging the result to the first
equation would eliminate u. If both a and
p are non-zero, multiplying the first and the second
equations by p and a, respectively, and subtracting
the second from the first will eliminate u2.
From this result, solving for u and plugging the result back
to one of the original equation would eliminate u completely.
Thus, you will obtain a polynomial p(x,y) = 0.
What does it represent? Why?
- The ellipse with center at (p, q), axes parallel to
the coordinate axes, and semi-major and semi-minor axis lengths
a and b has an equation
(x-p)2/a2
+ (y-q)2/b2 = 1
It can be parameterized with trigonometric functions by
x = a cos(t) + p and
y = b sin(t) + q. Please verify this
relation. Convert this trigonometric parameterization to a rational
one. Does your parameterization contain circles as special cases?
- Analyze the relationship between u and the two branches of
the hyperbola parameterized with the following:
x = a (1 + u2) / (2u)
y = b (1 - u2) / (2u)
Plot several points that correspond to different u will be
very helpful.