- Compute the curvature, normal vector or binormal vector of the
following parabola:
**f**(*u*) = (*u*, 1 +*u*^{2},*u*+*u*^{2}) - Consider the following two curve segments with the origin their
joining point:
**f**(*u*) = (*u*, -*u*^{2}, 0 )

**g**(*v*) = (*v*, 0,*v*^{2})*u*and*v*are in [-1,0] and [0,1], respectively, Are they*C*^{1},*G*^{1},*C*^{2}or*G*^{2}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) )*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. - The ellipse with center (
*p*,*q*), axes parallel to the coordinate axes, and semi-major and semi-minor axis lengths*a*and*b*has an equation(

It can be parameterized with trigonometric functions by*x-p*)^{2}/*a*^{2}+ (*y-q*)^{2}/*b*^{2}= 1*x*=*a*cos(*t*)*+ p*and*y*=*b*sin(*t*)*+ q*. Please verify this result. 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 +*u*^{2}) / (2*u*)

*y*=*b*(1 -*u*^{2}) / (2*u*)*u*will be very helpful. - We showed in
**Rational Curves**that a quadric polynomial parametric form can only represent a parabola through calculation. There are some minor flaws in the calculation. Please fill these gaps by answering the following questions:- We assume that
*a*and*p*are both non-zero. What would happen if*a*and*p*are zero? What curve will you get? - What if only one of
*a*and*p*, say*a*, is non-zero? What curve will you get? - In solving for
*u*, the denominator is*bp-aq*. What if*bp-aq*is zero? Note that*bp-aq*=0 is equivalent to*a/p=b/q*? What curve will you get?

- We assume that