Math 4530
Introduction to
Curves and Surfaces
Spring term, 2001

Homework Assignments

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Homework assignments (most recent first):
Bold-face problems are to be handed in, others are only recommended.


   Due Friday April 20
     Oprea, 7.3.2, 7.3.5, 7.3.7, 7.3.10, 7.3.18, 7.3.19
more to come? Notice, I have replaced 7.3.11 with 7.3.10. Also, on 7.3.10, please draw a picture of the second order Enneper surface. See WEreps.pdf or WEreps.mws for the automated procedure.


   Due Friday April 13
     Oprea, 7.1.2, 7.1.3, 7.1.5, 7.1.6, 7.1.7

   Grading: 5 points, 1 point each.


   Due Friday April 6
     The computer-aided computations of mean and Gauss curvature, Oprea 4.2.5 (see link below), as well as Oprea 4.3.3 are due at the beginning of class. They are 30% of your second midterm score - the rest of the midterm will be given in class on Friday.


   Due Friday March 30
     Oprea 4.2.1, 4.2.3, 4.2.5, 4.3.4 You only have to plot the surfaces in 4.2.5, don't bother computing H and K this week. Plotting is really easy in Maple, see e.g. minimals.pdf. As far as computing curvatures, you've already done enough of these computations by hand, and I didn't give you the routines to automate this task in time. Oprea has Maple versions of the automation commands on pages 119-121. I modified these procedures slightly to reflect our matrix way of understanding them, and have put them into a Maple file, compcurv.mws, and a .pdf version you can look at from your browser, compcurv.pdf. If you download the .mws file and open it from Maple, essentially all you have to enter is the parameterizing function, and then use the procedures in the file to compute first fundamental form, second fundamental form, matrix of differential map, Gauss curvature, and mean curvature. I will ask for your answers to this part of 4.2.5 next week, so you have time to play with the automation procedures.

     hwmarch30.pdf  solution to maple part of hw
     hwmarch30.mws 

   Grading: total = 6 points; 4.2.1 = 1 point, 4.2.3 = 1 point, 4.2.5 = 3 points (graphing only), 4.3.4 = 1 point.


   Due Friday March 23
     HW1-HW5 (from notes of Monday March 5)
     3.2: 3, 5, 6, 7, 8, 9a, 15, 17
     3.3: 2, 3, 5, 7, 17, 19, 20


   Due Friday February 23
     2.4:   1, 2, 3, 5, 9, 11, 13, 15, 17, 24
The following problems are postponed until next week:
     2.5:   5, 14
     2.6:   1, 2, 3


   Due Friday February 16
     2.3:   1, 2, 3, 4, 5, 8, 9, 12, 15, 16
     2.5:   HW1, HW2, HW3, HW4, HW5. These are from Wednesday Feb 14 notes, but HW3, 4, 5 are elaborations on #9, 11, 12 from section 2.5 of the text.


   Due Friday February 9
     2.2: 1, 2, 3, 4, 7, 8, 9, 10, 12, 13, 16, 17, 19.


   Due Friday January 26
   Postponed until Friday February 2, due to number of problems assigned. (Of course, now I might think of a couple more...)
     1.5:   1, 2, 4, 5, 6, 7, 8, 9, 10, 11, 12-14
     1.6:   1, 2, 3


  Due Friday January 19
     1.3:   2, 5, 6, 7, 8, 9 10
     1.4:   look over these problems, they review geometry in 3-space. You don't need to hand anything in from this section.
     Class problems:   There were 3 problems in Wednesday's class notes, having to do with the identity on line 11 of page 13. If we denote the left side of the equation (the dot product of the two cross products) by L(u,v,x,y), and the determinant on the right side by R(u,v,x,y), then the 3 problems are as follows:
     HW1:   It is true that L and R are multilinear, that is they are linear in each of their four arguments. Prove this fact for the first argument (the "u" argument) of each, using linearity properties of dot product, cross product and determinant which we already know. Recall, a function F is linear if F(u1+u2)=F(u1)+F(u2) and F(c*u1)=c*F(u1), for all vectors u1, u2 and scalars c.
     HW2: Prove that F=L and F= R both have the following "alternating" properties: F(u,v,x,y)=-F(u,v,y,x), F(u,v,x,y)=-F(v,u,x,y).
     HW3: Prove that F=L and F= R both have the following symmetry property: F(u,v,x,y)=F(x,y,u,v).


  Due Friday January 12
     1.2:   1, 2, 3, 4, 5
  Grading: 6 points possible, distributed as follows:
     1.2:   1-4 = 1 point each, 5 = 2 points.