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Exam 2 grading

Posted: Wed Nov 29, 2006 10:32 am
by goodwine
  • Problem 1: 5 points each for k1 through k4 and 5 points for x(t+dt). A purely computational error on an early step was -4 with -1 on each of the additional terms. Non-computational errors (not properly substituting for x+k1/2, for example) was -4 for each occurance. Since the answer was in the back of the book, simply writing the answer was worth no credit.
  • Problem 2: 6 points per choice/graph.
    • wrong answers were -6
    • no explanation of the answers were -2 each
    It was acceptable for the second order methods to say either method produced the result (this was the intention of the problem when I wrote it). It turns out that it was possible to distinguish the two methods since the Taylor series method would have zero slope for the first step. If you did this and justified it correctly, you received full credit. If you partially justified the correct 2nd order method (by saying it was similar to either Euler or 4th order RK, as the case may be) you received full credit. If you had the wrong second order methods but each one was for the other second order method, i.e., you switched the 2nd order RK and 2nd order Tayler, it was -3 each. If you had the correct 2nd order method with the right graph but an incorrect justification, it was -2 each (a lot of people tried this by saying something like "Taylor always underestimates" which isn't true).
  • Problem 3: the main point of this problem was to correctly compute df/dt for the two f's. Since almost no one did this correctly, you received full credit for doing it regardless of how complete the program was. Otherwise
    • 5 points for a wrong attempt at a program (many people tried to use the formula for when f depended only on one x variable -- that's wrong, and yes, I did address this in class)
    • 10 points for an incorrect forumla for df/dt but at least an indication that you can't use the formula for one x variable.
  • Problem 4:
    • 5 points for anything on the paper
    • 8 points for the correct setup for PI control for q(t)
    • the listed number of points for the rest but with -1 or -2 for computational errors.