Department of Mathematics
Bernhard G. Bodmann

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Math 1450H, Frequently Asked Questions:

  • Y. L. asks: I have a question about question 37 on ch 2.4 (p 118) and was wondering if you could explain it? I think I pretty much get the logic of most of the steps, but I was wondering how 1/2 a was chosen as a restriction ( for the step reasoning | x-a |< 1/2 a ) ?
    See additional explanations and two possible solutions to this problem.
  • J. R. contributed a neat diagram called the "house of calculus" that relates the behavior of a function with its derivatives and helps remember the derivative tests.
  • B. N. helped us memorize trig formulas.
  • B.N. asked whether the solution to 8.2 Exercise 1 (b) is correct as stated. This part of the problem asks to find the surface area of the object generated by revolving the graph of y=x4 about the y-axis. The standard form of the answer would be the integral of 2*pi*y1/4 (1+ y-3/2/16)1/2 from y=0 to y=1. Now substituting x=y1/4, dx=(1/4)y-3/4dy, and eliminating the y so that 4 x3 dx = dy then gives that the answer is indeed the integral of 2*pi*x*(1+16 x6)1/2 from x=0 to x=1. The deeper reason for this is that ds=(1+(dx/dy)2)1/2dy OR ds=(1+(dy/dx)2)1/2dx and the integral of the surface area is s(0)s(1) 2*pi*x(y(s))ds OR the integral s(0)s(1) 2*pi*x(s)ds.
  • For anyone who wants a bit more information than the terse explanation of a proof by induction in Stewart's book, see more illustrative examples.
Bernhard G. Bodmann,   University of Houston    ---    Last modified:  April 20 2007 - 13:48:50
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