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
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
OR the integral
- 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.