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*y^1/4 (1+ y^-3/2/16)^1/2 from y=0 to y=1. Now substituting x=y^1/4, 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 $\int$_s(0)^s(1) 2*pi*x(y(s))ds OR the integral $\int$_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.