Today:
- Announcements:
- Your first weekly assignment (to do by hand -- with help
of Mathematica, if you like!) is returned.
A couple of comments:
- If the problem says "graph", I need to see a graph.
- \((a+b)^2 \ne a^2+b^2\): if you got ~5.69706... on the second problem you probably made this error.
- You could have known that that answer was wrong for
another reason: the radius of the cross-sectional areas
(circles) is always greater than 1, so the volume of a cylinder
of radius 1, height 2, is \(2\pi > 5.70\).
- Some of you left your answers in forms that don't give a
sense of the magnitude. It's good to have a decimal answer, to
wrap your head round. So one answer evaluated to around 1.5
(but that wasn't made explicit). If you had made that explicit,
you might have determined that the answer was wrong.
- Also returned is your first group work. Again, you all got
perfects on that -- you were being evaluated on working
together, and you were stellar.
- Some of you didn't give me much work. Next time I hope
that you'll provide a clearer solution to as many
problems as you do, showing work of course.
- I noticed more of the algebraic error \((a+b)^2 \ne a^2+b^2\)....
- I also noticed some "dx dropping" -- integrals need a
differential (e.g. dx, du) in them, or they're
undefined!
- On at least one "tangent line" the slope was left in
terms of x. A tangent line is of the form \(mx+b\), where
m and b are numbers (no dependence on \(x\)); but better
still, put it in the form \(f'(x_0)(x-x_0)+f(x_0)\)!
- For problem 1: every time you learn a new derivative, you
also learn a new anti-derivative (i.e., you can solve a
new integral). So some of you wanted to use \(u=\frac{x^2}{4}+1\)
as your \(u\) substitution, maybe in hopes of using a log. Good
thinking: you just learned how to handle \(\frac{1}{u}\)! But
this looks more like the derivative of arctan, so add that to
your toolbag.
- On problem 2, some of you left
\(sin^{-1}\left(-\frac{1}{2}\right)\) in the tangent line --
formally correct, but think of that answer this way: "whose sine
is -1/2?" And the answer is....
- I'll put solution Mathematica files on the schedule and
assignments page, after you've attempted them.
- Your IMath homeworks are due tonight.
- Our first test is next week, Tuesday, and will cover chapter 6
(the problems will have a very "calc 1" flavor, but with some
new functions).
- Today we continue on with arc trig functions, but want to get to
section 6.8 as well -- indeterminate forms
- So here we go:
- Some Mathematica to get us started -- all you need is sine....
- 6.6: in particular, \(\cos^{-1}(x)\)
- 6.8 -- indeterminate forms
Website maintained by Andy Long.
Comments appreciated.