Day 8 in MAT229

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Announcements:
- I graded your Week 1 HW:
  - Key for Week 1 HW (#1, #2)
  - Comments on #1: You were to compute an average value (height) of the function on the interval; if you plot the function on this interval, you can see roughly what a typical height is -- this should be close to the height you calculated.
    Try to do reality checks whenever you can.
  - Similarly on the volume calculation: \[ f(x)=\frac{e^x+e^{-x}}{2}=\cosh(x) \]
    (The graph of $\cosh(x)$ is the shape of a hanging chain; the St. Louis arch is in the form of this shape, the catenary.)
    - the radii of the discs are at least 1 in radius, the interval is of length 2; so a minimum possible volume is a cylinder of radius 1 and height 2: volume $2\pi$;
    - a maximum is the radius of the highest point on the interval $[-1,1]$, $\approx 1.6$. So about $2\pi*2.56 \approx 16$.
    - So the answer has to be between $6 \lt V \lt 16$ (it's actually around 9). If your value isn't in this ballpark, you could benefit from this type of analysis!
      The skill of estimating is one that we should spend more time on in school.
- I graded your Lab 2.
  - Key for Lab 2 (.nb)
  - Comments: first of all, some general comments:
    - if you have questions, ask. I'm happy to help! I am frequently lonely during our class office hours.... I can meet outside of our class office hours; and you can always email with a Mathematica or mathematics question.
    - Graph, even if you're not asked for one! Is your tangent line really tangent? Can you see the inflection points that you claim exist? Why is there a maximum where you claimed there was a minimum?
    - And zoom in on the region of interest.
    1. You're looking for the tangent line at $x=0$; it doesn't hurt to graph the function and your "tangent line", to see if your "tangent line" is really tangent there!
    2. Some of you used "e" instead of "E" -- if you use the former, Mathematica has no idea what "e" is.
      You should be showing off a maximum and two inflection points in your final graph.
    3. A fair number of you defined g[x_]:=E^2x which Mathematica interprets as $e^2 x$ -- i.e. a linear function, with slope $e^2$. You need those parentheses to get $e^{2x}$: g[x_]:=E^(2x) Mathematica doesn't yet read minds....
      When you graphed these two functions, you might have noticed that one of them appeared linear -- it wasn't behaving like an exponential function. Make sure to engage your brain first....
    4. You were asked for exact numbers for the intersection points -- so use Solve, not NSolve.
      And the "area" requested was old-fashioned area, not signed area -- so you can use \[ |g(x)-h(x)| \] for your integrand to Mathematica, and that will take care of it!
    5. The volume calculation will look like \[ V=\int_0^{1.17456}\pi(q(x)^2-p(x)^2)\ dx \] A fair number of you tried \[ V=\int_0^{1.17456}\pi(q(x)-p(x))^2)\ dx \] Big difference! But the most important difference is that it's wrong...:(
- You've got a lot going this week (especially a whole bunch of IMath comes due).
- Exam 1 is upcoming, too. We're planning a practice exam, and then the real thing is going to be this weekend.
Today we get serious with section 6.8 -- indeterminate forms. So here we go: it's about limits:
- As noted before, the limit definition of the derivative is perhaps our most important example: \[ \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \] We'll use the alternative form of that definition today, to motivate L'Hôpital's rule \[ \lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a} \]
  Notice how I write "Hôpital": it was evidently originally spelled like a large dispensory of societal healthcare, but I don't like using that form because it may make mathematicians pronounce it incorrectly.... So "L'Hôpital" it is!
Materials:
- 6.8_indeterminate.nb
- 6.8_indeterminate.pdf
- Long yacks through the notes using Mathematica:
  - 6.8.InverseTrigReview
  - 6.8: indeterminate forms

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