Day 22, Mat128: Calculus A

Last Time

Next Time

Announcements:
- If you need to Zoom to class, here's the link. Old Zooms are here.
- For next time, Wednesday, please submit today's worksheet, and the preview for section 3.3.
- I'm declaring a general amnesty for homeworks -- worksheets that haven't been submitted, previews, etc.
  The important thing is that you try them. And submit them, so that I can see what you're doing!
  The upshot: even though the deadlines have passed on some of these, you can still submit them for some credit (I can't promise all, because that wouldn't be fair to those who already submitted, on time).
Last time: We looked at how to use the derivative to investigate the shape of a curve.
Information about the shape of the graph $y=f(x)$ comes from
- the function itself,
- its first derivative $f'(x),$ and
- its second derivative $f''(x).$
I was just grading your 2.6 (inverse) "worksheets", which involved you studying two functions, and I hoped that you'd compute their derivatives.
Those derivative calculations didn't happen, however, so let me take you through those now.
Today we talk a little more the extrema which we identified last time. One of the most important roles of the derivative is to identify those places where a function hits its maximum or minimum.
- This frequently goes by the name of "optimization" -- we're trying to find the best way to do something, and setting a particular parameter at a certain value yields the best result. Our job is to find that particular value.... and the derivative is frequently our tool of choice.
  If you've ever fiddled with the dial on a radio, trying to get a station to come in better, you understand: you're adjusting the position of a value along an axis to get the clearest signal -- the maximal amount of signal to noise.
  It's the same with a thermostat, perhaps -- only you're fighting someone else in the house. For your money, you'd like it a little warmer; but they'd like it a little cooler. (This is a more complicated optimization problem!)
- For this simpler type of optimization, we need this
  
  Definition 3.1.6.:
  We say that a function $f$ has a critical number (or critical point) at $x=c$ provided that $c$ is in the domain of $f$, and $f'(c)=0$ or $f'(c)$ is undefined.
- Finding and Characterizing Extrema:
  1. We are planning to build a rectangular pen for our dog. We only have 200 meters of fencing. What are the dimensions of the pen that will provide our dog the most area to play in?
    The area of the pen is $Area=width*length$, or $A=w*l$ -- but the two dimensions are not independent, since we know the perimeter: \[ 2(w+l)=200 \]
    Hence we conclude that \[ w+l=100 \] or that \[ l=100-w \]
    Therefore \[ A(w)=w*(100-w) \]
    Where is this function a maximum? Well, we don't even need calculus to solve this one -- we can use symmetry, and the fact that $A$'s graph is a parabola:
    
    The peak has to occur exactly half way between the roots (0 and 100), so when $w=50$. But this makes $l=50$ as well, indicating that the maximum area $A$ occurs when the pen is square, 50 meters by 50 meters (with an area of 2500$m^2$).
    But if we'd differentiated, sure enough we'd find that $w=50$ is the critical point: \[ A'(w)=100 - 2w = 0 \textrm{ when } w=50 \]
    Calculus and algebra both allow us to determine that the ideal rectangular pen for our critters is a square. (Although a more advanced form of calculus, called the calculus of variations, allows us to conclude that to maximize the area per perimeter, you should construct a circular fence -- a corral!)
  2. What if we can use an existing wall, e.g. the side of a long barn? What would the dimensions be? What do you think?
    Now the two dimensions are related by \[ 2w+l=200 \]
    so \[ l=200 - 2w=2(100 - w) \]
    Therefore \[ A(w)=2w*(100-w) \]
    No need to differentiate: this is just twice the previous equation, so it has the same critical point: \[ A'(w)=100 - 2w = 0 \textrm{ when } w=50 \] But now the length is 100m. You might think of this as two of the previous square pens glued together!
  3. The point is that these sorts of problems, when they can be formulated as functions, present opportunities for "optimization".
    Generally we seek the global maximum or the global minimum -- which we have found in the cases above. Because those functions were quadratics, we know that they have only one point on their graphs where the derivative is zero -- and no points where their derivatives don't exist.
    
    One other important consideration is the domain. In those problems, the dimensions had to be positive. So we could have constrained the solutions to be where both $w>0$ and $l>0$.
    This sets up the possibility for the maximum or minimum to be at the border (as is true in this case -- the global minimum is 0 -- if we run the fencing the whole length, so that it's all length, with $w=0$. This results in a fence that Fido won't like at all!
    Furthermore, however, we might have local "extrema" (maxes or mins), that, while larger than their neighbors, give Fido only a local advantage -- but a better solution could be found elsewhere in the domain:
  4. Now let's consider a more pathological function: that of the preview for this section:
    
    and this set of images showing (some of) the kinds of things that can go wrong (at least in the case of a continuous function -- if you introduce discontinuities, things can get uglier):
    
    Figure 3.1.5. From left to right,
    1. a function with a relative maximum where its derivative is zero;
    2. a function with a relative maximum where its derivative is undefined;
    3. a function with neither a maximum nor a minimum at a point where its derivative is zero;
    4. a function with a relative minimum where its derivative is zero;
    5. and a function with a relative minimum where its derivative is undefined.
  5. There are two important tests in this section which we can deduce from the graphic above:
    1. The first derivative test: Let $c$ be a critical number of continuous function $f$.
      - If $f^\prime$ changes sign from positive to negative at $c$, then $f$ has a local maximum at $c$.
      - If $f^\prime$ changes sign from negative to positive at $c$, then $f$ has a local minimum at $c$.
      - If $f^\prime$ does not change sign at $c$, then $f$ has neither a max nor a min at $c$.
    2. Second derivative test: Suppose $f^{\prime\prime}$ is continuous near $c$.
      - If $f^{\prime}(c)=0$ and $f^{\prime\prime}(c)>0$, then $f$ has a local minimum at $c$;
      - If $f^{\prime}(c)=0$ and $f^{\prime\prime}(c)<0$, then $f$ has a local maximum at $c$.
  6. Let's try out Activity 3.1.2 from this section, and then I'll turn you loose to work on your worksheet.
  7. Worksheet for 3.1
Links:
- A Chain Rule tutorial
- Our free, on-line textbook, called Active Calculus.

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