Day 10, Mat128: Calculus A

Last Time

Next Time

Today:

Announcements
- We're moving! We're going to be in MEP314 from now on. So let's go there, once we're all here.
- Your first exam is returned today. It went okay for most, but not all:
  A Key for the exam is available. Important points:
  1. Personally my biggest disappointment was that I don't believe that anyone mentioned the tangent line in the second part of page 1. Correct me if I'm wrong!
  2. We're still really struggling with "the most important definition in calculus". This from Day 04:
    The most important definition in calculus is the definition of the derivative (here is the derivative of $f$ at $a$):
    
    $f'(a)=\lim_{h\to 0}{\frac{f(a+h)-f(a)}{h}}$
Today:
- We'll review some parts of the exam, featuring some highlights from your papers.
- We'll review your worksheet from last time.
- We'll start talking about the second derivative (section 1.6).
  
  If a function's derivative is another function, does that function have a derivative?
  The derivative of the derivative of a function is called the second derivative of that function. And how do we interpret these "higher derivatives" in the context of a motion?
- If the derivative is good at showing where a function is increasing or decreasing (see the formal definition in section 1.6), the second derivative is important for showing the inflection of a function (its concavity).
- Let's think of a quadratic motion, e.g. the motion of an eraser thrown across the room. Let s be the height of the eraser:
  \[ s(t)=\frac{1}{2}gt^2+v_0t+h_0 \]
  Each of the coefficients has an important, intuitive role to play:
  - $h_0$ is the height at time $t=0$;
  - $v_0$ is the rate of change in height at time $t=0$;
  - $h$ is the acceleration (due to gravity) on the eraser.
  So \[ s'(t)=gt+v_0 \] and \[ s''(t)=g \] Of course we could continue on and define higher derivatives (third, fourth, etc.). The notation for the $n^{th}$ derivative is frequently $f^{(n)}(x)$: \[ s'''(t)\equiv s^{(3)}(t)=0 \]
  (The third derivative is called the "jerk", amusingly enough! There's no jerk in the flight of the eraser. You might feel some jerk if you're on a roller coaster....)
  We can continue, of course, and so we see that
  $s^{(n)}(t)=0$ for all $n>2$.
- Notations: Here are other notations for the derivative function and second derivative functions, if we denote $y=f(x)$:
  \[ f^\prime(x) = \frac{dy}{dx} = \frac{d}{dx}\left(f(x)\right)=\left(f(x)\right)'=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \] \[ f^{\prime\prime}(x) = f^{(2)}(x)=\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{d}{dx}\left( f(x) \right) \right) \]
  Several of these forms remind us that differentiation is itself a function (we call it an "operator", since it works on functions): it takes a function in its domain and returns another function in its range -- the derivative function.
  Notice also that the second derivative can be thought of as the derivative of the derivative: \[ f''(x)=\lim_{h\to 0}{\frac{f'(x+h)-f'(x)}{h}} \] Hence, all your table work (forward, backward, centered difference formulas) work.
- The two most important examples, to illustrate the role of the second derivative, are these poster children for concavity:
  - $f(x)=x^2$
  - $f(x)=-x^2$
- Let's start in on your Section 1.6 worksheet, with problem 4: $f(x)=x^3$.
  This is our poster child for inflection.

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