Day 9, Mat128: Calculus A

Last Time

Next Time

Today:

Announcements
- You have a new IMath assignment (more limit practice).
- We'll have a quiz over derivative functions today.
Last time:
- we continued discussing derivative functions, and, in particular, different ways of estimating (or calculating them): forward, backward, and centered differences.
- We started section 1.5, and the corresponding worksheet.
Today:
- We'll start by trying to wrap up that 1.5 worksheet.
- Then we'll move on to Section 1.6: the Second Derivative
  - We move beyond the first derivative $f'(x)$ (the derivative function to a function $f(x)$), and on to the second derivative of a function.
    If a function's derivative is a function, does that function have a derivative?
    The derivative of the derivative of a function $f$ is called the second derivative of $f$, and frequently denoted $f''$. 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 $t=0$;
    - $g$ is the acceleration (due to gravity) on the eraser.
    One can show that \[ s'(t)=gt+v_0 \] and that linear function has slope $g$, so \[ s''(t)=g \] Of course we could continue on and define higher derivatives (third, fourth, etc.). The notation for the $n^{th}$ derivative is frequently $f^{(n)}(x)$, where the parenthese around the "exponent" mean we're to understand that as differentiation: \[ s'''(t)\equiv s^{(3)}(t)=0 \]
    (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 $y=f(x)$:
    \[ f^\prime(x) = \frac{dy}{dx} = \frac{d}{dx}\left(f(x)\right)=\left(f(x)\right)'=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \] \[ f^{\prime\prime}(x) = f^{(2)}(x)=\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{d}{dx}\left( f(x) \right) \right) =\lim_{h \to 0} \frac{f'(x+h)-f'(x)}{h} \]
    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) still applies.
  - The two most important examples, to illustrate the role of the second derivative, are these poster children for concavity:
    - $f(x)=x^2$: $f'(x)=2x$, and $f''(x)=2$: $f''(x)>0 \longrightarrow$ concave up:
    - $f(x)=-x^2$: $f'(x)=-2x$, and $f''(x)=-2$: $f''(x)<0 \longrightarrow$ concave down:
    Activity 1.6.2 allows us to explore your preview exercise more deeply.
  - Let's start in on your Section 1.6 worksheet, with problem 4: $f(x)=x^3$.
    This is our poster child for inflection:
- Then we'll have a quiz over derivative functions.

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