Day 43, Mat128: Calculus A

Today:

• Announcements
  • Any questions left over from the exam? I hope that you've had a chance to look over key.
  • Your homework due from 3.4 is returned: graded p. 235, #9, 18, 23
    • #9 -- equal degrees, --> just look at coefficients of leading terms
    • #18
      • square roots may require absolute values
      • I picked up on another very serious algebraic problem on this problem: Some of you wrote $\sqrt{9x^6-x}=3x^3-\sqrt{x}$. That's the same disasterous reasoning as $(a+b)^2 \ne a^2+b^2$.

        This bad algebra must be driven from your mind. It means that you'll always be wrong, and give bogus answers.

    • #23 --> higher degree in numerator means infinite limit, but slant asymptote.
    • #29 and 19 -- bonus!
      • Conjugates on 19, and
      • a special limit on #29.

    General notes:
    • Don't be a limit dropper.
    • Limits are operators: hungry functions. They require something to operate on. Same for derivatives. Make sure you give them something to satisfy their appetites....

  • You have a new course that has appeared in Blackboard -- it's a summary "quiz" of calculus. If you'll take that, I'll give you an additional GOHF card as well as an additional GOQF card.

    Your score will not have any numerical impact on your grade -- it's just diagnostic. Simply do your best.

  • Your final is next week, Wednesday, at 1:00. Your final exam grade will replace your lowest hourly exam, provided the final is better.

  • A reminder that we'll have a review here Tuesday night, 7-9pm.

  • We'll have a review on Friday. Come with any questions. I'll take us on a swing over the content of the course, then open things up to questions.

• Section 3.9 -- anti-derivatives -- is about the art of "thinking backwards"
  • Whereas differentiation is essentially a mechanical exercise, anti-differentiation may require insight, and broad exposure to lots of different functions. One asks oneself:
    "Do I know a function whose derivative looks like this?"

  • Definition:
    

  • All anti-derivatives are the same up to a constant:
    

  • Differential Equations:
    • A differential equation is an equation linking a function and its derivative(s).

    • The simplest differential equations look like this:
      $\frac{dy}{dx}=g(x)$

    • Usually an initial condition (or conditions) will also be provided -- so that the solution ($y(x)$) has value $y=y_0$ at $x=x_0$.

    • Our job then is to find the function $y(x)$ which satisfies both
      • the differential equation, and
      • the initial condition(s).

    • Examples:
      • A simple example:
        1. $\frac{dy}{dx}=2x$
        2. $y(0)=1$
        Name that function! Answering the first question gets you the class of the solution; answering the second gets you the particular member of the class.

      • A spring satisfies Hooke's law:
        $F=-kx$

        where $k$ is the spring constant (characteristic of the particular spring). Newton said that

        $F=ma$

        If we put those two laws together, we get

        $\frac{d^2x}{dt^2}=\frac{-k}{m}x$

        For initial conditions, we should give initial position and initial velocity.

      • Gravity is essentially constant at the Earth's surface: we can use a differential equation to deduce the formula for the height h of an object as a function of time.

        The acceleration due to gravity at the earth's surface is often taken as approximately

        $g=-9.81m/s^2$

        where the minus sign indicates that gravity's inclination is to make your height decrease (to 0 at the center of the earth, if gravity could).

        Your force on the earth (or the earth's force on you!) is your weight, $m g$.

        Since

        $F=ma$

        If height is given by $h(t)$, then

        $m\frac{d^2h}{dt^2}=mg$

        or

        $\frac{d^2h}{dt^2}=-9.81$

        Once again, we should give initial position and initial velocity.

        Now you ask yourself this: do I know a function whose second derivative is a constant? And the answer is YES! -- a quadratic function. The specific quadratic function is

        $h(t)=\frac{-9.81}{2}t^2+v_0t+h_0$

        (Check and make sure!). Then, once you've specified $h_0$ and $v_0$, the motion of the projectile is fully determined.

  • This process turns out to be crucial for deciding what $f$ looks like, given $f'$ (or higher derivatives). Remember when we were looking at a graph of $f'$, and trying to guess what $f$ looks like? Well, we're not going to need to do that, once we know the anti-derivative.

  • Some examples:
    1. #3, p. 273
    2. 20
    3. 33
    4. 41

• Now let's look at a problem (or two) that illustrates why calculus is so important and useful:
  • #9, p. 257
  • #16