Day 24 in MAT229

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Today:

Announcements:
- Homework 7.7 is returned
  - 2 -- LRR is an overestimate, RRR is an underestimate; Trap is the average of those (but we can say more, right Mark?), and Mid is the other. How do we know what we know?
  - 28 -- I wrote Mathematica Routines (based on one I stole!) to do these. That's easier than doing them all by hand.
    What's the true value of the integral? Compare your answers to that, first of all, to see if you're in the ballpark.
    How does the error change as the number of subintervals doubles?
  - 30 -- For this one, I used the 1,4,2,4,2-step pattern.
    First of all, estimate the value of the area.
    Don't forget units (meter squared), and don't forget the endpoint values -- 0 at both ends.
- Please read section 10.2 for next time.
- A note about Friday's help session: Excel spreadsheets and approximate integration
Section 8.1: Arc Length
- What kinds of problems are we looking at in this section? Calculating the linear length of a curve (arc length).
- Let's consider arc length:
  
  ${L=\int{dL}\approx\sum_i\Delta{L_i}}$
  The usual trick is to discretize something continuous, and then pass to a limit:
  
  Now hopefully we can figure out how the formula will look:
  
  So what's the general formula?
- Examples:
  - #2, p. 567
  - #3, p. 567
  - #37, p. 568

Section 10.1: Parametric Equations

And now for something completely different....

A good introduction to parametric curves is given by ballistics. If we shoot a bullet into the air with speed v horizontally, and we neglect all forces but gravity, then the bullet will trace out a parabola (some bullets are larger than others):

Now how might we characterize the path of the bullet? The answer is a parametric curve, of the form C(t)=(x(t), y(t)).

${x(t)=vt}$		${y(t)=h-\frac{1}{2}gt^2}$
or	${c(t)=(vt, h-{\frac{1}{2}gt^2})}$
If we wish, we can solve for t in the equation for x and use that to eliminate the parameter t from the equation for y, hence getting an equation for the parabola traced out:
	${y=h-\frac{1}{2}g(\frac{x}{v})^2}$

According to this parameterization, where is the "bullet" at time t=0? In which direction is the motion occuring -- left to right, or right to left? (You have a parametric graphing mode on your calculator -- let's try it out.)

Orbits of planets in the heavens, movements of ants on a hill, a robotic arm in an assembly plant: all these can be described by parametric curves.

A couple of famous examples involving the cycloid:
- The Brachistochrone Problem (1696, solved by Newton in a single day after he heard the challenge)
- The Tautochrone Problem (1673)
Parametrizing a line:
Parametrizing a circle:
There are two distinctively different kinds of derivatives that we're going to have to consider here:
- There are the time-derivatives, which give us the actual speeds of the "particles" as they zip along in time. For the bullet, the speed will be
  
  ${s(t)=\sqrt{x'(t)^2+y'(t)^2}=\sqrt{v^2+(gt)^2}}$
- There is the slope of the tangent line to the curve traced out by the particle, which one can obtain using the time-derivatives:
  
  ${y'(x)=\frac{y'(t)}{x'(t)}=\frac{-gt}{v}=\frac{-gx}{v^2}}$
One of the most important applications of parametric curves is in the Bezier curve section. There are lots of important instances where we need to be able to design custom curves, and this is one of the most common ways of creating them. For example, you can use them to create your own "sigmac", using control points. (I had all my students create their own in numerical analysis.)
Examples:
- #1, p. 665
- #3
- #7
- #15

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