Day 58 in MAT229

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Announcements:
- Return of your homework:
  - #2
  - #28
  - #38
- It is time to evaluate the course. I'm sure that you've received emails to that effect....

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 design (particularly Bezier curves). There are lots of important instances where we need to be able to design custom curves, and Bezier curves 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
- Deriving the equations of the cycloid (p. 663)

Section 10.2: Calculus with Parametric Equations
- Arc Length and Speed
  - I already indicated above what the speed should be -- so I'm not going to spend much time on that.
    How can we use this formula, however? Suppose a particle has followed the parametric curve ${C(t)=(x(t),y(t))}$ : then we can compute how far the particle has travelled during the interval ${a\le{t}\le{b}}$ easily using the dirt formula, ${d=rt}$ (in its modified form ${\Delta{d}=r\Delta{t}}$ ).
    In this case, the rate is just the speed. So we compute the integral
    
    ${D\approx\sum_{i}^{}\Delta{d_i}=\sum_{i}^{}\sqrt{x'(t_i)^2+y'(t_i)^2}\Delta{t_i}}$
    
    or, in the limit,
    
    ${D=\int_{a}^{b}\sqrt{x'(t)^2+y'(t)^2}dt}$
    This is actually just a re-expression of the arc length formula:
    
    ${D=\int_{x_0}^{x_1}\sqrt{1+y'(x)^2}dx=\int_{x_0}^{x_1}\sqrt{1+\left(\frac{y'(t)}{x'(t)}\right)^2}dx}$ and then we factor out the x'(t), to get ${D=\int_{a}^{b}\sqrt{x'(t)^2+y'(t)^2}dt}$ where ${x(a)=x_0}$ and ${x(b)=x_1}$ .
    But arc length may be different from the distance the particle travelled: a particle can revisit many sections of the curve y(x) -- so once again we need to be careful to distinguish between the independent variable of interest (whether x or t).
  - Other calculations can be generalized. For example, the formula for surface area:
  - Examples:

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