| T.R | Title | User | Personal Name
 | Date | Lines | 
|---|
| 1152.1 |  | KOBAL::GILBERT | Ownership Obligates | Tue Nov 21 1989 16:58 | 13 | 
|  |     I assume that what you really want has all the 'drag' at B, not along
    the length of the string.  That being so, the differential equation
    for the curve is:
    
    	dy      - y
    	-- = ----------
    	dx   sqrt(1-y�)
    
    There's gotta be some better way.
    
    FWIW:  A hanging chain forms a catenary.  If you cut the catenary at
    its lowest point, the cut end will trace out the above tractrix
    (from "Mathematical Snapshots").
 | 
| 1152.2 | Cat & Mouse | DEC25::ROBERTS | Reason, Purpose, Self-esteem | Wed Nov 22 1989 11:20 | 17 | 
|  |     RE: 1152.1
    "I assume that what you really want has all the 'drag' at B, not along
    the length of the string."
    Quite right.
    Another way to express the problem is:
    Mickey Mouse leaves his house and takes a walk due east. Sylvester the
    cat, initially one unit north of Mickey, stalks him, remaining exactly
    1 unit distant. At any instant, Sylvester is always moving directly
    toward Mickey. Where is Sylvester when Mickey is t units away from
    home?
    /Dwayne
 | 
| 1152.3 | Looked it up | AKQJ10::YARBROUGH | I prefer Pi | Wed Nov 22 1989 14:58 | 10 | 
|  | The equation of a tractrix symmetric about the y-axis and asymptotic to the 
x-axis is
	 2                         2  2  2
	x  = (a arccosh(a/y)-sqrt(a -y ))
where a>0 is the height of the curve above the x-axis at x=0.
(From Bronshtein & Semendyayev, 'Handbook of Mathematis', p.85). 
Yeah, it's messy.
Lynn 
 | 
| 1152.4 |  | DEC25::ROBERTS | Reason, Purpose, Self-esteem | Wed Nov 22 1989 16:33 | 4 | 
|  |     Is anyone able to derive it?
    
    /Dwayne
    
 | 
| 1152.5 | Derive by showing it works? | VMSDEV::HALLYB | The Smart Money was on Goliath | Wed Nov 22 1989 16:34 | 1 | 
|  |     Should you be able to differentiate .3 and get .1 ?
 | 
| 1152.6 |  | AITG::DERAMO | Don't stop short of the peak! | Wed Nov 22 1989 17:02 | 6 | 
|  | >>    Should you be able to differentiate .3 and get .1 ?
	I believe so, with a = 1.  If not, one of them is
	probably wrong.
	Dan
 | 
| 1152.7 |  | KOBAL::GILBERT | Ownership Obligates | Wed Nov 22 1989 17:09 | 2 | 
|  |     Yes, that fits the equation in .1 nicely.  Note that it's most easily
    done with x treated as a function of y.
 |