| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 915.1 | Messy answer | SSDEVO::LARY | One more thin gypsy thief | Mon Aug 08 1988 20:10 | 34 |
| If the ladder hits the floor at (x,0) and the wall at (0,y) then:
2 2
x + y = 10
xy = x + y (or, "the line between (x,0) and (0,y) passes through (1,1)")
we are looking for arctan(y/x) = arctan(y-1) from the second equation
Now, adding twice the second equation to the first:
2 2
x + y + 2xy = 10 + 2(x+y)
2
(x+y) - 2(x+y) - 10 = 0; solving the quadratic in x+y we get:
x+y = 1 + sqrt(11) (keeping it in the 1st quadrant)
Similiarly, subtracting twice the second equation from the first:
2
(x-y) = 10 - 2(x+y) = 8 - sqrt(11), x - y = (+ or -) sqrt(8 - sqrt(11))
There are two solutions, depending on whether we use the positive or negative
square root, but the two solutions merely interchange x and y and subtract the
angle from 90 degrees. Taking the positive square root,
y = (1 + sqrt(11) - sqrt(8 - sqrt(11)))/2
the angle is arctan((sqrt(11) - sqrt(8-sqrt(11)) - 1)/2)
which is closed form, but not very clean...
|
| 915.2 | | LISP::DERAMO | Daniel V. {AITG,LISP,ZFC}:: D'Eramo | Mon Aug 08 1988 23:31 | 17 |
| If the ladder hits the floor at (x,0) and the wall at
(0,y) then
2 2 2
x + y = 10 (not 10, as in .-1)
Now, "the line between (x,0) and (0,y) passes through
(1,1)" translates to
x - 0 x - 1
----- = ----- or x = y(x - 1) or x + y = xy
0 - y 0 - 1
as in .-1 is correct. But combining these two is still
a mess. :-)
Dan
|
| 915.3 | | ISTG::GOKHMAN | Boris the Bear | Tue Aug 09 1988 15:05 | 8 |
| Combining the two is just as simple as in .1 and yields
x+y=xy=Sqrt(101)+1
Substitution of y=Sqrt(101)+1-x into the second eqaution gives an
easy answer.
boris
|
| 915.4 | blush, stammer | SSDEVO::LARY | One more thin gypsy thief | Tue Aug 09 1988 16:27 | 6 |
| > 2 2 2
> x + y = 10 (not 10, as in .-1)
hmn, er, ah, a-HMNN, well, yes, as a matter of fact, it does seem so, er, ah...
Richie
|
| 915.5 | | LISP::DERAMO | Daniel V. {AITG,LISP,ZFC}:: D'Eramo | Tue Aug 09 1988 18:57 | 14 |
| re .3
>> Substitution of y=Sqrt(101)+1-x into the second eqaution gives an
>> easy answer.
So what was the angle? :-)
When I said it was messy I didn't mean it wasn't closed
form, but that it was an inverse trig function of something
with one or two square roots. I couldn't see right off
how it would simplify, to something like a rational number
time pi radians.
Dan
|
| 915.6 | | ELWOOD::CHINNASWAMY | | Fri Aug 12 1988 15:53 | 7 |
| reply
If we use trig, the answer comes out to be
theta = (1/2)inverse sine((1+sqrt(101))/50)
|