For all t in the interval 0≤t≤1, a thread is pinned from the point (0, t) on the y-axis to the corresponding point (1-t, 0) on the x-axis.
Under what curve do the threads lie?
y = x*t/(t-1) + t Eqn 1
What we want is the locus of points where the above line at value t intersects with the same line valued at t+dt.
y = x*(t+dt)/((t+dt)-1) + (t+dt)
x*t/(t-1) + t = x*(t+dt)/((t+dt)-1) + (t+dt)
x*t/(t-1) = x*(t+dt)/((t+dt)-1) + dt
multiply by (t-1)((t+dt)-1)
x*t*(t+dt-1) = x*(t+dt)*(t-1) + dt*(t-1)(t+dt-1)
xt^2 + xt*dt - xt = x*(t^2-t+t*dt-dt) + dt*(t^2+t*dt-t-t-dt+1)
any term of dt^2 is assumed to be 0
-xt = x*(-t-dt) + (t^2*dt+t*(dt^2)-2t*dt-(dt^2)+dt)
0 = -x*dt + (t^2*dt+0-2t*dt-0+dt)
x*dt = (t^2*dt-2t*dt+dt)
x = t^2-2t+1
x = (t-1)^2 Eqn 2
y = x*t/(t-1) + t substitute for x using Eqn 2
y = ((t-1)^2)*t/(t-1) + t
y = (t-1)*t + t
y = t^2 Eqn 3
Parametric solution:
{x = (t-1)^2 , y = t^2}
https://www.desmos.com/calculator/u2wq5dvhfy
Eliminate the parameter:
x = (t-1)^2 = t^2 - 2t + 1
x = y - 2√y + 1
2√y = y - x + 1 square both sides
4y = y^2 + x^2 + 1 + 2(y-x-xy)
4y = y^2 + x^2 + 1 + 2y-2x-2xy
y^2 + x^2 - 2xy - 2y - 2x + 1 = 0 is the desired equation.
(x-1)^2 + (y-1)^2 - 2xy = 1 (alternate form)
Which I think could be either a hyperbola or a parabola.
I think a 45 degree counterclockwise rotation transformation would probably answer this question.
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Posted by Larry
on 2025-01-03 11:45:36 |
Solution
