Finding the intersection envelope


We want to find the formula for the intersection between any two adjacent lines in our bridge.

Lines

Let's look at a line that intersects the x axis at a certain point t. Its function, yt(x), is 0 at t, and 1t at 0. yt(t)=0 yt(0)=1t. The general equation for such a line is yt(x)=(1t)x1tt. Now let's look at a line which intersects the x axis at some point t and the line just after it, which intersects the x axis at a point we call t+Δt. The curve we are after is the one we'd get if there were infinitely many chords. In other words, as the spacing of our chords on the x axis, Δt, approaches 0, the point in which our two lines intersect approaches a point on our envelope. Let's first find the intersection point of our two lines by setting yt(x)=yt+Δt(x). This gives (1t)x1tt=(1tΔt)x1tΔtt+Δtx1tt=Δtx1tΔtt+Δtx(1t)(t+Δt)=xt(1tΔt)Δtt(t+Δt)x=t2+tΔt As Δt0 we get x=t2. Putting this in yt(x) we get yt(t2)=(1t)t2(1t)tyt(t2)=(1t)2. So we now know that our curve is defined as all the points that satisfy (x,y)=(t2,(1t)2) for all t in [0,1]. So the curve we are after is y(x)=(1x)2.

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