Finding the intersection envelope

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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, $y_ t(x),$ is 0 at $t,$ and $1-t$ at 0.

  \[ y_ t(t)=0 \]    
  \[ y_ t(0)=1-t. \]    

The general equation for such a line is

  \[ y_ t(x)=(1-t)-x\frac{1-t}{t}. \]    

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 $t+\Delta t.$

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, $\Delta t,$ 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

  \[ y_ t(x)=y_{t+\Delta t}(x). \]    

This gives

  $\displaystyle (1-t)-x\frac{1-t}{t} $ $\displaystyle = $ $\displaystyle (1-t-\Delta t)-x\frac{1-t-\Delta t}{t+\Delta t} $    
  $\displaystyle -x\frac{1-t}{t} $ $\displaystyle = $ $\displaystyle -\Delta t -x\frac{1-t-\Delta t}{t+\Delta t} $    
  $\displaystyle -x(1-t)(t+\Delta t) $ $\displaystyle = $ $\displaystyle -x \cdot t(1-t-\Delta t)-\Delta t \cdot t(t+\Delta t) $    
  $\displaystyle x $ $\displaystyle = $ $\displaystyle t^2 + t\cdot \Delta t $    

As $\Delta t \rightarrow 0$ we get

  \[ x=t^2. \]    

Putting this in $y_ t(x)$ we get

  $\displaystyle y_ t(t^2) $ $\displaystyle = $ $\displaystyle (1-t)-t^2\frac{(1-t)}{t} $    
  $\displaystyle y_ t(t^2) $ $\displaystyle = $ $\displaystyle (1-t)^2. $    

So we now know that our curve is defined as all the points that satisfy

  \[ (x,y) = (t^2, (1-t)^2) \]    

for all $t$ in [0,1]. So the curve we are after is

  \[ y(x) = (1-\sqrt{x})^2. \]    

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