Math Challenge, Experts Only?

Question

The parabola y = x² is rolled on the x-axis, so that the curve tilts as it rolls. After some rolling, there will be a gap between the vertical line x = 0 and the parabola, because the vertical line x = 0 no longer intersects the parabola. How large can this gap get?

Simpler problem: Prove that this gap will appear.

torpedo, imagine that the parabola is solid above the curve, so it can be rolled on the "floor" of the x-axis. If this was a circle instead, you'd have no problem visualizing this.

suesysgoddess, you have the right idea, it involves the arc length of the parabola, as it maintains contact with the x-axis. Yes, obviously, the x-axis is tangent to the rolling parabola at all times.

suesysgoddess, very interesting link with informative graphs, which shows that the focus of a rolling parabola traces out Cosh(x). And that should have given you a clue of what to expect. The gap should be roughly half width of Cosh(x) minus 1/4, but the half-width of Cosh(x) increases indefinitely, as Quadrillerator pointed out. The graphs of your link shows this trend clearly.

Quadrillerator, you made a slight mistake at the beginning, so that your equation for the gap was given as:

ln|2z + √(1 + 4z²)|)/4 - z / (2√(1 + 4z²))

when it really should have read:

ln|2z + √(1 + 4z²)|)/4 - (√(1 + 4z²))/8z

Oddly enough, the limiting case for the 2nd term is 1/4 in both cases. The 1st term diverges extremely slowly, so that after the parabola has rolled a distance of 10^12, the gap has increased to only 3.55. Another expression for this 1st term is

(ArcSinh(2x))/4

which corroborates with the "appearance of hyperbolic functions" mentioned in suesysgoddess' link about rolling parabolas.

Answer 1

Consider a point Z on the parabola (z, z²).
The distance from O = (0,0) to Z along the parabola is given by
I(z) = ∫[from 0 to z of] dx √(1+(2x)²)

Now we consider the line tangent to the parabola at Z. We are looking for the distance along this line to a line perpendicular to it that is also tangent to the parabola. The slope of that line is given by -1/(2z) so that the point of tangency is given by 2x = -1/(2z) or Z' = (-1/(2z), 1/(4z²)). So the equation of this second tangent line is (y-1/(4z²)) = - (x + 1/(2z)) / (2z) or y = -x/(2z). Let's find Q, the intersection of this line with the first line of (y-z²) = 2z(x-z) or y = 2zx - z²
-x /(2z) = 2xz - z²
x(4z² + 1) = 2z³
x = 2z³ / (4z² + 1) and
y = -z² / (4z² + 1)

Now the distance d from Q to P is given by:
d² = (z - 2z³ / (4z² + 1))² + (z² + z² / (4z² + 1))²
= (z(2z² + 1) / (4z² + 1))² + (2z²(2z² + 1) / (4z² + 1))²
= (1+4z²) * (z(2z² + 1) / (4z² + 1))²
So
d(z) = √(1+4z²) * z(2z² + 1) / (4z² + 1)
= z(2z² + 1) / √(4z² + 1)

We want to find the critical points of
d(z) - I(z) so we take a derivative and set it equal to 0.
d(d(z) - I(z))/dz =
d'(z) - √(1 + 4z²)
and we use logarithmic differentiation
=d(z) * [1/z + 4z / (2z² + 1) - 4z / (4z² + 1) ] - √(1 + 4z²) = 0
So
z(2z² + 1) / √(4z² + 1) * [(6z²+1)/(z(2z² + 1)) - 4z / (4z² + 1) ] = √(1 + 4z²)
Now cross multiply to clear the square root:
z(2z² + 1) * [(6z²+1)/(z(2z² + 1)) - 4z / (4z² + 1) ] = (1 + 4z²)
z(2z² + 1) * [6z²/(z(2z² + 1)) - 4z / (4z² + 1) ] = 4z²
z(2z² + 1) * [2z²/(z(2z² + 1)) - 4z / (4z² + 1) ] = 0
So (hurrah for logarithmic differentiation)
2z²/(z(2z² + 1)) = 4z / (4z² + 1)
1/(2z² + 1) = 2 / (4z² + 1)
4z² + 1 = 4z² + 2
Hmmm, this isn't working for me:

Looking at the actual integral, I get:
I(z) = (2z√(1 + 4z²) + ln|2z + √(1 + 4z²)|)/4
while
d(z) = z√(1 + 4z²)/2 + z / (2√(1 + 4z²))
So we should be
looking at
I(z) - d(z)
= ln|2z + √(1 + 4z²)|)/4 - z / (2√(1 + 4z²))

At z=0 this difference is 0. However, d(I(z)-d(z))/dz
= (2+4z/√(1 + 4z²))/(2z + √(1 + 4z²)) - (2z/√(1 + 4z²)) * [1/x - 4z/(1 + 4z²)]
= 2/√(1 + 4z²) - 2/√(1 + 4z²) + 8z² / (1 + 4z²)^1.5
= 8z² / (1 + 4z²)^1.5
which is always positive.
Furthermore, since the first term (the ln) in I(z) - d(z) is unbounded while the second goes to -1/4:

The size of the gap grows monotonically and becomes arbitrarily large.

Answer 2

Okay, I see it. Since the graph of the function is increasing and the slope is also increasing at a constant rate, we should be able to 'roll' this graph forever but at a decreasing rate.. okay.. well.. it makes sense to me at least.

I found this wonderful link that illustrates your problem, but of course doesn't answer it.

I'll keep working on this. Very intriguing question.

Well, here is my hypothesis. I believe that the limit for this roll is when the directrix of the parabola is vertical. That is, it can continue rolling (the amount of rotation is decreasing as it continues to 'roll') infinitely.. and at the same time the directrix of the parabola is becoming increasingly vertical... but it can never become vertical.

So, my answer is the gap has a limit of 1/4 for the graph of y = x^2.

Answer 3

what do you mean when you say rolled,,,, is it spun about the x axis,,,???