The largest area: Part I

This adventure started when I attempted a harmless looking problem from the Facebook group “Actually Good Maths Problems”. The answer, which any motivated high school calculus student could find, is a sort of joke or prank on the solver, and I enjoyed it enough that I sent the problem to my good friend David Galea.

I thought that would be the end of it.

But then, as mathematicians often do, we started asking each other questions. This lead us down a rabbit hole as the problem morphed into an intriguing riddle with surprising results, which we still haven’t fully understood.

Let’s get into it.

Draw two parabolas, $y=x^2$ and the same parabola shifted up one unit, $y=x^2 + 1$ . Construct a tangent line somewhere on the upper parabola at a general point $x=t$ , and consider the area bounded below by the tangent, and above by the bottom parabola $y=x^2$ .

The problem is to find the tangent point that maximises this area, and to find that maximal area.

Noting that the derivative for the curve $y=x^2 +1,$ evaluated at the tangent point is $2t$ , and that the tangent goes through the point $(t, t^2 + 1)$ , the formula for the tangent line is given by

$y - (t^2 + 1) = 2t(x-t),$

or, simplifying,

$y = 2tx-t^2+1.$

The area between this tangent and the lower parabola is now a function of this parameter $t$ (subtract the bottom curve from the tangent line and integrate):

$A(t) = \displaystyle \int_{a(t)}^{b(t)} 2tx - t^2 + 1 - x^2 \ \mathrm{d}x,$

where $a(t)$ and $b(t)$ are the left and right intersection points respectively, i.e. the two solutions of

$x^2 = 2tx - t^2 + 1.$

This is a quadratic, with the solutions $x = t \pm 1$ , so $a(t) = t - 1$ and $b(t) = t + 1$ , since the left intersection point has to be smaller than the right.

Now, we have an expression for $A(t)$ , the area we are trying to maximise. So we have to take a derivative with respect to $t$ . We could try to evaluate the integral first, but it’s easier and more interesting to use Liebniz’s rule for differentiating integrals (we need this, since the limits themselves depend on $t$ ). If we let the integrand be $f(x,t) = 2tx - t^2 + 1 - x^2$ , then

$\begin{aligned}\displaystyle \frac{\mathrm{d}A}{\mathrm{d}t} &= \displaystyle \int_{t-1}^{t+1} \frac{\partial}{\partial t} f(x,t) \ \mathrm{d}x + f(t+1,t) \frac{\mathrm{d}}{\mathrm{d}t}(t+1) - f(t-1,t) \frac{\mathrm{d}}{\mathrm{d}t}(t-1), \\ &= \displaystyle \int_{t-1}^{t+1} 2t - 2x \ \mathrm{d}x + f(t+1,t) - f(t-1,t), \end{aligned}$

It’s not too hard to show that $f(t+1,t) - f(t-1,t) = 0$ , and the integral also evaluates to $0$ .

Thus we have the fact that $A'(t) = 0$ and therefore, the area under the curve is a constant, independent of $t$ !

If we let $t = 0$ , then we can see that the area is

$A = \displaystyle \int_{-1}^1 1 - x^2 \ \mathrm{d}x = \frac{4}{3}.$

So we have our answer. The area is the same no matter where we place the tangent, and it’s $4/3$ .

As I said, a mildly amusing result. Don’t believe it? Try it here, drag the tangent point and see the area remain the same.

The next chapter in the story was David and I wondering whether the area was also constant when we generalise the problem to higher dimensions.

What we found was extremely surprising, I hope you’ll join us to explore this problem further in the next post.

Stay curious!

One response to “The largest area: Part I”

A problem of paraboloids – The Calculus of Explanations

August 15, 2024 at 5:32 pm

[…] we will consider the 3D analogue of the problem mentioned in the previous post, and first define the two paraboloids (the 3D analogue of […]

LikeLike

The Calculus of Explanations