*Aidan Strong*

What curve has the property that a bead placed anywhere on it will slide to the bottom in the same time, independent of where it’s placed? This question is called the **Tautochrone problem**, and it was first solved by Christiaan Huygens in 1659.

Huygens showed the solution was a special type of curve called a **cycloid**. This is the curve that a fixed point on the edge of a rolling circle traces out.

So how on earth did Huygens manage to solve the problem? His proof was published in his book Horologium Oscillatorium in 1673, and it’s done geometrically:

He showed that, if the cycloid is traced out by a circle with radius r, then the time taken for the bead to slide down it from any starting position is given by the equation

Where ‘g’ is the acceleration due to gravity.

We’ll come back to the Tautochrone in a bit, but let’s first look at another problem of a similar flavour, called the **Brachistochrone problem**. This problem was posed by Johann Bernoulli in the journal Acta Eruditorum,* *in 1696. This problem asks ‘which curve between any two points will a bead slide down in the shortest time possible’.

This problem attracted solutions from many of the ‘superstars’ of maths, including Leibniz, L’hopital, Jacob Bernoulli (Johann’s brother), and Newton. In fact, Bernoulli sent the challenge directly to Newton, and after receiving it Newton stayed up all night to solve it, sending his solution back to Bernoulli anonymously in time for the next post! Immediately after reading his solution, Bernoulli said that he “recognises a lion from his claw mark”, and knew that it must’ve been Newton (or maybe it just had a return address, who knows?).

Anyway, here is the solution which Newton sent to Bernoulli:

Newton’s solution states that the shortest path is (*drumroll*) a cycloid! Newton didn’t initially provide any proof, however after another mathematician David Gregory asked him for an explanation, Newton was kind enough to share his method. So how did he do it? Well, its rather complicated, and relied on the following diagram:

This solution however is surprising for a number of reasons! It shows that the quickest path can include an uphill portion towards the end in which the bead would be slowing down – intuitively, this is because the cycloid is the best compromise between building up enough speed initially, maintaining this speed across the curve, and minimising the length of the curve. What’s even more surprising however, is that it’s the exact same solution as the Tautochrone problem! Bernoulli thought so too, writing at the end of his own solution of the Brachistochrone problem:

*“Before I end I must voice once more the admiration I feel for the unexpected identity of Huygens’ tautochrone and my brachistochrone. I consider it especially remarkable that this coincidence can take place only under the hypothesis of Galileo, so that we even obtain from this a proof of its correctness. Nature always tends to act in the simplest way, and so it here lets one curve serve two different functions, while under any other hypothesis we should need two curves.”*

(The ‘hypothesis of Galileo’ here refers to the fact that a velocity of a falling object varies only with distance fallen, and is independent of mass.)

The brachistochrone and tautochrone problems are far more than just a beautiful coincidence of mathematics, however – they are the first examples of what would become a new area of mathematics called **calculus of variations**, pioneered by mathematicians Euler and Lagrange.

To understand this a little more, let’s consider how long a bead would take to slide along any curve from point A to point B.

Now in school, we all learnt the formula:

So we just need to know the speed and the distance. To find the speed v, we’ll use the principle of conservation of energy on the bead, which states that the kinetic energy of the bead is equal to the gravitational potential energy:

Now to find the distance, we’ll imagine zooming into the curve really closely, so the curve looks like it’s a straight line contained in a rectangle. Then, we can apply Pythagoras’ theorem ‘a^{2} + b^{2} = c^{2}’:

This means that the time the bead takes to travel across the really small section of the curve is ds/v, and so the total time to travel across the curve is given by:

(If you have not met calculus or integration, don’t panic! The ‘squiggly S’ symbol simply means we are summing over all of the little ‘ds’ pieces of the curve.)

So to get back to the main point of what on earth we’re trying to do here – this formula gives the total time for a bead to travel along any curve starting at A and ending at B. So to solve the Brachistochrone problem, we need to find a curve which minimises this equation.

This is where the aforementioned Calculus of variations comes into play! In particular, to find the curve that minimises the time, we can solve the Euler-Lagrange equation, which is a special differential equation discovered in the 1750s, when Euler and Lagrange were studying the Tautochrone! Plugging the above formulas into the Euler-Lagrange equation eventually gives the correct solution of the cycloid.

If you’ve not had enough cycloid action for today, do not fear – click here for a fantastic real life demonstration of the Brachistochrone and Tautochrone problems.

I love this problem Steven strogatz and grant Sanderson have a video about this that features a different and fairly new geometric solution that I think was inspired by some other guy. I’m sorry I don’t remember his name. Very nice thanks.

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