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Ask most people the fastest slide between two points and they say a straight line, because it is shortest. In 1696 Johann Bernoulli showed why that answer is wrong, and the search for the right one started a new branch of mathematics. This post covers who Bernoulli was, what the brachistochrone problem asks, and a worked example you can check with Class 11 physics.
Who was Johann Bernoulli?
Johann Bernoulli was born in Basel on 27 July 1667 and died there on 1 January 1748, according to his MacTutor biography. His father wanted him in the family spice business, and at 15 he worked in it for a year without success. At the University of Basel he studied medicine but spent his time reading Leibniz's new papers on calculus with his older brother Jacob, the subject of our post on Jacob Bernoulli and the law of large numbers.
In Paris he taught calculus to the Marquis de l'Hôpital, who paid him well and later published the first calculus textbook (1696) based on those lessons. MacTutor says the rule students call l'Hôpital's rule is therefore Bernoulli's result; proof came only in 1922, when a copy of Bernoulli's course turned up in Basel. In 1695 he took the chair of mathematics at Groningen, and later succeeded Jacob at Basel. His son Daniel is the subject of our post on Daniel Bernoulli and Hydrodynamica.
What is the brachistochrone problem?
"Brachistochrone" comes from Greek for "shortest time." In June 1696 Bernoulli posed it in the journal *Acta Eruditorum*: a bead slides without friction from a point A to a lower point B that is not directly below it. Which curve gets it there fastest?
MacTutor says Leibniz persuaded him to extend the deadline so foreign mathematicians could try, and that five solutions arrived, including those of Johann, Jacob and Leibniz. The MacTutor page on the cycloid adds that Newton and de l'Hôpital also solved it, that Bernoulli published his own solution in 1697, and that the problem was a starting point for the calculus of variations. Galileo had earlier given an incorrect answer.
The answer is a cycloid, the curve traced by a point on the rim of a rolling wheel, turned upside down.
Why is the straight line slower?
On a steep start the bead picks up speed early and then covers most of the distance fast. A straight ramp has a gentle constant slope, so the bead spends a long time moving slowly near the top. The cycloid trades a little extra length for a lot of early speed.
Energy conservation tells you something surprising: the speed at B is the same on every frictionless path, v = √(2gh), because only the drop h matters. What differs is how quickly the bead reaches high speed along the way.
Worked example: cycloid against straight ramp
Take g = 9.8 m/s² and a cycloid made by a wheel of radius r = 1 m. From its top cusp to its lowest point it drops 2r = 2 m and moves sideways πr = 3.142 m.
Speed at the bottom (both paths). v = √(2 × 9.8 × 2) = √39.2 = 6.26 m/s.
Straight ramp length. L = √(3.142² + 2²) = √13.87 = 3.724 m.
Acceleration along the ramp. a = g × (drop ÷ length) = 9.8 × 2 ÷ 3.724 = 5.263 m/s².
Time on the ramp. From L = ½at², t = √(2L/a) = √(7.448 ÷ 5.263) = √1.415 = 1.190 s.
Time on the cycloid. For a cycloid starting from rest at its cusp, the standard result is t = π√(r/g) = 3.142 × 0.3194 = 1.004 s.
The cycloid wins by about 0.19 s, roughly 16% faster, even though it is the longer path.
There is a neat check. In a short note in the *Philosophical Transactions* for 1697, Newton proved that the time down the straight chord AB and the time along the cycloid arc are in the ratio AB : AC, where AC is the perpendicular from A to the curve's normal at B. Here the normal at the lowest point is vertical, so AC is the horizontal distance 3.142 m. The ratio 3.724 ÷ 3.142 = 1.185 matches 1.190 ÷ 1.004 = 1.185.
Common mistakes in incline problems
Using a = g on a slope. Along a frictionless incline a = g sin θ, here 9.8 × 2/3.724.
Using height instead of length. The kinematics equation uses distance along the ramp, L, not the drop.
Thinking the faster path gives a higher final speed. Both reach 6.26 m/s; the question is only time.
Forgetting the square root. t = √(2L/a), not 2L/a.
The JEE (Main) 2026 syllabus covers the work-energy theorem and motion in a straight line; the cycloid result itself is beyond it. Check the syllabus for your exam year. Our work-energy theorem problems guide has more practice.
If you want to see why the curved path wins rather than take it on trust, Fermi's Simulation Lab has interactive simulations for ideas that are easier to see than read.
What can a student take from Johann Bernoulli?
Bernoulli was quarrelsome, and MacTutor records long feuds with his brother and even his son. But his challenge problems show a useful habit: pose a question so clearly that the strongest minds of your time can test themselves on it. Writing a problem precisely is half of solving it.
Practice routine for this week:
Repeat the example for a wheel of radius 0.5 m. Both times should shrink by a factor of √2, so check that the ratio stays 1.185.
Compare a straight ramp with a path that drops 2 m vertically and then runs 3.142 m flat. Which is faster, and why?
Explain in two lines why the final speed is the same on all three paths.
Try step 2 first: it shows on paper why an early drop helps.
