Uploaded October 2015 | Updated September 2026, 2 hours ago
Another viral problem that's flying around the internet, from the Higher Mathematics exam in Scotland.
BBC Article about it: bbc.com/news/uk-scotland-34476699
I did talk a lot (some detailed stuff). If you're impatient and just want the solution:
- For part a, it starts at 4:44
- For part b, it starts at 6:27
The explanation video why x=8 is the minimum point: youtu.be/iHfXfioskSA
The links to all the videos I referred to:
- youtube.com/watch?v=bP-LwfsgjKA (Pythagoras)
- youtube.com/watch?v=6WGKARihjQc (Differentiation)
- youtube.com/watch?v=0euuGmA5sxk (Chain Rule)
Here's an alternate solution (credits to user Mandolinic, who I'll quote directly);
"You don't need calculus to solve this problem at all! You just need to know some basic physics - optics, refractive index, critical angle, and the speed of light in a transparent medium. You also need to know that a ray of light always travels along a path of minimum time. If we could somehow replace the croc by a light photon, and replace travelling in water and on land by refraction through two different media, we can just apply some standard physics to find the minimum time.
When a ray of light hits the interface between two transparent media of differing refractive indices at just the right angle (the "critical" angle) it follows the line of the interface. It also takes the shortest time possible to do it. This is just the path the croc follows when it swims at an angle across the water, and then travels along the bank to the zebra. So by analogy with the refractive case, you just need to work out the critical angle, which is calculated from sin-1(n1/n2) where n1 and n2 are the two refractive indices.
From inspection of the formula we get relative speeds of 4 and 5, and the croc is 6 metres from the bank.
Refractive index n is defined as the speed of light in the medium/speed of light in vacuum, and when you divide one index by another, the speed of light in a vacuum cancels out. So the critical angle turns out to be either sin-1(5/4) or sin-1(4/5). But sin-1(5/4) doesn't exist, so the critical angle must be sin-1(4/5). We could now use a calculator or tables to work out this angle, but we don't need to, because if we have sin-1(4/5), we have a triangle whose hypotenuse is 5, and whose opposite side is 4. Most of us will immediately recognise this as a 3-4-5 triangle. The croc is travelling along the hypotenuse of a 3-4-5 triangle.
The croc is 6 metres from the river bank and it travels along a 3-4-5 triangle, so it must travel 8 metres along the edge of the bank, and 10 metres in the water.
And that's enough to solve the problem. We know the value for x which gives a minimum is 8, and we can plug that back into the formula to get the time.
No need for complicated differentiations at all - just good old basic physics."
Another viral problem that's flying around the internet, from the Higher Mathematics exam in Scotland.
BBC Article about it: bbc.com/news/uk-scotland-34476699
I did talk a lot (some detailed stuff). If you're impatient and just want the solution:
- For part a, it starts at 4:44
- For part b, it starts at 6:27
The explanation video why x=8 is the minimum point: youtu.be/iHfXfioskSA
The links to all the videos I referred to:
- youtube.com/watch?v=bP-LwfsgjKA (Pythagoras)
- youtube.com/watch?v=6WGKARihjQc (Differentiation)
- youtube.com/watch?v=0euuGmA5sxk (Chain Rule)
Here's an alternate solution (credits to user Mandolinic, who I'll quote directly);
"You don't need calculus to solve this problem at all! You just need to know some basic physics - optics, refractive index, critical angle, and the speed of light in a transparent medium. You also need to know that a ray of light always travels along a path of minimum time. If we could somehow replace the croc by a light photon, and replace travelling in water and on land by refraction through two different media, we can just apply some standard physics to find the minimum time.
When a ray of light hits the interface between two transparent media of differing refractive indices at just the right angle (the "critical" angle) it follows the line of the interface. It also takes the shortest time possible to do it. This is just the path the croc follows when it swims at an angle across the water, and then travels along the bank to the zebra. So by analogy with the refractive case, you just need to work out the critical angle, which is calculated from sin-1(n1/n2) where n1 and n2 are the two refractive indices.
From inspection of the formula we get relative speeds of 4 and 5, and the croc is 6 metres from the bank.
Refractive index n is defined as the speed of light in the medium/speed of light in vacuum, and when you divide one index by another, the speed of light in a vacuum cancels out. So the critical angle turns out to be either sin-1(5/4) or sin-1(4/5). But sin-1(5/4) doesn't exist, so the critical angle must be sin-1(4/5). We could now use a calculator or tables to work out this angle, but we don't need to, because if we have sin-1(4/5), we have a triangle whose hypotenuse is 5, and whose opposite side is 4. Most of us will immediately recognise this as a 3-4-5 triangle. The croc is travelling along the hypotenuse of a 3-4-5 triangle.
The croc is 6 metres from the river bank and it travels along a 3-4-5 triangle, so it must travel 8 metres along the edge of the bank, and 10 metres in the water.
And that's enough to solve the problem. We know the value for x which gives a minimum is 8, and we can plug that back into the formula to get the time.
No need for complicated differentiations at all - just good old basic physics."
