Cow-culus and Elegant Geometry - Numberphile

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Zvezdelina Stankova with two solutions (one messy, one beautiful) to a problem involving an injured cow. Extra footage at: • Cow-culus v Geometry (... --- More links & stuff in full description below ↓↓↓
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Опубликовано:

28 сен 2024

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Комментарии : 536

@numberphile Год назад

Extra footage at: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-aW7R1U6FMLs.html

@utsavthakur6879 Год назад

I'm from India. We get these question in our coordinate geometry part and IIT-JEE Entrance exam. I knew it!!!!! We learn this in application of reflection of point in 2D plane.

@СвятославГлуздов-ч5ч Год назад

Area and length have 1st order continuity. Therefore, it is possible to make a parallel transfer of the gradient of the hypotenuse line. This action is equivalent to flipping triangle AFX about the vertical axis. Therefore, we get the similarity of triangles. This result follows from the double ratio of four points. Therefore angle FXA is equal to angle CXB.

@utsavthakur6879 Год назад

Hey I have a question in which I find difficult to believe the answer 1. Suppose that circle of equal diameter are packed tightly in "n" rows inside equilateral triangle then Lim (Area of n circles/Area of equilateral triangle) n-> ♾️ Is pi/2(root3)

@TheSucread Год назад

What Professor Zvezda is explaining fantastically in this video is the standard exercise in undergraduate physics courses when Fermat's principle is introduced in optics. To be completely true, the path taken by light (weighted by an index of reflection, so called optical path) needs to be stationary (which typically is minimal). There are so cool physical examples when light takes a non-minimal path, e.g. max path in optical fiber with gradient of index of refraction, saddle points for light reflective from a concave mirror.

@jagatiello6900 Год назад

This problem turned out to be somewhat similar to that of calculating the length of a 1-turn helix over a cylinder. The reflection trick also reminded me of the method of images (in electrostatics).

@trucid2 Год назад

Another way to think about the weighted problem is if the farmer was on the other side of the river and needed to swim across. Assuming no current, and given farner's speeds over land and water, where should he get out of the river on the other side?

@olivialuv1 Год назад

Non-minimal paths gets me thinking about geodesic differentiations & geodesic spaces

@Triantalex 8 месяцев назад

??.

@animeshbaranawal2781 Год назад

I need to give credit to Brady. What makes these videos even more enjoyable are the excellent and more so human questions posed by Brady to the professors. Big fan of all his channels!

@mina86 Год назад

FYI, what Dr. Stankova hinted towards the end was Principle of least action which is a way to think about problems in physics.

@methatis3013 Год назад

Its actually Fermat's principle when it comes to light. Light takes the path of least time, not least distance

@firstnamelastname307 Год назад

I think it is a geodesic in an extremely curved space : a folded piece of paper along the river

@ar_xiv Год назад

And the dog water frisbee problem is equivalent to light changing medium like from air to glass (or water)

@alexwang982 Год назад

@@firstnamelastname307 it is

@TAP7a Год назад

@@methatis3013 that is an application of the principle of least action, yes

@patrickthegoat Год назад

As someone who struggled with finding math interesting at school, I can only say it’s videos like this one that have made me fall in love with the beauty of mathematics as an adult. Thank you!

@Narokkurai Год назад

I remember actually solving this one geometrically in middle school. I figured that there were two "sub-optimal optimizations": one where the farmer minimized his walk to the water and then went directly to the cow, and another where he minimized his walk to the cow and took a long route to the water. I drew the intersection of these paths and then supposed that dropping the perpendicular from that point would give me the point on the river with the shortest path possible, without ever needing to actually solve for X. I didn't know how to prove my strategy, however it was correct.

@hps362 Год назад

That is a fascinating observation. Some maths voodoo right there.

@marpaub Год назад

What you describe is looking for the intersection of the lines FB and AC. The points (0|2) and (4|0) lie on FB and the points (0|0) and (4|6) on AC. Therefore, FB is equivalent to Y = -1/2 * X + 2 and AC is Y = 3/2 * X. Set -1/2 * X + 2 = 3/2 * X to find the intersection. It follows that X = 1.

@fahrenheit2101 Год назад

The fact that it worked is astonishing.

@glowingfatedie Год назад

"without ever needing to solve for X" uhh... but solving for X is the problem as given. At any rate, your insight about the perpendicular from X was spot-on, and solving the similar triangles, you do get AX=1 and XB=3.

@wtspman Год назад

Brilliant insight. What you describe is the same as a crossed ladders problem.

@marklonergan3898 Год назад

"skip to 9:30". Me: (wondering who comes to numberphile to skip the hard maths)

@bengt-goranpersson5125 Год назад

I really like Zvedelina. 11:37 "We are the bosses." Unironically one of the best math advice I've ever heard to be honest.

@NoActuallyGo-KCUF-Yourself Год назад

It's what I tell my students all the time, "because I said so" - it's a postulate, I can do whatever I want.

@MaddHatter Год назад

For the 0 width assumption, I think it makes more sense to say "we don't care about the middle of the river because you don't go to the middle, you go to the shore line, the shoreline is 0 width" I know it is trivial, but makes more sense to me.

@ArawnOfAnnwn Год назад

That makes sense for the farmer on the same side, but not for the hypothetical one on the other side. Cos if the river had a width, then sure he could fill his bucket at the shoreline too, but he'd still have to swim across the river to get where the first farmer would start his walk back from. It makes the two farmers non-identical if the river has a width.

@PowerChannel88 Год назад

The zero width assumption was made for the phantom farmer analogy. The width of the river doesnt matter any way to the problem, but if someone visualizes it with the phantom farmer they might be thinking they need to be at the other shore. And when the width is zero both shores are the same.

@MaddHatter Год назад

For both comments, I agree I guess I was poorly communicating that we can assume 0 width as an analog for the mirror person. Not that the river is 0 width...

@PendragonDaGreat Год назад

I think the fact that we're also dealing with distances measured in kilometers and most river widths are measured in the tens to maybe hundreds of meters (I mean obviously not all, portions of the Amazon are over 10km wide) allows us to ignore it. Also obviously the reflection occurs on the near shoreline, and not the river centerline.

@gustafcarstam5578 Год назад

@@ArawnOfAnnwn its a phantom farmer. he floats above the water like any phantom would. He only needs to reach the shoreline on the other side of the river at the same spot as the real farmer reaches the shoreline. There is no point in them meeting up somewhere in the river.

@DS-xh9fd Год назад

Once on a physics midterm, we were asked to use Fermat's principle to prove that the angle of reflection equals the angle of incidence for mirrors. We were expected to use calculus, but I wrote the geometric proof. The grader gave me almost no points for my solution, but I took the test to the teacher later and managed to get my score adjusted.

@carultch Год назад

One possibility is that the author of the exam, was trying to see if you understood the methods taught in the class. So I can understand why you wouldn't get credit for a geometric proof like this. I could understand this being the case for this problem appearing on a Calculus exam, but not on a Physics exam.

@johnchessant3012 Год назад

I've seen this problem before in "The Art and Craft of Problem Solving" by Paul Zeitz (an absolute must-read for anyone interested in contest math problems). So to me, the shortcuts used in the algebraic solution are really elegant too; it's easy to just multiply everything out and plug-and-chug, but knowing what can save some effort there is an art in and of itself.

@mokopa Год назад

It's really worth watching all the way through to the end where it gets quite more thought-provoking, philosophical even, than just an interesting mathematical problem.

@firstnamelastname307 Год назад

agreed ... two assumptions ... axiom or law ... that not seem to matter because same ... is mind blowing

@SmileyMPV Год назад

I watch numberphile to hear mathematicians answer questions like "what happened to the width of the river" with "who cares"

@vanhavirta Год назад

Makes sense, since the farmer gets the water from the shore (the edge of the river). That has no effect on the reflection aspect 😊

@poulanthrope Год назад

I didn't think about it in terms of either method, I kind of split the difference. I thought about turning them into similar triangles, and since they were 2 and 6 on one side, that reduces to 1 and 3, which add to 4, so the farmer should go to 1 since A and B are 4 apart. It was intuitive, quick, and correct. My thoughts on 'proof' would be to take it to extremes. If you're shifting the point the farmer reaches left or right, it's shortening one hypotenuse but lengthening another. It's intuitive for the transition point between lengthening and shortening the path to happen when both triangles are the same scaled shape (scaled step increases on either side would have the same effect on both hypotenuses, so the path wouldn't be shortening on one side and lengthening on the other), and since it's longest at A and B, that makes the path shortest when the triangles are similar. I guess my justification takes longer, but the solving took almost no time at all.

@karlvyh Год назад

My first thought was similar triangles as well. I felt the need to justify why they are the shortest, so I imagined the farmer on the opposite side of the river, which makes identical triangles to the given problem and turn his path into a straight line. From a physics perspective, light takes the minimal path so our path should reflect at the river, giving similar triangles.

@Son_Of_Atreides Год назад

I was also thinking about similar triangles, but subconsciously. I was actually thinking about a light ray reflecting on a mirror and the equal angles of the reflection.

@skilletpan5674 Год назад

I was thinking about ratios. 2+6 is 8 and 2 is 1/4 of 8. 1/4 of 4 is 1 so I assumed that was the anwser.

@wingracer1614 Год назад

I had three ideas and that was one of them. My two other ideas were very wrong.

@krishna8976 Год назад

I'm glad you kept the whole video. It's why your videos are amazing!

@zergbergerdelemon9634 Год назад

Zvezda's videos are always insanely inciteful

@frankjohnson123 Год назад

When I knew there was an elegant solution I immediately thought that triangle FAX should be similar to triangle CBX but didn't realize why until the explanation. Nice problem!

@bandana_girl6507 Год назад

The connection to light can actually solve a lot of the other issues that come up, potentially even turning them back in to geometry problems

@m.h.6470 Год назад

My absolute first thought was: The river is a mirror and the path to take is a light. Which means, either the path from the origin to the river or the path from the river to the cow can be mirrored on the river to give a straight line. This gives a big rectangular triangle with 8 and 4 as the legs. Or, if you put it in a coordinate system, the hypotenuse becomes a line with the gradient of 8/4 = 2 (or -2, if the cow path is mirrored). Now you only need to calculate the intersection with the x axis. Given that the origin is 2 away from the x axis and the gradient is also 2, the line will intersect after 1 unit. As such, x = 1

@ganymedemlem6119 Год назад

I'm very happy that I thought up aa different way to do it. My initial thought was that the minimum distance would have both triangles have the same angles at each point so I thought I would try to solve for the angle to get the length of the hypotenuse. Glad to know I was on the right track!

@Retroist2024 Год назад

The mirroring method is also used in physics such as static electricity, a grounded plane acts as a mirror that reflects charges above

@raychi871 Год назад

No words can describe how amazing this video is. THANK YOU

@MrPeloseco Год назад

Big fan of Numberphile, but bigger fan of Zvedelina! ❤️

@StuMas Год назад

It felt like any point between A and B might be the same but, the phantom farmer was simply genius.

@phizc Год назад

The "Dog on beach fetching frisbee in sea" problem is an analog for refraction. The index of refraction is how much slower the dog swims than runs. I watched one of the Richard Feynman lectures on YT where he showed the same thing, but in his example it was a lifeguard and a drowning person. Cool stuff. The light follows the path that minimizes time.. Well, probabilistically.. It was about quantum mechanics after all..

@xyz.ijk. Год назад

I love the Professor's work and lament that I didn't have such an extraordinary teacher growing up. (Notwithstanding that an extraordinary 5th grader in Bulgaria is not the same here in the U S. because Europe is about four or five years ahead of us in almost every subject. It's painful how far behind we here are here in the U.S.)

@StaticMotions Год назад

Clear, concise, and brilliant explanation

@vsalt69 Год назад

A physical way to implement this: The farmer has a helper walk along the riverbank holding a mirror parallel to the riverbank. The moment the farmer can see the cow in the mirror, their helper is at the point on the riverbank the farmer should walk to. Although in the physical world, the farmer would probably want to minimize the length of their walk while carrying water...

@carultch Год назад

Right, you might introduce two different speeds for the farmer to walk. One when carrying the bucket, and one when walking unladen. I came up with an example, where the farmer can run 10 km/hr with an empty bucket, but can only travel at 5 km/hr with a full bucket, using the numbers given in this problem. Empty bucket distance to run: d1 = sqrt(2^2 + x^2) Empty bucket time to run: t1 = d1/v1 Full bucket distance: d2 = sqrt(6^2 + (4-x)^2) t2 = d2/v2 Objective equation: T = t1 + t2 T = sqrt(4 + x^2)/v1 + sqrt(x^2 - 8*x + 52)/v2 T = sqrt(4 + x^2)/10 + sqrt(x^2 - 8*x + 52)/5 The solution occurs at x=1.84 km. There is an exact expression for it, but it is complicated. Perhaps there is a special case of given speeds, that make it easy to solve, but I haven't explored them. You probably could solve it with Snell's law and refraction theory, as a shortcut to solving it with Calculus.

@kayleighlehrman9566 Год назад

I wonder how much more devious the question becomes if the farmer walks slower with a full bucket than with an empty bucket?

@deltalima6703 Год назад

The cow is a sphere, why doesnt the farmer just roll it over to the water?

@deltalima6703 Год назад

The stream is downhill from the cow (or it would be a lake), so all the cow needs is a little nudge in the appropriate direction to get there.

@carultch Год назад

It would turn from a reflection problem, into a refraction problem. Instead of aiming for the point where his angle of approach equals his angle of reflection, like we do when he walks the same speed, we'd solve it as follows. Consider the phantom farmer on the opposite side of the river, who walks to point x, at speed v1. The phantom farmer then meets the real farmer, and the real farmer walks at speed v2 to the cow. We'll call the angle of approach for the phantom farmer, phi1. and we'll call the angle the real farmer continues, phi2. Both angles measured perpendicular to the river. Recall Snell's law: n1*sin(phi1) = n2*sin(phi2) Since n's are inversely proportional to speed, this means: sin(phi1)/v1 = sin(phi2)/v2 Now we would construct the triangle where phi1 and phi2 come in this proportion.

@carultch Год назад

For an example that simplifies nicely, consider the same setup as we had with the original problem, except the cow is 2*sqrt(3) kilometers (approx 3.46 km) from the river, and the farmer's speed when carrying the full bucket is 1/sqrt(2) of his speed (approx 70.7%) when carrying the empty bucket. To recap the other data, the farmer starts 2 km away from the river, and the x-position of the cow is 4 km. The farmer approaches the river on a 45 degree angle of approach to the x-position of 2 km. Using Snell's law: sin(45 deg) = sin(phi2)*sqrt(2) Solving for phi2, we get phi2 = 30 degrees. This is consistent with a triangle that is 2 km on its base, and 2*sqrt(3) km on its height. Total distance travelled = 2*sqrt(2) km + 4 km = 6.83 km

@joajoajpedroj9253 Год назад

I had a professor give very similar examples last semester and between classes talked to him about alternative solutions. His response was that he was giving that simple example and wanted us to use the calculus solution because anything more would be horrible for someone learning basic calculus but sometimes impossible with "5th grade" solutions.

@mathematicacivilis Год назад

Great example of how a change in perspective, by deforming the problem in this case, is many times key to gain new insights. As always, Professor Stankova does a marvelous job guiding us through all the intricacies of both solutions. One reflection over the geometric solution (pun intended): This is a small variation of the solution presented in the video. Extend segment BC to the other side of the river to point P, such that, F'P is parallel to AB. We'll have F'P = AB = 4 AF' = BP = 2 and CP = BC + BP = 6 + 2 = 8 Now, consider the two triangles XBC and F'PC. They are obviously similar (2 equal internal angles). Therefore: (4-x)/6 = 4/(6+2) => 4-x = 3 => x = 1 We can generalize this approach to obtain: x = AB/(BC + AF) * AF and AB-x = y = AB/(BC + AF) * BC or x = cotan(α) * AF and AB-x = y = cotan(α) * BC , where α = angle(CF'P)

@sgtreckless5183 Год назад

Zvedelina is honestly one of the S tier Numberphile presenters.

@sgtreckless5183 Год назад

@@JohnPretty1 Like top tier, one of the bests!

@trizgo_ Год назад

4:04 me: *sips coffee and listens intently*

@Michael9W Год назад

There is also a mechanics solution. It leads to equal angles as well. Imagine a rope sheave which can move along the river freely and rubber that goes from the farmer to the cow through the sheave. The sheave will stop where the total length of the ruber is minimal and it will be where ruber goes to farmer and ruber goes to cow on the same angle to the river. (and it will work for curved river as well)

@firstnamelastname307 Год назад

exactly! an ellipse tangent to the river

@NAT0P0TAT0 Год назад

its crazy how so many people in the comments (myself included) just intuitively knew that the shortest path would have both triangles being the same angle seeing the example with one triangle flipped upside down makes it super obvious but before that 'proof' we just somehow knew

@HAL-oj4jb Год назад

For anyone who wants to know about the problem with the dog and the frisbee: it is a similar problem but instead of reducing the distance, you have to reduce the time of the path (which is the same for the farmer because his speed is constant), which leads to the the laws of Snellius, which also descripe how light bends when it passes through glass or water. Allegedly, dogs know this intuitively and actually take the fastest paths in these kinds of situations, but I'm not sure how legible this is.

@belagoblyos7822 Год назад

In this case: how the light reflects (and not bends).

@Deejaynerate Год назад

This one was really fun. I figured out the geometric solution very quickly when I realized that the shortest path would be the one where both triangles had congruent angles, then used trig identities (tan theta = opposite over adjacent) to calculate the rest. However, seeing the calculus solution in action has also reminded me of how effective and precise it can be. Overall, this was a great video, reminds me of a problem I took from calculus one involving a cow looking at a sign.

@FreudeAnEtwasFinden Год назад

Great Video! Thank you Zvedelina! 🙂❤ Could you please comment again on the axioms of nature and equivalent mathematical concepts or structures? ( 18:20 - 18:55 ) It would be very interesting to know what else you think about it. ...maybe with another example?

@TitusRex Год назад

Use the reflection of one of the points across the river and draw a straight line.

@josiasreis Год назад

"the phantom farmer" sounds like a cool superhero

@SuperYoonHo Год назад

Thank you!

@stulora3172 Год назад

Concerning the dog-frisbee problem: I remember reading that some mathematicians actually tested this with dogs and it turns out, they instinctively run and swim in along the optimal path!

@Richardincancale Год назад

So the cow / bucket story = reflection and the dog / frisbee story = refraction. Are there animal analogues for all physics problems?

@Shildifreak Год назад

But wouldn't be carrying a full bucket be more exhausting than carrying an empty bucket? :D (Nevermind it was mentioned at the end.)

@numberphile Год назад

I did discuss this with Zvezda - can’t recall if it was in camera. You can easily start inserting extra variables such as this. Escalates quickly I’m sure.

@LA-MJ Год назад

It's refraction instead of reflection

@kappasphere Год назад

@@LA-MJ Even the original method was to imagine a case of refraction instead of reflection - but with a refractive index of 1. What changes in the case of being slower after passing "through" the river is that the refractive index is exactly the ratio between the farmer's original walking speed and the speed that he has when carrying a bucket. So you don't even need a new analogy of light, it's still the exact same analogy but with different numbers.

@carultch Год назад

For an example that works out nicely, consider the same numbers, but place the cow at 3.46 km from the river. Suppose the farmer with a full bucket can walk at 1/sqrt(2) of his speed with the empty bucket. The optimum point turns out to be x=2 km. The farmer approaches the optimum point at a 45 degree angle of incidence, and walks away at a 30 degree angle of reflection.

@NoActuallyGo-KCUF-Yourself Год назад

Yes! I love Prof Skankova!

@ribal3269 Год назад

I never click as fast as when I see a Numberphile video with professor Zvezda in the thumbnail!!

@quill444 Год назад

My initial thoughts after three minutes into the puzzle: Since walking "up river" is uphill, and since carrying a bucket full of water is much more difficult than walking with an empty bucket, it seems possible that it might be energy efficient for the farmer to walk straight upstream, not just near to Point B, but perhaps even a bit farther! Then, once the bucket is filled, his walk to the cow (under more duress from the full bucket) is all downhill. You must account for the fact that walking with a full bucket is inordinately more difficult than walking with an empty bucket! - j q t -

@samcalder6946 Год назад

Real farmer would probably drive his 4x4 directly from F to C, then shoot the cow to put it out of it's misery.

@dave190785 Год назад

I swear I got this straight away using a vertical line of symmetry and similar triangles. Never got anywhere close a numberphile question that quickly.

@johnsmith1474 Год назад

Actually the light, in moving toward & reflecting from the water, takes every possible path. I need to reopen Feynman's "QED" to review the very simple and elegant explanation.

@oldcowbb Год назад

i'm glad calculus is finally getting some love from this channel lately

@advaykumar9726 Год назад

Take the position of the farmer (0,2) and takes its image and join the image and the coordinate of the cow which is (4,6) The point where the line intersects the x axis is the required point The equation of the line will be y=2x-2 putting y=0, we get x=1 Hence the point is (1,0)

@Nikolas_Davis Год назад

I have a simpler calculation for finding X. To get to the cow, the phantom farmer walks 4km "horizontally" and 6+2=8km "vertically" (perpendicular to the river). So, he covers half the distance horizontally that he covers vertically, and that holds for any part of the distance, since he walks on a straight line. To get to the river, he has to walk 2km vertically, so when he gets there he will have walked 1km horizontally. Therefore, X = 1km.

@samcalder6946 Год назад

Immediate thought at 1:52 ... reflect the smaller triangle around the X axis, form an effective 8 by 4 triangle, IE. a 2:1 gradient, so go up by 2 to reach X=1 Reassuring the instincts are still dialled in after all these years.

@only20frickinletters Год назад

The dog and frisbee gives a good intuition for how light refracts, but another is to consider one side hitting the water and slowing down first, making the light turn like a canoe paddled harder on that side.

@darreljones8645 Год назад

For the record, the distance the farmer travels in the optimal path is sqrt(5) + sqrt(45), or about 8.944km. If the farmer didn't have to go to the river, the shortest path to the cow is sqrt(32), or about 5.657km.

@bscutajar Год назад

Same exact thing happens with refraction. The path when crossing two media at an angle is also the shortest one. Edit: the dog and frisbee problem is exactly what I am talking about

@lucasmoulin1886 Год назад

Great video ! The geometric approach can also be done using Thales's theorem to find the value of x, instead of similar triangles.

@Lolwutdesu9000 Год назад

You can also solve for matching the gradients of F'X and XC. You're essentially working out the 'coordinates' of point X. You get the same result. Simpler IMO than creating triangles.

@paulf5351 Год назад

It's simpler for me if the river is a mirror. You can aim for the reflected image in in the mirror. Just like bank shots in pool and off the backboard in basketball.

@stevewolfe6096 Год назад

A physical solution. Consider the river as a rod with a frictionless ring Run a (frictionless) string from the farmer through the ring to the cow and pull tight. The ring will slide to the point where the force from each segment is equal i.e. the angles are equal >> the two right triangles are similar.

@AndyHolt2 Год назад

Another potential solution is taking that reflection idea for the straight line and then y=mx+c The line goes through (0,-2) and (4,6) You can derive the gradient from that and then the crossing is x where y=0

@Rick.Fleischer Год назад

Such a beautiful simplification.

@egdiroh Год назад

The slope of F’C is easy to compute so the intersection, without triangle comparison

@JimEckhardt Год назад

Prof zvedelina reminds me of Gru. I love it!

@Kwezi_42 Год назад

04:58 “We will humanely get rid of those twos”❤

@jeremydavis3631 Год назад

I helped a 10th-grade geometry student with this problem. Of course my first thought was to use calculus, but this student didn't know any calculus, so that wouldn't have helped at all. We actually solved it using the way light reflects off a surface and my knowledge that light always takes the fastest path. The student didn't already know that, so it was a bit of a leap, but doable. A few hours later, the reflection popped into my head, and I couldn't believe how obvious the solution had suddenly become.

@VenSensei Год назад

Wonderfully elegant.

@phasm42 Год назад

That was really elegant.

@stapler942 Год назад

I'm just impressed we didn't have to assume any spherical cows here.

@Your2ndPlanB Год назад

I think there's a similar 'geometric' solution to the dog-frisbee problem: if we draw the beach/water boundary as the x-axis, we simply scale all y-distances on the "water" side by _c_ , where _c_ is the speed on the sand divided by the speed in the water. Then for similar reasons, the optimal path will be given by the straight line drawn to the "phantom" scaled frisbee. And again, this corresponds to a light beam going from one medium to another, and having some angle of refraction :)

@AB-Prince Год назад

my mind immediately went to creating two similar triangles.

@postmodernist1848 Год назад

What we generally do at school with these minimising/maximising tasks is solve f'(x) = 0 and find when it's positive or negative. This way you can find the points of maximum and minimum when the sign changes from negative to positive (which is a minimum point) and vice versa (which is a maximum point)

@SuperYoonHo Год назад

I knew that from the great arthur t benjamin's ttc course!

@gsurfer04 Год назад

Excellent example of visual calculus

@anon6514 Год назад

Calculus is like heavy machinery. It's really powerful but can take a while to get going.

@michimedi Год назад

One could have just ignored the square roots and minimize not the distance but the squared distance. So much easier to solve.

@ominollo Год назад

Zvedza is great!

@jukmifggugghposer 7 месяцев назад

The elegant solution is oddly reminiscent of the method of images in electromagnetism. You replace the real farmer (charge) with a phantom farmer in another region, generally by a reflection, and a lot of what you can say about the phantom farmer also applies to the real one. And thinking in terms of the phantom farmer can radically simplify otherwise intimidating problems.

@antonspitfire7789 Год назад

Звезда молодец. Очень интересно.

@marpaub Год назад

The solution using triangles is unnecessarily complicated. On the straight line F'C lie the two points (0|-2) and (4|6). From this the equation of the straight line is Y = 2 * X - 2. The equation for the abscissa is Y = 0. If you set the two equations equal, the result for the intersection point is 0 = 2 * X - 2. From this follows immediately X = 1.

@vitorschroederdosanjos6539 Год назад

once i saw the simple solution i was laughing for literaly 10 seconds. amazing

@digitig Год назад

I think the farmer would indeed aim for that point if they are a decent pool or snooker player.

@digitig Год назад

Oh - and the point someone made about going slower carrying buckets of water? Spin.

@Netherfiend Год назад

Great video! I thought about this by putting the farmer on the y axis(on the other side of the river), and the river on the x axis. Then I can get the slope from (y2 - y1)/(x2 - x1) or [6- (-2)]/[4-0] = 8/4 = 2. Then plugging into grade school slope formula y=mx+b (m as slope, b as y-intercept), 0 = 2x+(-2) -> 0 = 2x-2 -> 2 = 2x -> x=1.

@advpmishra Год назад

Now I realise that COW-CULUS was a pun ...

@sukritdubey2630 Год назад

This reminds me of a problem shared by our high school teacher once : There is an ant sitting at one corner of a room's floor, it has to reach the diagonally opposite corner on the ceiling of the room, so what is the minimum possible distance for the travel? One observes that a similar approach of "finding the apparent straight line" works again for an elegant solution.

@SergeMatveenko Год назад

Another geometrical way would be to solve first for same distances to the river from the farmer and from the cow and then prove that the similar path with equal angles is the solution for the general case as all cases have their place on this path.

@Joris1111111 Год назад

So, if I'm the farmer, I have to go in a straight line towards the Fantom Cow. I hope there's a giant mirror placed at the river. 🤣

@MrMeszaros Год назад

Whooo nice. My thought was that a simplification could be if the farmer and the cow would be both 6km away from the river. Then it kind of follows that he should go symmetrically.

@ricardo.mazeto Год назад

Very interesting problem, but there's an easier solution. Flip the farmer F to the other side, F', and turn the paper 90 degrees. Now the river is parallel to the y axis, and the line FF' is the x axis. The deltas of the line CF' are now the rise over run, the parameters, of the function CF', -4 and 8. Evaluate the function at -2, the river. Done.

@finlandtaipan4454 Год назад

If I were the rancher, I would not take the shortest path. I would take the fastest path. Walking with buckets full of water is slower than walking with empty buckets. Therefore, I would fill the buckets at point Y instead of point X. To find point Y, I would use the indexes of refraction (reciprocals of the velocities) for each segment of the journey. I hope I will find the formulae in part 2.

@ayushsingh5517 Год назад

actually, to tell the truth, I knew the geometrical solution. I am in class 7 and learned this method while studying the reflection of light. Btw, ur videos are amazing!

@rvcmathguy Год назад

Maybe I missed it somewhere in the hundreds of comments, but I’m surprised no one mentioned Elvis the Calculus Dog whose owner Tim Pennings found was using a path remarkably close to the theoretical optimal path when chasing a ball thrown into the water.

@jonathanc8845 Год назад

The width of the river also doesn't matter because you only need to measure the reflected farmer's distance from the edge of the river nearest the cow.

@JochenBoy Год назад

18:00 "The farmer is doing exactly the same thing as the light is doing" - does that imply that the farmer is bright?

@fahrenheit2101 Год назад

Saw the first 2 minutes, and realised that it's exactly the same as a MAT problem I did once. My guess is you reflect and draw a straight line. (btw I adore that method, so I hope I'm right) Edit: Finished the video, and yep, I was right. And I was also right in seeing that it was just like light (it reminded me of annoying ray diagrams I had to draw for Physics a few years back, so that's why I guessed it'd be the same)

@Neil-ii3dp Год назад

Sorry y'all. If I'm the farmer I would minimize the distance I have to walk with the full bucket, even if it meant walking farther overall. Even 6 km is a long way to walk carrying a full bucket!

@micronalpha Год назад

Truly exquisit!

@grzegorznaskret4453 Год назад

I would walk straight to point b, take the water and walk the shortest path with water to the caw 😂

@ArcheoLumiere Год назад

Paused at 2:18, theorized solution below the cut, all done in my head. The shortest distance between two points is a straight line, but we need to go a distance between three points, the simple way to deal with this is to flip line AF over the river, creating point F' and create a new point F", which is 4km away from the farmer parallel to the river, and 8km away from the cow. At this point, you're essentially looking for a self similar triangle to a right triangle with base 4km (AD, or F'F") and height 8km (F"C, aka AF+CD), where the known height is 2km (AF), ergo, your distance along the river (AX, or x) is 1km.