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.
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.
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)
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!
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.
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).
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?
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!
The familiar calculus problem with the dog and the Frisbee in the water gives the same answer as Snell's law of refraction. All of these things turn into Lagrangians and the principle of least action - which feels more fundamental than most of the laws we learn in high-school physics.
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.
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.
"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.
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.
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.
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.
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...
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.
@@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.
Fun little exercise: After the Farmer has filled the bucket of water it obviously is more heavy so it gets more exhausting the earlier he fills it up. So actually we should account for that increase of energy he needs. So how does X change if you assign different weights to the the two paths? Say walking with the empty bucket costs 1 energy per unit length. Walking with the full bucket costs 2 energy per unit length (it's a big bucket lol). Try to minimize the total energy as a function of X. Edit: 21:10 same thought haha
Another way to modify the problem is set conditions that the farmer can run (say) 3 times faster with an empty bucket than walk with a full bucket and the solution criterion is minimize the time to get water to the cow.
Just take a physical environment, where light take speed 1 before entering the water and has speed 0.5 in a water. Then use laser to figure out where you direct it to hit the cow. In this test the cow should be in a river.
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.
As a math teacher I will definitely show this problem and the solutions to my pupils when I get to teach calculus-type stuff. Absolutely fascinating! Greetings from Germany!
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.
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.
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.
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.
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.
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.
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!
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!
I met these concepts studying Fermat's principle in physical optics at Uni, it was also introduced with a farmer but with a meadow and a ploughed field. Thanks
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)
Also: The ellipse with focal points F and C tangent to the river determines X. And we then also see that for any ellipse, the angles made by any tangent line and lines from focal points to point of contact are equal. And: To estimate X having river and cow in sight, farmer might want to seek advice from a billiard player if one happens to be available. Note: The problem at hand is old and attributed to Heron. And: The path can be seen as geodesic in an extremely curved space, in this case : the paper folded along the river.
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.)
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.
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.
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..
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)
From calculus, I found that if 'a' and 'b' are the two distances to the river and the distance to be traveled along the river is 'c', then x = a * c / (a ± b), (the sign of b is whatever satisfies 0 < x < c), which works out in this example as x = 2 * 4 / (2 + 6) = 8 / 8 = 1. I tried to find a useful geometric interpretation for that equation, without much success. What I did was begin with the drawing as in 2:36 and extend AF upward to a point P, and extend BC upward to a point Q such that PQ is parallel to AB and AP = BQ = AF + BC. Now draw the rectangle APQB. Draw a line through X parallel to AF. Draw a line through F parallel to AB. These two lines divide rectangle APQB into into four non-congruent rectangles. The equation implies that the length "x" is the solution when the upper left subrectangle and the lower right subrectangle are equal in area! What's the geometric justification for this?
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.
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.
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)
I solved it easily in a different way: Imagine the two parts of the way being a rope that is wrapped around a boat on the river. Now you pull both ends to shorten the string. Of course the boat stops moving if the angle between river and the two string parts is the same. So you have 2 congruent triangles that are also found in the video.
Paused at 3:49. Ok, when I heard there was a trick I checked to see if just going in a straight line one of the two trips would work. It gives 2+ sqrt(52) about 9.2111, and sqrt(20)+6 about 10.4721. Then I checked the midpoint, sqrt(8)+sqrt(40) about 6.3246. Ok, niether straight path gives better than just the midpoint, so maybe it's the midpoint? Straight. Straight. The shortest distance between two points is a straight line, right? Oh! What if the cow was on the other side of the river! Then we could make a straight line! And the farmer's path doesn't change lengths if you just flip directions, so it works. Now just find the slope. He goes down 2 and then "down" 6 more for -8 rise, and runs by 4, so the slope is -2. Where is he on the river after going down 2? 1 away from his starting point. x=1 QED Booyah.
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.
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...
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.
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
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.
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.
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
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.
I think it’s even simpler to extend the large triangle down to F’. You end up with a base of 4, height of 8. Then the small triangle’s ratio is proportional: 4:8 = (4-x):6. x must equal 1. Apologies if someone else in comments pointed this out.
I would reflect the point F to the opposite side of the river line (water surface is reflective, isn't it?) and connect the resulting point F' and point C with straight line. The resulting triangles are similar in scale 2:6=1:3. So, x+3x=4 and x=1.
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.
Yes, the calculus way works when it's not all consistent throughout the problem. Like when different materials are involved that light moves through, etc. The calculus way does what calculus always does - it lets you solve a problem where conditions along the solution trajectory are changing.
Re Brady’s question on instinctually finding X. I’m imagining a scenario where the farmer can see the cow and river, but have no measurements. I think it’s solvable via bearings and perspective (as the farmer’s vision is vertically above the plane) The direction to point A is determinable (the river should have left-right symmetry when looking at it). The bearings FA and FC can then be drawn from a point on the ground. The location and bearing of point B can be determined by imagining a line from C to the river, and using the same perspective technique to determine the perpendicular from F to that BC line. That line can be extended to point C’ which is determinable also using perspective. Then the farmer just heads towards C’.
I think the question is more human than this. As in, the farmer could instinctually A: want to get to the water as fast as possible, B: minimize total distance, C: minimize distance walking with a full bucket, D: not care at all and just wing it. So technically, this problem becomes a social experiment when you ask that question lol.
I guess I'm describing how I'd go about it, though yes... I would sort of wing it and not do the exact measuring. Exact methods wouldn't be realistic in practice as the plane would likely not be perfectly flat, and the river not perfectly straight. However, when looking around, I would make a crude mental effort to estimate where the point C' would be and head for it.
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 -
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
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
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.
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?
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!
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
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.
19:20 No, the farmer wouldn't necessarily get to (or near) the right point by "instincts". People frequently end up "getting off course", getting lost, going the wrong way, and so on. Instincts can't deal with stuff like this very well, mostly because of the scale. If it were a small distance where you could see the entire thing at once, you might get close, but once you're on the ground and going long distances, seeing only a bit at a time, things get off track pretty quickly and easily.
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.
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.
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 :)
Very nice geometrical solution, but I personally find it more fun if the problem was approached functionally - find a function f(x) = ax + b connecting phantom farmer and a cow. Or! Knowing that the light takes a shortest path, give farmer a laser pointer and tell him to point it at the river in. a way that a cow is illuminated by the laser :)
Another simple solution (I think). Imagine the diagram is a side view into a basketball court. The shortest distance will be the one that bounces the ball right up to the goal (assuming no gravity). This bounce point will have the same angle either side of a horizontal line straight upwards, otherwise the ball would miss the goal.
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I once wrote a term paper on Euclidean geometry (ruler and compass construction), so I immediately came up with the simple solution. This is how everyone used to do division. Too bad the answer is an irrational number, otherwise I would have solved it all in my head without needing a calculator. :D You can even explain it basically without any maths: The farmer spends some time going sideways and some time going towards the river and back. The optimal path is the one where those times are proportionally split, so since one quarter of the vertical travel time is towards the river and three quarters are back up, one quarter of the time should be spent on the way down and three quarters of time on the way up. Moving at constant speed sideways means he should move one quarter of the way to the right while he goes towards the river. Almost sounds trivial when described like that, but you have to have that idea first (or recognise it from an earlier problem). ;) With the mirroring solution, you don't even need to find the point where the farmer reaches the river to get the optimal length, it's just one big triangle.
I would like the solution for the Stairs "rolling" at the wall and floor what is the velocity relations of the ends. Or the problem of will a "car hit the curb when descending at an edge (thing strain level, edge, then, straigh down at some angle -- maximum angle that the car clears?)
my first thought was ratio of distance to river and distance in between should be same to allow shortest path. but i don't can't think of how to prove its the shortest path.
7:29 I see something rare while appling the square root; from what i know SQRT(A^2) is +A and -A: (-A) * (-A) = A^2 and also (+A) * (+A) = A^2 If you apply square root, a +/- must be added. Better shown with and example: The square root of 4 is 2, but it is also -2, since (-2)*(-2)=4. Using sqare roots gives two results, positive and negative, not just one. What am i missing? Why you did not put such +/-?
In quantum mechanics the fact that "light takes the shortest path of reflection" it's because it's more statistically likely. Particles can take wacky paths that don't match straight trajectories (only when the wave function is allowed to "spread", that is, it never collapses to a point)
I would think the farmer would run somewhere between the proper point and the middle of the river/line segment. If he's in a hurry, he'd probably think "I'm going between two places, I need to stop halfway between." I had a hunch it was about similar triangles, but yup I immediately started on the calculus to check if my measurement was right. Duh, Zvezda's reflection solution is so clever!
Extra footage at: th-cam.com/video/aW7R1U6FMLs/w-d-xo.html
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.
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.
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)
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!
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.
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).
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?
Non-minimal paths gets me thinking about geodesic differentiations & geodesic spaces
??.
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!
This is brilliant. Thank you, professor Zvezda! 😊
The thanks so nice, you gave it twice.
this is my favourite "gotcha" moment in any Numberphile so far. Well done, Prof. Stankova!
FYI, what Dr. Stankova hinted towards the end was Principle of least action which is a way to think about problems in physics.
Its actually Fermat's principle when it comes to light. Light takes the path of least time, not least distance
I think it is a geodesic in an extremely curved space : a folded piece of paper along the river
And the dog water frisbee problem is equivalent to light changing medium like from air to glass (or water)
@@firstnamelastname307 it is
@@methatis3013 that is an application of the principle of least action, yes
The familiar calculus problem with the dog and the Frisbee in the water gives the same answer as Snell's law of refraction. All of these things turn into Lagrangians and the principle of least action - which feels more fundamental than most of the laws we learn in high-school physics.
Feynman described action beautifully in his Messenger Lectures.
"skip to 9:30".
Me: (wondering who comes to numberphile to skip the hard maths)
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.
That is a fascinating observation. Some maths voodoo right there.
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.
The fact that it worked is astonishing.
"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.
Brilliant insight. What you describe is the same as a crossed ladders problem.
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.
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.
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.
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...
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.
@@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.
Always ecstatic to see a new Zvezda video!
I really like Zvedelina. 11:37 "We are the bosses."
Unironically one of the best math advice I've ever heard to be honest.
It's what I tell my students all the time, "because I said so" - it's a postulate, I can do whatever I want.
Fun little exercise:
After the Farmer has filled the bucket of water it obviously is more heavy so it gets more exhausting the earlier he fills it up. So actually we should account for that increase of energy he needs.
So how does X change if you assign different weights to the the two paths?
Say walking with the empty bucket costs 1 energy per unit length.
Walking with the full bucket costs 2 energy per unit length (it's a big bucket lol).
Try to minimize the total energy as a function of X.
Edit: 21:10 same thought haha
Another way to modify the problem is set conditions that the farmer can run (say) 3 times faster with an empty bucket than walk with a full bucket and the solution criterion is minimize the time to get water to the cow.
Don't forget to account for water evaporating from the bucket which varies due to the temperature differnces at different parts of the day.
Just double the 6 km to 12 km and you get the solution. (I guess)
Just take a physical environment, where light take speed 1 before entering the water and has speed 0.5 in a water. Then use laser to figure out where you direct it to hit the cow. In this test the cow should be in a river.
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.
agreed ... two assumptions ... axiom or law ... that not seem to matter because same ... is mind blowing
As a math teacher I will definitely show this problem and the solutions to my pupils when I get to teach calculus-type stuff. Absolutely fascinating!
Greetings from Germany!
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.
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.
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.
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.
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.
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.
I had three ideas and that was one of them. My two other ideas were very wrong.
Zvezda's videos are always insanely inciteful
Professor Zvezda is a FABULOUS educator!
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.
I'm glad you kept the whole video. It's why your videos are amazing!
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!
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!
This is brilliant. Thank you, professor! 😊
This is really interesting, showing that rethinking the problem can reduce the complexity of solving.
I met these concepts studying Fermat's principle in physical optics at Uni, it was also introduced with a farmer but with a meadow and a ploughed field. Thanks
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)
I watch numberphile to hear mathematicians answer questions like "what happened to the width of the river" with "who cares"
Makes sense, since the farmer gets the water from the shore (the edge of the river). That has no effect on the reflection aspect 😊
No words can describe how amazing this video is. THANK YOU
Also: The ellipse with focal points F and C tangent to the river determines X. And we then also see that for any ellipse, the angles made by any tangent line and lines from focal points to point of contact are equal.
And: To estimate X having river and cow in sight, farmer might want to seek advice from a billiard player if one happens to be available.
Note: The problem at hand is old and attributed to Heron.
And: The path can be seen as geodesic in an extremely curved space, in this case : the paper folded along the river.
Clear, concise, and brilliant explanation
Great way of explaining the solution! Thank you. If I may, the shortest path between 2 points is always a straight line.
Love the videos with Zvedelina!
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.)
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.
Big fan of Numberphile, but bigger fan of Zvedelina! ❤️
Zvezda is always a treat to watch
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.
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..
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)
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
The mirroring method is also used in physics such as static electricity, a grounded plane acts as a mirror that reflects charges above
From calculus, I found that if 'a' and 'b' are the two distances to the river and the distance to be traveled along the river is 'c', then x = a * c / (a ± b), (the sign of b is whatever satisfies 0 < x < c), which works out in this example as x = 2 * 4 / (2 + 6) = 8 / 8 = 1.
I tried to find a useful geometric interpretation for that equation, without much success. What I did was begin with the drawing as in 2:36 and extend AF upward to a point P, and extend BC upward to a point Q such that PQ is parallel to AB and AP = BQ = AF + BC. Now draw the rectangle APQB. Draw a line through X parallel to AF. Draw a line through F parallel to AB. These two lines divide rectangle APQB into into four non-congruent rectangles. The equation implies that the length "x" is the solution when the upper left subrectangle and the lower right subrectangle are equal in area! What's the geometric justification for this?
Zvedelina is honestly one of the S tier Numberphile presenters.
@@JohnPretty1 Like top tier, one of the bests!
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.
It felt like any point between A and B might be the same but, the phantom farmer was simply genius.
i'm glad calculus is finally getting some love from this channel lately
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.
I never click as fast as when I see a Numberphile video with professor Zvezda in the thumbnail!!
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)
exactly! an ellipse tangent to the river
I solved it easily in a different way: Imagine the two parts of the way being a rope that is wrapped around a boat on the river. Now you pull both ends to shorten the string. Of course the boat stops moving if the angle between river and the two string parts is the same. So you have 2 congruent triangles that are also found in the video.
Paused at 3:49. Ok, when I heard there was a trick I checked to see if just going in a straight line one of the two trips would work. It gives 2+ sqrt(52) about 9.2111, and sqrt(20)+6 about 10.4721. Then I checked the midpoint, sqrt(8)+sqrt(40) about 6.3246. Ok, niether straight path gives better than just the midpoint, so maybe it's the midpoint? Straight. Straight. The shortest distance between two points is a straight line, right? Oh! What if the cow was on the other side of the river! Then we could make a straight line! And the farmer's path doesn't change lengths if you just flip directions, so it works. Now just find the slope. He goes down 2 and then "down" 6 more for -8 rise, and runs by 4, so the slope is -2. Where is he on the river after going down 2? 1 away from his starting point. x=1 QED Booyah.
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.
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...
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.
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
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.
In this case: how the light reflects (and not bends).
I wonder how much more devious the question becomes if the farmer walks slower with a full bucket than with an empty bucket?
The cow is a sphere, why doesnt the farmer just roll it over to the water?
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.
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.
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
Just apply Snell's law
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.
I think it’s even simpler to extend the large triangle down to F’. You end up with a base of 4, height of 8. Then the small triangle’s ratio is proportional: 4:8 = (4-x):6. x must equal 1.
Apologies if someone else in comments pointed this out.
Such a beautiful simplification.
I would reflect the point F to the opposite side of the river line (water surface is reflective, isn't it?) and connect the resulting point F' and point C with straight line. The resulting triangles are similar in scale 2:6=1:3. So, x+3x=4 and x=1.
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.
"we can try that with a real dog and see what happens" - was REALLY hoping this was going to be in the Extra Footage
Professor Stankova is awesome!
Yes, the calculus way works when it's not all consistent throughout the problem. Like when different materials are involved that light moves through, etc. The calculus way does what calculus always does - it lets you solve a problem where conditions along the solution trajectory are changing.
Re Brady’s question on instinctually finding X. I’m imagining a scenario where the farmer can see the cow and river, but have no measurements. I think it’s solvable via bearings and perspective (as the farmer’s vision is vertically above the plane) The direction to point A is determinable (the river should have left-right symmetry when looking at it). The bearings FA and FC can then be drawn from a point on the ground. The location and bearing of point B can be determined by imagining a line from C to the river, and using the same perspective technique to determine the perpendicular from F to that BC line. That line can be extended to point C’ which is determinable also using perspective. Then the farmer just heads towards C’.
I think the question is more human than this. As in, the farmer could instinctually A: want to get to the water as fast as possible, B: minimize total distance, C: minimize distance walking with a full bucket, D: not care at all and just wing it.
So technically, this problem becomes a social experiment when you ask that question lol.
I guess I'm describing how I'd go about it, though yes... I would sort of wing it and not do the exact measuring. Exact methods wouldn't be realistic in practice as the plane would likely not be perfectly flat, and the river not perfectly straight. However, when looking around, I would make a crude mental effort to estimate where the point C' would be and head for it.
You're option C though is probably smarter than option B. That didn't occur to me.
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 -
Real farmer would probably drive his 4x4 directly from F to C, then shoot the cow to put it out of it's misery.
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
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
Use the reflection of one of the points across the river and draw a straight line.
Great video ! The geometric approach can also be done using Thales's theorem to find the value of x, instead of similar triangles.
4:04 me: *sips coffee and listens intently*
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.
Hang a chain between F & C and then see where the chain touches between A & B. Shortest distance is the curve shape the chain makes??
Yes! I love Prof Skankova!
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?
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!
I've got x = 1 and x = -2 , the second solution is where the short hypotenuse overlaps with the long one.
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
Thank you!
17:32 I believe this is a specific case covered by Snell's theorem/ law.
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.
19:20 No, the farmer wouldn't necessarily get to (or near) the right point by "instincts". People frequently end up "getting off course", getting lost, going the wrong way, and so on. Instincts can't deal with stuff like this very well, mostly because of the scale. If it were a small distance where you could see the entire thing at once, you might get close, but once you're on the ground and going long distances, seeing only a bit at a time, things get off track pretty quickly and easily.
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.
So the cow / bucket story = reflection and the dog / frisbee story = refraction. Are there animal analogues for all physics problems?
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.
The slope of F’C is easy to compute so the intersection, without triangle comparison
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 :)
Very nice geometrical solution, but I personally find it more fun if the problem was approached functionally - find a function f(x) = ax + b connecting phantom farmer and a cow. Or! Knowing that the light takes a shortest path, give farmer a laser pointer and tell him to point it at the river in. a way that a cow is illuminated by the laser :)
Another simple solution (I think). Imagine the diagram is a side view into a basketball court. The shortest distance will be the one that bounces the ball right up to the goal (assuming no gravity). This bounce point will have the same angle either side of a horizontal line straight upwards, otherwise the ball would miss the goal.
I once wrote a term paper on Euclidean geometry (ruler and compass construction), so I immediately came up with the simple solution. This is how everyone used to do division. Too bad the answer is an irrational number, otherwise I would have solved it all in my head without needing a calculator. :D
You can even explain it basically without any maths: The farmer spends some time going sideways and some time going towards the river and back. The optimal path is the one where those times are proportionally split, so since one quarter of the vertical travel time is towards the river and three quarters are back up, one quarter of the time should be spent on the way down and three quarters of time on the way up. Moving at constant speed sideways means he should move one quarter of the way to the right while he goes towards the river. Almost sounds trivial when described like that, but you have to have that idea first (or recognise it from an earlier problem). ;)
With the mirroring solution, you don't even need to find the point where the farmer reaches the river to get the optimal length, it's just one big triangle.
I would like the solution for the Stairs "rolling" at the wall and floor what is the velocity relations of the ends. Or the problem of will a "car hit the curb when descending at an edge (thing strain level, edge, then, straigh down at some angle -- maximum angle that the car clears?)
my first thought was ratio of distance to river and distance in between should be same to allow shortest path. but i don't can't think of how to prove its the shortest path.
18:00 "The farmer is doing exactly the same thing as the light is doing" - does that imply that the farmer is bright?
7:29 I see something rare while appling the square root; from what i know SQRT(A^2) is +A and -A:
(-A) * (-A) = A^2 and also (+A) * (+A) = A^2
If you apply square root, a +/- must be added.
Better shown with and example: The square root of 4 is 2, but it is also -2, since (-2)*(-2)=4.
Using sqare roots gives two results, positive and negative, not just one.
What am i missing? Why you did not put such +/-?
x is between 0 and 4, which means both x and (4-x) are positive
In quantum mechanics the fact that "light takes the shortest path of reflection" it's because it's more statistically likely. Particles can take wacky paths that don't match straight trajectories (only when the wave function is allowed to "spread", that is, it never collapses to a point)
omg, Zvedelina Stankova! Yes! She's so great!
At first I thought: Well light takes the quickest path and it also reflects on a surface. So it must be the same angle in and out. -> 9:30
I would think the farmer would run somewhere between the proper point and the middle of the river/line segment. If he's in a hurry, he'd probably think "I'm going between two places, I need to stop halfway between."
I had a hunch it was about similar triangles, but yup I immediately started on the calculus to check if my measurement was right. Duh, Zvezda's reflection solution is so clever!