@@nunkatsu "and as their cherished memories transformed into a curse, stinging them with a moment in time they can never return to, they parted ways at a rate approximately 5.099 feet per second. More accurately it was the square root of 25 feet per second as he was walking due north at 5 feet per second and she--"
In all honesty if I were that teacher I would have made up that story just to mess with the kids. There would be a piece of the story in every assignment and then the grand finale would be on the test!! 😁
Problem will solve itself. Because if they keep going straight from the direction in which they started moving, and somehow manage to keep the same rates (on average) over time, then with a bit of luck they'll meet eachother again at the same place after 4 years and 59 days.
I made a similar question for students that was rather funny. A dude leaves his house to visit his “family,” traveling straight North at some rate;his girl finds out he is actually heading to his lil boothang’s house(he’s cheating on her), so 1 hour after he leaves, all of his stuff is in a moving truck headed East at a certain rate cuz she kickin his ass out(at least she had the courtesy to secure a vehicle for his stuff instead of just dumping it all on the sidewalk). The question is how fast is the distance between this man and his PS5 increasing 30 minutes after the girl sends the truck away with his stuff?
As another comment thread here shows, they are not unnecessary constraints. They rule out the people who thought that this "place" is the interior of a toroidal spaceship, for example, because the question requires days and rain. And a pair of flatlanders on an infinite 2D plane is ruled out by the requirements of two perpendicular walkable directions and something being walked on rather than in. The seemingly trivial 2s time makes the roughly spherical geometry of a planet and relativistic effects negligible. No commenter has gone for a 3D hyperbolic space as I write this, but that is no doubt just a matter of time. Have you _met_ mathematicians? They will assume that the question is set in two otherwise disconnected spaces that only intersect at the origin, or a taxicab distance metric (ruled out by the pedestrians), if you do not add seemingly irrelevant stuff to the question. (-:
@@JdeBP But you have to abstract out the irrelevant detail. I can tell you how to run a perfect chicken-farm, just as long as you assume perfectly spherical chickens laying perfectly spherical eggs.
I don't think enough value is placed on the fact that it was a rainy day. A more interesting problem would have been how fast could they each have moved on a sunny day, and would the boy cry as hard? Taylor Swift mentions a lot about rain in her songs. All this should be part of the calculus.
"I'll never let you see the way my broken heart is hurting me I've got my pride, and I'll know how to hide all my sorrow and pain I'll do my crying in the rain." - The Everly Brothers, 1962
Since the Earth is approximately spherical, the correct answer should be slightly smaller than the square root of 26, decreasing as they get further apart due to the curvature of the Earth's surface (assuming the distance between them is measured in a straight line right through the Earth). A special case is at or near the South pole, where the girl would be spinning around very fast, each part of her body spinning Eastwards around the South pole, or running in very small circles; and the boy would move away from her ar 5 feet per second initially, dwindling to 0 when he nears the North pole (which would take approximately 25 months). To be precise, as the Earth is not a perfect sphere bus slightly flattened though the effects of it spinning, the boy would be farthest away from the girl at several hundred kilometers away from the North pole and from there on the distance would decrease slightly. At the North pole the girl would be spinning in the other direction, but also Eastwards.The boy would not know what to do as he can't get any further North and the girl is in the way. It would be an awkward kind of separation. The devil is in the details with these problems.
@@joostvanrens I assume she spins around by moving her feet. The parts of her body that are at or near the Earth's and her own rotational axis would not move Eastwards fast enough, but the outer parts might move faster so on average the atoms in her body would move Eastwards at exactly 1 foot per second. Of course all of this is hypothetical. I don't think it is well advised to do this as part of ending a relationship. Also, the boy might complain about having to go to the South pole just to discover he's part of a breakup and a dancing dervish act. I mean, she could also just have told him instead of going through all of that?
It doesn't say that they met, and separated, on this planet. It could've been on a spaceship with perfectly flat floors, with compass directions arbitrarily assigned for easy navigation.
@@SgtSupaman On the South pole she'd still be spinning round even after just 2 seconds and the boy would distance himself from her ar only 5 feet per second. I would run, but hey.
The teacher likely intended the problem to be solved the following way: call the horizontal distance traveled by the girl is x, the vertical traveled by the boy y, and then we have x^2 + y^2 = z^2, where z is the distance between them. Take the derivative with respect to time and simplify to get x(x') + y(y') = z(z'). Find x, y, and z at 2 seconds as in the video and then solve for z'. Since the rate of change is constant the methods shown in the video are perfectly valid.
Yes, this is the general method for solving this kind of problem. In the general situation, z(t) = sqrt(x(t)^2 + y(t)^2) and the simple approach would require using the chain rule on that formula. But it's much simpler to use implicit differentiation. The second approach of the video is only viable because x' and y' are both constants.
I greatly appreciate how this video focused solely on the math of the problem and completely ignored the likely reason it went viral in the first place. XD Seriously, there's a story behind that word problem. I wonder what it is.
2:03 NO let's NOT! Time is irrelevant here since there's no acceleration ever mentioned except instantaneous at t=0. After that speed and direction are fixed vectors and COMPLETELY independent of time!
Huh? But he's trying to find the distance at time t not the velocity at time t. Yes, velocity will not change since there is no acceleration, but the distance most definitely will as he's shown in his solution.
@alfredoprime5495 easy for the general person like us - I'm absolutely appalled that Presh feel for it hook line and sinker in this video - TWICE!!!! (2 different methods here) I'm deducting more marks for BOTH answers here than I would for some who used the correct method but got the wrong answer - excellent value in showing AND MARKING correct working. Presh's method here - especially that table in method 1 that ended up with 3 out of 3 IRRELEVANT columns - is 100% wrong! Correct answer or not! GOOD GRIEF!!!
Not sure what the point was in determining the distances at various times. Neither of their vectors or velocities are changing, so their relative velocity is also constant.
Sometimes questions are written with a lot of unnecessary information and data so the student is challenged to extract only the relevant data and do the correct calculation.
This is a related rates problem. The way to solve is to use implicit differentiation on Pythagorean theorem a²+b²=c² with respect to t which gives 2a(da/dt)+2b(db/dt)=2c(dc/dt). At t=2, a=10 b=2 c=2√26 da/dt = 5 db/dt = 1 Solve for dc/dt
The problem does not say which way the boy is running! It just says he's due north. It also never says that they're walking or running at a constant speed. It just gives their current speed, and without saying how long ago they started separating. Finally, with the assumptions of constant speed and boy moving due north, the solution is too easy, and doesn't require "differential calculus." Terribly worded problem.
Yours is the best criticism of the wording of this problem. The writer of the math problem likely got caught up in the emotional memory of an actual personal experience and forgot how to math. His relationship problem is as unsolvable as his math riddle! 😂
Although, the theme might have something to do with that. Either they mixed it up real quick unable to focus, or it isn't a real problem someone gave out for classes.
It doesn't say that the boy is running due north, but that he is due north. If he is running at an angle so that he remains due north of the girl, then his velocity becomes the hypotenuse of the triangle. The answer here would be 2(sqrt 6) or about 4.9 ft/sec.
@@kdemetterstudents could relate to?If I got that question on my 11th grade math final I’d be running in tears out the classroom(because my bf had dumped me a day prior) So glad that’s a year ago now
@@oreo_6206 Then you agree that students can relate to it :-) I didn't say everyone would like it. It could indeed suck for someone who was just dumped. Relatable applies to all emotions. My point is : this teacher was trying to make content students can relate to, rather than the ultra boring content you typically have. A teacher that tries to make the lesson/exam more fun and exciting is trying to be a better teacher than one who just sticks to safe, boring lesson content. Trying something new is riskier of course. It can misfire, and achieve the opposite effect. But the teacher can learn from that and become a better teacher. Whereas the one that just sticks to the boring safe stuff never grows. A school should be an environment where everyone ( teachers and students) can experiment, make mistakes and learn. That means you may sometimes unintentionally offend someone. That's ok. The alternative (unfortunately most schools and classes today) creates a sterile place where nothing can be risked out of fear of causing offense. That's a place where no real learning can happen, where all creativity and fun is crushed. That produces obedient slaves, not resilient, creative and flourishing human beings.
I would argue it's not really a "trick question", because it's really just testing the student's ability to determine reasonableness of the solution (and to recognize when they are working with a linear result). If you think about it geometrically, each point in time is describing increasing _proportional triangles._ That means that if each of the two sides is increasing linearly over time, the third side must also be increasing linearly too. Therefore it actually makes intuitive sense that the answer should be the same regardless of what point in time is used.
There is another possible solution to this problem, depending on your interpretation of the wording. The problem does NOT say the boy is running north. It says that he IS due north. That can be interpretted as not being an indication of DIRECTION, but an indication of LOCATION-that is, he is traveling 5ft/s, and is maintaining a location due north of THE GIRL This would mean that after one second, the girl is one foot east of start, and the boy has traveled 5 feet on a trajectory that he is now due north of her location. This means that on the triangle Presh draws, the 1 foot is still the base, but the 5 feet is now the trajectory-meaning the other leg, which is the distance they are separated (and speed they are separating) is sqrt(24). I agree that the problem’s author most likely meant that the boy is RUNNING due north-but that’s not what he said. He said the boy IS due north.
Well, if you're ready to admit that girl emits her own electromagnetic field by which we can define "north" side relative to her, then yeah, it may be a solution...
@@lazyvector what? north relative her means toward the north pole from her, just like north from the point where they started would be toward the north pole from that point. how does making it due north from her instead of the point where they started mean north is suddenly not defined by the north pole?
This one made me laugh - in all other similar problems by Presh, the extraneous information given turns out to be highly relevant, so instead of just solving it (easy), I was left sitting there wondering how the facts that the boy was crying and that it's a rainy day etc. would end up influencing the answer.
How fast are they separating is a very interesting question, since the type of separation was not specified. There are multiple answers: Physically -> [26^1/2] feet/sec or ~3.48mph. Though his run is more of a walk, at 3.41 mph, and her walk is more of a crawl at just over 2/3 of a mile per hour. Emotionally, Financially, Socially -> Unable to answer from the information given
This is an example of question you could give after students read one page of an introductory calculus text. For a linear function the derivative is the slope.
This reminds of this question that's in my module for JEE Advanced: Once upon a time in the Lush Green romantic village of the punjab, there were two lovers: Soni and Mahiwal. They were deep in love but the society was against them as they belonged to different communities. So they had to meet secretly. Soni and Mahiwal are living on the same side of river bank 3 km apart. The river flows with a velocity 2.5 km/hr and is 3km wide. Both of them have a boat each which can travel with a velocity of 5 km/hr in still water. On the first day they decide to meet on the same bank as they live. They start at the same time. Soni travels upstream and Mahiwal travels downstream to meet each other. On the second day they decide to cross the river to meet on the other side of the bank. Mahiwal rows the boat at an angle of 90° to the river flow. Q.20: What is the time for which they row the boat till they meet on the first day? Q.21: On the second day, what is the angle at which Soni should row the boat (with respect to the river flow) to reach the same point as Mahiwal on the other bank?
1:34 isn't time an unnecessary dimension here? Just follow Pythagoras theorem and hypotenuse will be in speed - what the question asks, right? Asking for the speed after 2 seconds is irrelevant since speed here is uniform (COMPLETELY independent of time since there's no acceleration mentioned in the question). Let's not complicate the solution by adding irrelevant time.
I was wondering that too. Though I guess the aim is to demonstrate the student can construct d(t) = t√(5²+1²), differentiate it and solve for t = 2. Because the question can be further tweaked to make the girl walk in a sine wave pattern along the easterly axis and now you (I think) have to use calculus (haven't done it but I imagine the rate would have a cosine in it somewhere 😂)
My attempt: Okay, so the boy is heading due north and the girl is heading due east, those are perpendicular directions, and if we are measuring the distance between two points on perpendicular lines we are creating a right triangle. At 0 seconds, the triangle's height and base are 0 ft, so the distance between the boy B and the girl G is 0. At 1 second the triangle's height is 5 feet, and the base is 1 foot (we can say point B is 5 feet north and point G is 1 foot east). The distance between B and G is the hypotenuse of this right triangle. So good old Pythagoras tells us the length of the hypotenuse of a right triangle is the square root of the sum of the height squared and the base squared, so BG = square root of (B^2 + G^2). So BG (the distance between Boy and Girl) is square root of (5^2 + 1^2), square root of 26 is 5.0990195... so BG is about 5.099 feet. At 2 seconds, the same formula holds, but the triangle is now bigger, because B is 10, and G is 2. So BG is now square root of (10^2 + 2^2), square root of 104 is 10.19804.... if we want we could build a table of these values with one row per second which would look like this (T is *time*, number of elapsed seconds): T, B, G, BG 0, 0, 0, 0 1, 5, 1, 5.099 2, 10, 2, 10.198 3, 15, 3, 15.297 4, 20, 4, 20.396 But something interesting happens if we look at the difference between each value of BG and the previous value of BG (that is BG(t) minus BG(t-1)), you always get 5.099. The distance between B and G is growing at the rate of 5.099 ft/sec no matter what second you pick, so I think the answer is at 2 seconds they are separating from each other at 5.099 ft/sec. Now to unpause and see how wrong I am! 🙂
_Teacher sipping coffee talking to colleague in staff room_ "Your questions enter their mind, my questions enter their souls..." Students... "Did you figure it out?" "Yeah...but is the guy okay after the breakup??"
Ive seen a bipartite graph theory question where a professor is trying to figure out how to great groups to perfectly split up friends so that no one was friends with anyone in their group.
I just wonder how many of the students got distracted by the story and missed the fact that the question didn't ask "how far they are separated" but "how FAST they are separatING"
The only problem I can see here is the fact that the two persons are not moving on top of a straight plane area, but on the surface of sphere. That would give a whole different meaning to the directions they are walking and the angle between their movement vectors. But I very much doubt the teacher thought about that .
They might have thought about that, then realised that we're dealing with two people who are a smidgen over 10 feet apart on a sphere that's approximately a bazillion feet across.
The problem never said the boy was running due north only that he was due north and crying. It was very specific that the gril was walking east. How do we know that the boy is not running toward the girl to try and get back together?
The fact that the question specifies that they are separating. Don't underestimate the necessity to the problem statement of all of the seemingly irrelevant chaff in the question. (-:
Unanswerable because the boy’s position is due north but we do not know what direction he is running. If it said he is running due north that would work but it doesn’t it says is due north and running. Further they are both moving at a constant velocity therefore this is a trigonometry problem, not a calculus problem. Or it could say the boy is moving due north. That would also work.
This is clearly to test your mathematical skills in the face of emotional turmoil provided that you happen to be going through a break up when taking that exam.
basic kinematics can be solved in a single step with relative velocity, the relative velocity of boy wrt to girl is (5i-1j)ft/sec or in magnitude sqrt(26)ft/s therefore the boy and the girl are separating at the rate of sqrt(26)ft/s. would've been a more fun problem if they added acceleration for both or one of the person
I notice that you can answer this with vector components. The line between former lovers defines a direction. If you take the component his 5 ft/s along the line and add it to her component of 1 ft/s along the line, you get 4.9029 + 0.1961, which is 5.0990
Legend is obviously not a mathematician. They twain are in a non-Euclidean geometry with positive curvature and walking in straight lines. (They cannot be on an infinite flat Euclidean plain, because the question specifies that the "place" has days.) Straight lines _always meet at two points_ in such geometries, unless they are co-linear, which the question _also rules out_ by implying that they are perpendicular. They will meet again in the future, over and over if they do not stop walking in their straight lines.
Assuming we don't take the curvature of the earth into consideration, nor the fact if they are traveling on the same plane (neither going uphill or downhill) after 1 second they'd have been √(5²+1²) feet apart , so approximately 5.1 feet. Since they're both traveling at a constant speed, every second they will move 5.1 feet further away from each other. Not sure why the problem asks for how fast they are separating from each other after 2 seconds, the time is irrelevant and the question might as well be "how fast are they separating from each other". But then again, this is quite a weird problem anyway.
The solving is really overcomplicated. Since all velocities are constant, you can take Vboy - Vgirl (the hypotenuse of the right triangle with legs 5 ft/s and 1 ft/s) to get the answer, √26 ft/s.
A physics professor told me that in another class he teaches, there was a problem about a girl being kidnapped and how to find the rate of the van she was in or something along those lines
The boy isn’t even running: He’s traveling at 5ft/sec, which means he’s going at 300ft/min or 18000ft/hr. We know a mile is 5280ft long, so dividing 18000ft by 5280ft leaves us with about 3.41mph (miles per hour). The average walking speed of an adult (which he’s likely to be, unless they got together when they were younger than 10, and if that’s the case how on earth did they last that long together?) is about 3mph, with a range going as high as 4mph and as low as 2.5mph. So 3.41 would be a regular walk, or at most a light speed walk. In short, according to my calculations, that man isn’t even running, just a light speed walk in the rain. Additionally: Why the crap is the girl walking so slowly?! She’s moving at 3600ft/hr (1ft/sec = 60ft/min = 3600ft/hr) which is 0.68mph! That’s far below the average walking speed!!! Heck, the average army crawling speed is 0.5meters/sec, and with 1m being about 3.28ft, that would be 1.64ft/sec!!! She’s literally walking slower than someone army crawling!!! Who walks that slowly?! Especially since SHE’S the one who broke up with HIM!!! Like, I know breakups are hard, but if you’re gonna be so devastated I could beat you in a race where you’re walking and I’m army crawling maybe you need to rethink some life choices.
I was half expecting that we gonna need to take account the curviture of the earth and take the closest distance in the 3D space rather from closest earth's surface path
The separation speed will not be constant if the curvature of the Earth is significant compared too the distances travelled. Fortunately it isn't if we are willing to approximate sqrt(26) as 5.099.
to solve the problem (in this specific scenario) you don't even have to find each length, you only need to use the pythagorean theorem to get sqrt (26)
Just noticed that if second 1 makes the first triagnle, second 2 extends this triangle 4x (add one to the top, one to the side, and one fills the empty space in the middle), 3d second expands it 9x (build extension simmilary as in 2s), 4th second expands it 16x. So in each case the number of triangles building the big one is t^2. That would be a nice visualization of that case
I looked at the thumbnail and just read the first sentence of the problem. Then I shifted my eyes up towards the name of the course and physically jumped when I saw “Differential Calculus” 😭
Next question : assuming they are walking on a sphere with a circumference of approximately 24,901 miles , how long would it take them to meet again ? Next next question : assuming they meet up again, what is the probability that they will stay together. Assume the boy brings chocolates or flowers and calculate for both
@@kdemetter Timespan for girl to complete ½ of the circumference of spherical Earth: [(½ circumference) × (24,901 miles per circumference) × (5,280 feet per mile) ÷ (1 ft/sec)] ÷ (3,600 seconds per hour × 24 hours per day) = (760.86388888...) days Timespan for boy to complete 2½ of the circumference of spherical Earth: [(2½ circumference) × (24,901 miles per circumference) × (5,280 feet per mile) ÷ (5 ft/sec)] ÷ (3,600 seconds per hour × 24 hours per day) = (760.86388888...) days ==> They'll encounter each other at the _antipode_ of their break-up location in about 761 days.
@@kdemetter Assuming they are both keeping a constant rate of moving, and that they keep moving straight on after heading in the direction when they started moving: Timespan for girl to complete ½ of the circumference of spherical Earth: [(½ circumference) × (24,901 miles per circumference) × (5,280 feet per mile) ÷ (1 ft/sec)] ÷ (3,600 seconds per hour × 24 hours per day) = (760.86388888...) days Timespan for boy to complete 2½ of the circumference of spherical Earth: [(2½ circumference) × (24,901 miles per circumference) × (5,280 feet per mile) ÷ (5 ft/sec)] ÷ (3,600 seconds per hour × 24 hours per day) = (760.86388888...) days ==> They'll encounter each other at the _antipode_ of their break-up location in 760 days, 20 hours and 44 minutes, _approximately_ .
@@kdemetter Assuming they are both keeping a constant rate of moving, and that they keep moving straight on after heading in the direction when they started moving: Timespan for girl to complete ½ of the circumference of spherical Earth: [(½ circumference) × (24.901 miles per circumference) × (5.280 feet per mile) ÷ (1 ft/sec)] ÷ (3.600 seconds per hour × 24 hours per day) = (760.86388888...) days They boy completes 2½ of the circumference at 5 times the girl's moving speed, hence also arrives at the antipode of their break-up location in the same number of days.
If you want to make the question more interesting, have the girl leave first, and the boy leave 5 seconds later. Let t=0 be the point when the BOY starts moving and evaluate at t=2.
My brain simply went "the rates are in separate axes, so the hypotenuse is the rate we are looking for". There was no need to calculate distance traveled or time in this problem. It's just simple use of the Pythagorean theorem.
I was going to mention that it depends how close to the North pole they are, but I see that's already been covered. So instead I'll postulate that perhaps they met at a gym 8 long years ago, and maybe the boy is running at 5 ft/sec on a treadmill. Hence why the direction he is running doesn't matter, and we're only told he "is due north". They're therefore separating at the speed the girl is walking away, 1 ft/sec.
This and the comments taught me something that i might have missed in geometry is i did learn this (i suck with triangle hypotensuses), i never knew that the hypotenuse remains an equal value in root form when scaled proportionally (idk how common knowledge it is, i just never thought of it like that and im sleepy so idk)
Here is another way to doing it ---> x is distance walked by the boy y is the distance walked by the girl after 2 seconds which is 10ft and 2ft as 5ft/s and 1ft/s are there speed mention in the question by Pythagoras theorm we can right x^2+y^2=z^2 so on differentiating the equation wrt time we get 2x(dx/dt)+2y(dy/dt)=2z(dz/dt) cancel 2 on both the sides and we get x(dx/dt)+x(dy/dt)=z(dz/dt) x=10ft y=2ft z=sqrt(10^2 + 2^2) (dx/dt)=5ft/s (dy/dt)=1ft/s Put everything into differential equation and we get dz/dt=52/sqrt(104) which approximates to 5.099ft/s 💔.
![I.N "HALLUCINATION" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/n5B5q1Hwt_U/mqdefault.jpg)
There's making your relationship issues other people's problem, and then there's making your relationship issues other people's calculus problem.
The other possibility is that the calculus teacher was bored or writing a romance novel
I love this lmfao
It's more interesting
lol
Find the "x" be like
Teacher: Math is everywhere.
Student: It can't possibly be everywh-
Teacher: *Math is everywhere.*
Why did I read this as "Teacher: meth is everywhere Student:it can't possibly be everywh- Teacher: *meth is everywhere*
@@shafaayraaj9196 xd lol
@@shafaayraaj9196 at this point. it is probably interchangeable
@@sayorancodeWAIT WHA-
@@shafaayraaj9196
you’re onto something ❌
you’re *on* something ✅
Clearly the answer to the question is "Not fast enough."
who hurt brk
I'm pretty sure the dude is accelerating in real time. He just starts out slow
@@EliteCameraBuddy who is "brk" ?
@@samanthakarunarathna4838 brk means bark, he is just furry
@@idkwhatwritehere000I thought brk was a misspelling of bro
i am more interested in the passive aggressive backstory of this exam writer.
The teacher should start writing romantic novels
@@nunkatsu "and as their cherished memories transformed into a curse, stinging them with a moment in time they can never return to, they parted ways at a rate approximately 5.099 feet per second. More accurately it was the square root of 25 feet per second as he was walking due north at 5 feet per second and she--"
In all honesty if I were that teacher I would have made up that story just to mess with the kids. There would be a piece of the story in every assignment and then the grand finale would be on the test!! 😁
@@Thetruthgirl that idea actually is pretty sick
@@Thetruthgirl that’s actually a really good way of teaching. The only issue is finding a topic they’d find interesting
"How can we solve this problem?" Therapy, definitely therapy.
Demn bro 💀
Get drunk! Play Zelda!
Yes, a better help promotion would be appropriate on this one
Problem will solve itself.
Because if they keep going straight from the direction in which they started moving, and somehow manage to keep the same rates (on average) over time, then with a bit of luck they'll meet eachother again at the same place after 4 years and 59 days.
I came down to the comments to make this same joke. Well played.
In my experience, neither boys nor girls who have just broken up with each other run or walk in an exactly straight line.
Personally, I’ve always preferred zigzags
Assume the penguin is a cylinder
Yes, I prefer following a hyperbolic trajectory given by xy = c² in a coordinate system where the point of break up is the origin
@@IntrepidFC
Ah yes, because what we need is a zig-zag lined triangle. This teacher was going far too easy on them with a straight-lined triangle.
I have a feeling the guy was doing parkour as he was running to to relieve stress
Question ❌
Break-up story ✅
Story ❌
Strory ✅
Valuable lesson on real life scenarios: to be given unnecessary information and be obliged to use inappropriate method.
This is like info dumping in writing stories
“We don’t have to worry about negative distances here” Jeez that was smooth
I made a similar question for students that was rather funny. A dude leaves his house to visit his “family,” traveling straight North at some rate;his girl finds out he is actually heading to his lil boothang’s house(he’s cheating on her), so 1 hour after he leaves, all of his stuff is in a moving truck headed East at a certain rate cuz she kickin his ass out(at least she had the courtesy to secure a vehicle for his stuff instead of just dumping it all on the sidewalk).
The question is how fast is the distance between this man and his PS5 increasing 30 minutes after the girl sends the truck away with his stuff?
Cruel! lol
lmao, those problems are the kinds of problems that make math tests enjoyable.
LOVE IT!😂😂😂😂😂
That's the way 😂👍
Hell yeah.
Bro got a lit too personal 💀
That was not the bro, bro.
@@BKNeifert Being 'bro' is not about the gender, bro.
@@u2bear377 Ah, okay.
@@BKNeifert you forgot to add bro, bro
Sis got a little too personal
As an engineer, I'm quite professional in clipping bs and analyzing only the necessary constraints
Me too
As another comment thread here shows, they are not unnecessary constraints. They rule out the people who thought that this "place" is the interior of a toroidal spaceship, for example, because the question requires days and rain. And a pair of flatlanders on an infinite 2D plane is ruled out by the requirements of two perpendicular walkable directions and something being walked on rather than in. The seemingly trivial 2s time makes the roughly spherical geometry of a planet and relativistic effects negligible. No commenter has gone for a 3D hyperbolic space as I write this, but that is no doubt just a matter of time.
Have you _met_ mathematicians? They will assume that the question is set in two otherwise disconnected spaces that only intersect at the origin, or a taxicab distance metric (ruled out by the pedestrians), if you do not add seemingly irrelevant stuff to the question. (-:
The necessary constraint: if you want to get laid put your calculator away.
@@JdeBP But you have to abstract out the irrelevant detail. I can tell you how to run a perfect chicken-farm, just as long as you assume perfectly spherical chickens laying perfectly spherical eggs.
@@someonespadre ???
I don't think enough value is placed on the fact that it was a rainy day. A more interesting problem would have been how fast could they each have moved on a sunny day, and would the boy cry as hard? Taylor Swift mentions a lot about rain in her songs. All this should be part of the calculus.
rainy day = floor is slippery = friction is negligible
@@cloverisfan818I love this reply 😂😂
"I'll never let you see
the way my broken heart is hurting me
I've got my pride, and I'll know how to hide
all my sorrow and pain
I'll do my crying in the rain."
- The Everly Brothers, 1962
Irony, like rain on your break-up day.
Brakeup = calculus -> physics
Since the Earth is approximately spherical, the correct answer should be slightly smaller than the square root of 26, decreasing as they get further apart due to the curvature of the Earth's surface (assuming the distance between them is measured in a straight line right through the Earth).
A special case is at or near the South pole, where the girl would be spinning around very fast, each part of her body spinning Eastwards around the South pole, or running in very small circles; and the boy would move away from her ar 5 feet per second initially, dwindling to 0 when he nears the North pole (which would take approximately 25 months).
To be precise, as the Earth is not a perfect sphere bus slightly flattened though the effects of it spinning, the boy would be farthest away from the girl at several hundred kilometers away from the North pole and from there on the distance would decrease slightly.
At the North pole the girl would be spinning in the other direction, but also Eastwards.The boy would not know what to do as he can't get any further North and the girl is in the way. It would be an awkward kind of separation.
The devil is in the details with these problems.
At the south pole the girl couldn't be walking east
@@joostvanrens I assume she spins around by moving her feet. The parts of her body that are at or near the Earth's and her own rotational axis would not move Eastwards fast enough, but the outer parts might move faster so on average the atoms in her body would move Eastwards at exactly 1 foot per second.
Of course all of this is hypothetical. I don't think it is well advised to do this as part of ending a relationship. Also, the boy might complain about having to go to the South pole just to discover he's part of a breakup and a dancing dervish act. I mean, she could also just have told him instead of going through all of that?
It doesn't say that they met, and separated, on this planet. It could've been on a spaceship with perfectly flat floors, with compass directions arbitrarily assigned for easy navigation.
Well then, I guess it's a good thing she asked specifically about their separation speed at 2 seconds so that none of this could come into play.
@@SgtSupaman On the South pole she'd still be spinning round even after just 2 seconds and the boy would distance himself from her ar only 5 feet per second. I would run, but hey.
The teacher likely intended the problem to be solved the following way: call the horizontal distance traveled by the girl is x, the vertical traveled by the boy y, and then we have x^2 + y^2 = z^2, where z is the distance between them. Take the derivative with respect to time and simplify to get x(x') + y(y') = z(z'). Find x, y, and z at 2 seconds as in the video and then solve for z'. Since the rate of change is constant the methods shown in the video are perfectly valid.
Yes, this is the general method for solving this kind of problem. In the general situation, z(t) = sqrt(x(t)^2 + y(t)^2) and the simple approach would require using the chain rule on that formula. But it's much simpler to use implicit differentiation. The second approach of the video is only viable because x' and y' are both constants.
Nah the teacher is just crying over there
Running vertically? That's an impressive trick if you can do it.
Is it 12 ft
Yeah this is definitely intended to be a related rates calc 1 problem
Mitochondria is the powerhouse of the cell
damn, so it's international
The REAL answer
and it's false , the mitochondria is more than the powerhouse of the cell . It's a cell inside a cell
"How do we solve this problem?" Well, when did the relationship dissolve? Was there yelling involved?
Oh! Not *that* problem.
thats important too
The real problem here is using feet/second as a unit
You are right they should be using parsecs per nanoseconds.
@terryendicott2939 no. They should have been using lightyear per millenia
They should have been using Kelvin
Exactly. Real schools have been teaching only metres, litres and kilos for more than years now.
Not American enough! Should've been football field per eagle screech
I greatly appreciate how this video focused solely on the math of the problem and completely ignored the likely reason it went viral in the first place. XD
Seriously, there's a story behind that word problem. I wonder what it is.
Teachers make up fun scenarios all the time to better engage the students
@@Un1234lthe probably correct answer
The teacher mentally keeping track of each others speed, distance, compass directions and time during that pivotal moment.
2:03 NO let's NOT! Time is irrelevant here since there's no acceleration ever mentioned except instantaneous at t=0. After that speed and direction are fixed vectors and COMPLETELY independent of time!
Huh? But he's trying to find the distance at time t not the velocity at time t. Yes, velocity will not change since there is no acceleration, but the distance most definitely will as he's shown in his solution.
@alfredoprime5495 why when it's NEVER asked for?
@@fifiwoof1969 ugh! you're right. Classic mistake of not reading the actual question.
@alfredoprime5495 easy for the general person like us - I'm absolutely appalled that Presh feel for it hook line and sinker in this video - TWICE!!!! (2 different methods here)
I'm deducting more marks for BOTH answers here than I would for some who used the correct method but got the wrong answer - excellent value in showing AND MARKING correct working. Presh's method here - especially that table in method 1 that ended up with 3 out of 3 IRRELEVANT columns - is 100% wrong! Correct answer or not!
GOOD GRIEF!!!
@@fifiwoof1969 he acknowledged the speed does not change with time, he just didn't want a 30 second video.
Not sure what the point was in determining the distances at various times. Neither of their vectors or velocities are changing, so their relative velocity is also constant.
Giving the time to be 2 seconds was already pointless in the original question.
Right on. Just a vector diagram of velocity would solve this, 5 north, 1 east, root 26 hypotenuse and you're done.
Why wouldn't you just add the vector velocities and then take the length of that vector? That will only take you ten seconds, if you're slow.
Just to show that the rate of change of distance is constant
Sometimes questions are written with a lot of unnecessary information and data so the student is challenged to extract only the relevant data and do the correct calculation.
Mathematicians' attempt on teaching an English class.
that is amazing story writing though, including the speed they are moving adds a whole new dimension of heartbreak
“…Where we solve the world’s problems, one video at a time”
This problem sort of gave the outro a new meaning
Didn't expect them to solve the problem rather than explaining the question
This is a related rates problem. The way to solve is to use implicit differentiation on Pythagorean theorem a²+b²=c² with respect to t which gives
2a(da/dt)+2b(db/dt)=2c(dc/dt).
At t=2, a=10 b=2 c=2√26
da/dt = 5
db/dt = 1
Solve for dc/dt
The problem does not say which way the boy is running! It just says he's due north. It also never says that they're walking or running at a constant speed. It just gives their current speed, and without saying how long ago they started separating. Finally, with the assumptions of constant speed and boy moving due north, the solution is too easy, and doesn't require "differential calculus." Terribly worded problem.
Yours is the best criticism of the wording of this problem. The writer of the math problem likely got caught up in the emotional memory of an actual personal experience and forgot how to math. His relationship problem is as unsolvable as his math riddle! 😂
And what if the ground they’re walking on is not flat, would t that affect the distance they are separating?
North is up, π = e = 3, and sin(x) = x
Although, the theme might have something to do with that. Either they mixed it up real quick unable to focus, or it isn't a real problem someone gave out for classes.
Whoever made this question surely had to be speaking from experience
It doesn't say that the boy is running due north, but that he is due north. If he is running at an angle so that he remains due north of the girl, then his velocity becomes the hypotenuse of the triangle. The answer here would be 2(sqrt 6) or about 4.9 ft/sec.
Good observation. It states he is due north, but not which way he is running.
we can assume he is running north also
The teacher didn't have to go this hard for an exam question, lol!
It's from a calculus class. This was probably one of the easier ones.
@@Gruuvin1 Sounds like its a fakeout, given that the answer is constant
I suppose it could be worse but I'm kind of intrigued by horrible and morally reprehensible math problems.
MCQ. The teacher is a:
1) genius
2) troll
3) comedian
4) attention seeker
I think he was just trying to make a problem that the students could relate to.
5) romance novelist
6) All of the above
@@kdemetterstudents could relate to?If I got that question on my 11th grade math final I’d be running in tears out the classroom(because my bf had dumped me a day prior)
So glad that’s a year ago now
@@oreo_6206 Then you agree that students can relate to it :-)
I didn't say everyone would like it. It could indeed suck for someone who was just dumped. Relatable applies to all emotions.
My point is : this teacher was trying to make content students can relate to, rather than the ultra boring content you typically have.
A teacher that tries to make the lesson/exam more fun and exciting is trying to be a better teacher than one who just sticks to safe, boring lesson content.
Trying something new is riskier of course. It can misfire, and achieve the opposite effect. But the teacher can learn from that and become a better teacher. Whereas the one that just sticks to the boring safe stuff never grows.
A school should be an environment where everyone ( teachers and students) can experiment, make mistakes and learn.
That means you may sometimes unintentionally offend someone. That's ok.
The alternative (unfortunately most schools and classes today) creates a sterile place where nothing can be risked out of fear of causing offense.
That's a place where no real learning can happen, where all creativity and fun is crushed.
That produces obedient slaves, not resilient, creative and flourishing human beings.
i dont do math for higher studies, but this got me hooked till the end.💀
I would argue it's not really a "trick question", because it's really just testing the student's ability to determine reasonableness of the solution (and to recognize when they are working with a linear result).
If you think about it geometrically, each point in time is describing increasing _proportional triangles._ That means that if each of the two sides is increasing linearly over time, the third side must also be increasing linearly too. Therefore it actually makes intuitive sense that the answer should be the same regardless of what point in time is used.
There is another possible solution to this problem, depending on your interpretation of the wording. The problem does NOT say the boy is running north. It says that he IS due north. That can be interpretted as not being an indication of DIRECTION, but an indication of LOCATION-that is, he is traveling 5ft/s, and is maintaining a location due north of THE GIRL This would mean that after one second, the girl is one foot east of start, and the boy has traveled 5 feet on a trajectory that he is now due north of her location. This means that on the triangle Presh draws, the 1 foot is still the base, but the 5 feet is now the trajectory-meaning the other leg, which is the distance they are separated (and speed they are separating) is sqrt(24).
I agree that the problem’s author most likely meant that the boy is RUNNING due north-but that’s not what he said. He said the boy IS due north.
the only interesting point anyone in the comments made.
Well, if you're ready to admit that girl emits her own electromagnetic field by which we can define "north" side relative to her, then yeah, it may be a solution...
The boy is due north, not from the girl but from the location of the break-up.
@@lazyvector what? north relative her means toward the north pole from her, just like north from the point where they started would be toward the north pole from that point. how does making it due north from her instead of the point where they started mean north is suddenly not defined by the north pole?
@@yurenchu we know that is the interpretation used to get the answer described in the video, did you even read the comment?
Looks like malicious compliance, when you actually solve the problem mathematically and completely ignoring the backstory xD
Imagine if you solved the problem and told the teacher that the last problem was just as easy as her break-up?
This one made me laugh - in all other similar problems by Presh, the extraneous information given turns out to be highly relevant, so instead of just solving it (easy), I was left sitting there wondering how the facts that the boy was crying and that it's a rainy day etc. would end up influencing the answer.
It's awesome how he can explain without bursting into a laughter 👌👌
How fast are they separating is a very interesting question, since the type of separation was not specified. There are multiple answers:
Physically -> [26^1/2] feet/sec or ~3.48mph. Though his run is more of a walk, at 3.41 mph, and her walk is more of a crawl at just over 2/3 of a mile per hour.
Emotionally, Financially, Socially -> Unable to answer from the information given
I would run at a speed of 10 m/s if I had a problem involving imperial units! ❤
Not for more than 10 seconds or so - only a handful of men have ever run that fast, and only over 100m.
@craftsmanwoodturner exactly
OK, Mr. Bolt.
This is an example of question you could give after students read one page of an introductory calculus text. For a linear function the derivative is the slope.
I lwk thought this was related rates at first.
This reminds of this question that's in my module for JEE Advanced:
Once upon a time in the Lush Green romantic village of the punjab, there were two lovers: Soni and Mahiwal. They were deep in love but the society was against them as they belonged to different communities. So they had to meet secretly. Soni and Mahiwal are living on the same side of river bank 3 km apart. The river flows with a velocity 2.5 km/hr and is 3km wide. Both of them have a boat each which can travel with a velocity of 5 km/hr in still water. On the first day they decide to meet on the same bank as they live. They start at the same time. Soni travels upstream and Mahiwal travels downstream to meet each other. On the second day they decide to cross the river to meet on the other side of the bank. Mahiwal rows the boat at an angle of 90° to the river flow.
Q.20: What is the time for which they row the boat till they meet on the first day?
Q.21: On the second day, what is the angle at which Soni should row the boat (with respect to the river flow) to reach the same point as Mahiwal on the other bank?
Q20: They meet after 18 minutes on the first day.
Q21: Soni should row at an angle of 30° upstream with respect to the river flow.
I used chatgpt lool is it correct?
before listening to the complete question who thought there would be a role of those 8 years for evaluating the answer
Her parents wanted her to be a teacher, but that doesn't stop her inner wattpadder from adapting
For the record I am still traveling at 5ft/s and still crying.
now see what does your life have to be like to see a break up and think "i wanna make a related rates problem"
1:34 isn't time an unnecessary dimension here? Just follow Pythagoras theorem and hypotenuse will be in speed - what the question asks, right? Asking for the speed after 2 seconds is irrelevant since speed here is uniform (COMPLETELY independent of time since there's no acceleration mentioned in the question). Let's not complicate the solution by adding irrelevant time.
I was wondering that too.
Though I guess the aim is to demonstrate the student can construct d(t) = t√(5²+1²), differentiate it and solve for t = 2.
Because the question can be further tweaked to make the girl walk in a sine wave pattern along the easterly axis and now you (I think) have to use calculus (haven't done it but I imagine the rate would have a cosine in it somewhere 😂)
My attempt:
Okay, so the boy is heading due north and the girl is heading due east, those are perpendicular directions, and if we are measuring the distance between two points on perpendicular lines we are creating a right triangle. At 0 seconds, the triangle's height and base are 0 ft, so the distance between the boy B and the girl G is 0. At 1 second the triangle's height is 5 feet, and the base is 1 foot (we can say point B is 5 feet north and point G is 1 foot east). The distance between B and G is the hypotenuse of this right triangle. So good old Pythagoras tells us the length of the hypotenuse of a right triangle is the square root of the sum of the height squared and the base squared, so BG = square root of (B^2 + G^2). So BG (the distance between Boy and Girl) is square root of (5^2 + 1^2), square root of 26 is 5.0990195... so BG is about 5.099 feet.
At 2 seconds, the same formula holds, but the triangle is now bigger, because B is 10, and G is 2. So BG is now square root of (10^2 + 2^2), square root of 104 is 10.19804.... if we want we could build a table of these values with one row per second which would look like this (T is *time*, number of elapsed seconds):
T, B, G, BG
0, 0, 0, 0
1, 5, 1, 5.099
2, 10, 2, 10.198
3, 15, 3, 15.297
4, 20, 4, 20.396
But something interesting happens if we look at the difference between each value of BG and the previous value of BG (that is BG(t) minus BG(t-1)), you always get 5.099. The distance between B and G is growing at the rate of 5.099 ft/sec no matter what second you pick, so I think the answer is at 2 seconds they are separating from each other at 5.099 ft/sec.
Now to unpause and see how wrong I am! 🙂
You weren’t
So congrats :D
Honestly, that question would get rid of some stress during an exam lol
_Teacher sipping coffee talking to colleague in staff room_
"Your questions enter their mind, my questions enter their souls..."
Students...
"Did you figure it out?"
"Yeah...but is the guy okay after the breakup??"
We can assume there's no friction of the air tho right
That's already in there. They are going the speed they are going.
Ive seen a bipartite graph theory question where a professor is trying to figure out how to great groups to perfectly split up friends so that no one was friends with anyone in their group.
I just wonder how many of the students got distracted by the story and missed the fact that the question didn't ask "how far they are separated" but "how FAST they are separatING"
1:11 MAKE IT TO SCALE
The only problem I can see here is the fact that the two persons are not moving on top of a straight plane area, but on the surface of sphere. That would give a whole different meaning to the directions they are walking and the angle between their movement vectors. But I very much doubt the teacher thought about that .
They might have thought about that, then realised that we're dealing with two people who are a smidgen over 10 feet apart on a sphere that's approximately a bazillion feet across.
Then the latitude would matter.
I missed the fact that one is going north and one is going east. I would never have figured this out. That's why I will be working until I die.
It's oddly specific but it also oddly doesn't sound like they are basing it off anything currently happening to them at that moment
When you need pythagoras to solve your breakup, its joever
The problem never said the boy was running due north only that he was due north and crying. It was very specific that the gril was walking east. How do we know that the boy is not running toward the girl to try and get back together?
The fact that the question specifies that they are separating. Don't underestimate the necessity to the problem statement of all of the seemingly irrelevant chaff in the question. (-:
Unanswerable because the boy’s position is due north but we do not know what direction he is running. If it said he is running due north that would work but it doesn’t it says is due north and running. Further they are both moving at a constant velocity therefore this is a trigonometry problem, not a calculus problem. Or it could say the boy is moving due north. That would also work.
This is clearly to test your mathematical skills in the face of emotional turmoil provided that you happen to be going through a break up when taking that exam.
Now calculate this without assuming the earth is flat.
The earth is a sphere, and thus the answer depends on time.
basic kinematics can be solved in a single step with relative velocity, the relative velocity of boy wrt to girl is (5i-1j)ft/sec or in magnitude sqrt(26)ft/s therefore the boy and the girl are separating at the rate of sqrt(26)ft/s. would've been a more fun problem if they added acceleration for both or one of the person
Find the x ❌
Find the ex ✅
They are separating at the same speed as the beginning. There's no further acceleration cited.
I notice that you can answer this with vector components.
The line between former lovers defines a direction. If you take the component his 5 ft/s along the line and add it to her component of 1 ft/s along the line, you get 4.9029 + 0.1961, which is 5.0990
That is the most distracting math problem I’ve ever seen
Now that we've solved this problem, the boy and the girl can rest easy cuz now they know the speed at which they were seperating
Sadly, they had been separating from each other at a constant rate for months prior to the start of the word problem…😢
Wow, how creative and fun
Legend has it that the boy is crying until this day.
Legend is obviously not a mathematician. They twain are in a non-Euclidean geometry with positive curvature and walking in straight lines. (They cannot be on an infinite flat Euclidean plain, because the question specifies that the "place" has days.) Straight lines _always meet at two points_ in such geometries, unless they are co-linear, which the question _also rules out_ by implying that they are perpendicular. They will meet again in the future, over and over if they do not stop walking in their straight lines.
Calculating distance on a spheroid would have not resulted in a constant speed, but flat earthers do have a simpler life.
I- ladder leaning against wall problems have evolved since I was in high school it seems.
Assuming we don't take the curvature of the earth into consideration, nor the fact if they are traveling on the same plane (neither going uphill or downhill) after 1 second they'd have been √(5²+1²) feet apart , so approximately 5.1 feet. Since they're both traveling at a constant speed, every second they will move 5.1 feet further away from each other. Not sure why the problem asks for how fast they are separating from each other after 2 seconds, the time is irrelevant and the question might as well be "how fast are they separating from each other". But then again, this is quite a weird problem anyway.
5ft/sec? This guy’s running as fast as most people walk.
The solving is really overcomplicated. Since all velocities are constant, you can take Vboy - Vgirl (the hypotenuse of the right triangle with legs 5 ft/s and 1 ft/s) to get the answer, √26 ft/s.
That was kind of beautifully written, not gonna lie
For anyone curious, they are separating at a rate of square root of 26 ft/sec
Story of my life...
A physics professor told me that in another class he teaches, there was a problem about a girl being kidnapped and how to find the rate of the van she was in or something along those lines
I'm tired of solving Math problems, but I think he deserves help
The boy isn’t even running:
He’s traveling at 5ft/sec, which means he’s going at 300ft/min or 18000ft/hr. We know a mile is 5280ft long, so dividing 18000ft by 5280ft leaves us with about 3.41mph (miles per hour).
The average walking speed of an adult (which he’s likely to be, unless they got together when they were younger than 10, and if that’s the case how on earth did they last that long together?) is about 3mph, with a range going as high as 4mph and as low as 2.5mph. So 3.41 would be a regular walk, or at most a light speed walk.
In short, according to my calculations, that man isn’t even running, just a light speed walk in the rain.
Additionally:
Why the crap is the girl walking so slowly?! She’s moving at 3600ft/hr (1ft/sec = 60ft/min = 3600ft/hr) which is 0.68mph! That’s far below the average walking speed!!! Heck, the average army crawling speed is 0.5meters/sec, and with 1m being about 3.28ft, that would be 1.64ft/sec!!! She’s literally walking slower than someone army crawling!!! Who walks that slowly?! Especially since SHE’S the one who broke up with HIM!!!
Like, I know breakups are hard, but if you’re gonna be so devastated I could beat you in a race where you’re walking and I’m army crawling maybe you need to rethink some life choices.
This is just so f***ing funny!You made my day!!
Literally "Mind Your Decision" question
I was half expecting that we gonna need to take account the curviture of the earth and take the closest distance in the 3D space rather from closest earth's surface path
I like to think that the teacher will forget to change this question next year and one student will see it and go "Oh hey, I watched a video on this!"
The separation speed will not be constant if the curvature of the Earth is significant compared too the distances travelled. Fortunately it isn't if we are willing to approximate sqrt(26) as 5.099.
to solve the problem (in this specific scenario) you don't even have to find each length, you only need to use the pythagorean theorem to get sqrt (26)
"so how can we solve this problem" got me
Just noticed that if second 1 makes the first triagnle, second 2 extends this triangle 4x (add one to the top, one to the side, and one fills the empty space in the middle), 3d second expands it 9x (build extension simmilary as in 2s), 4th second expands it 16x. So in each case the number of triangles building the big one is t^2. That would be a nice visualization of that case
I looked at the thumbnail and just read the first sentence of the problem. Then I shifted my eyes up towards the name of the course and physically jumped when I saw “Differential Calculus” 😭
Teacher decided to make the question a little bit too relatable for the students because they couldn't understand it clearly
Tests always have something out of pocket as a problem
the teacher when they wrote this question: 🔥🔥🔥✍️
Next question : assuming they are walking on a sphere with a circumference of approximately 24,901 miles , how long would it take them to meet again ?
Next next question : assuming they meet up again, what is the probability that they will stay together. Assume the boy brings chocolates or flowers and calculate for both
@@kdemetter Answer to first "next question": about 761 days.
@@yurenchu Show your work :-)
@@kdemetter Timespan for girl to complete ½ of the circumference of spherical Earth:
[(½ circumference) × (24,901 miles per circumference) × (5,280 feet per mile) ÷ (1 ft/sec)] ÷ (3,600 seconds per hour × 24 hours per day) = (760.86388888...) days
Timespan for boy to complete 2½ of the circumference of spherical Earth:
[(2½ circumference) × (24,901 miles per circumference) × (5,280 feet per mile) ÷ (5 ft/sec)] ÷ (3,600 seconds per hour × 24 hours per day) = (760.86388888...) days
==> They'll encounter each other at the _antipode_ of their break-up location in about 761 days.
@@kdemetter Assuming they are both keeping a constant rate of moving, and that they keep moving straight on after heading in the direction when they started moving:
Timespan for girl to complete ½ of the circumference of spherical Earth:
[(½ circumference) × (24,901 miles per circumference) × (5,280 feet per mile) ÷ (1 ft/sec)] ÷ (3,600 seconds per hour × 24 hours per day) = (760.86388888...) days
Timespan for boy to complete 2½ of the circumference of spherical Earth:
[(2½ circumference) × (24,901 miles per circumference) × (5,280 feet per mile) ÷ (5 ft/sec)] ÷ (3,600 seconds per hour × 24 hours per day) = (760.86388888...) days
==> They'll encounter each other at the _antipode_ of their break-up location in 760 days, 20 hours and 44 minutes, _approximately_ .
@@kdemetter Assuming they are both keeping a constant rate of moving, and that they keep moving straight on after heading in the direction when they started moving:
Timespan for girl to complete ½ of the circumference of spherical Earth:
[(½ circumference) × (24.901 miles per circumference) × (5.280 feet per mile) ÷ (1 ft/sec)] ÷ (3.600 seconds per hour × 24 hours per day) = (760.86388888...) days
They boy completes 2½ of the circumference at 5 times the girl's moving speed, hence also arrives at the antipode of their break-up location in the same number of days.
If you want to make the question more interesting, have the girl leave first, and the boy leave 5 seconds later. Let t=0 be the point when the BOY starts moving and evaluate at t=2.
My brain simply went "the rates are in separate axes, so the hypotenuse is the rate we are looking for".
There was no need to calculate distance traveled or time in this problem. It's just simple use of the Pythagorean theorem.
I was going to mention that it depends how close to the North pole they are, but I see that's already been covered. So instead I'll postulate that perhaps they met at a gym 8 long years ago, and maybe the boy is running at 5 ft/sec on a treadmill. Hence why the direction he is running doesn't matter, and we're only told he "is due north". They're therefore separating at the speed the girl is walking away, 1 ft/sec.
LOL! Genius!
(By the way, the girl is walking at 1 ft/sec .)
@yurenchu Oops. Not sure where I got 2 from. I'll fix. Thanks.
That's rough buddy.
This and the comments taught me something that i might have missed in geometry is i did learn this (i suck with triangle hypotensuses), i never knew that the hypotenuse remains an equal value in root form when scaled proportionally (idk how common knowledge it is, i just never thought of it like that and im sleepy so idk)
Here is another way to doing it ---> x is distance walked by the boy y is the distance walked by the girl after 2 seconds which is 10ft and 2ft as 5ft/s and 1ft/s are there speed mention in the question by Pythagoras theorm we can right x^2+y^2=z^2 so on differentiating the equation wrt time we get 2x(dx/dt)+2y(dy/dt)=2z(dz/dt) cancel 2 on both the sides and we get x(dx/dt)+x(dy/dt)=z(dz/dt)
x=10ft y=2ft z=sqrt(10^2 + 2^2) (dx/dt)=5ft/s (dy/dt)=1ft/s
Put everything into differential equation and we get
dz/dt=52/sqrt(104) which approximates to 5.099ft/s 💔.