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.
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.
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 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?
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
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.
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.
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 💔.
2:13 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!!!
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.
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.
The couple should not get disheartened because the boy is longing for the girl by following a line of longitude and the girl is reconsidering by following a line of latitude, and it's possible that at some point in the future their paths will cross and they will meet again as the they initially move apart from each other, but due to the spherical geometry periodically move closer to one another.
It's not only possible, it is required by the "non-parallel" postulate of spherical geometry (as long as we hand-wave the radius of the planet a bit and don't assume point particle boys and girls with zero extents). (-:
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.
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.
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 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?
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.
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.
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. (-:
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! 🙂
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.
Are they meeting at the south pole? If so moving away East/West is stationary. And the answer is 5'/s without account for the curvature of the Earth, rotation of yhe Earth orbit of the Earth or orbit of the Sun around the Milky Way or travel of the Milky Way from the "centre" of the universe. The existential crisis of "breaking up" is somewhat lost if after 8 years they are at the South pole of an utterly insignificant planet in a vast universe.
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! 😂
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
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 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.
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.
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 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
Let me try to solve it before watching the video (with accuracy of 5 digits after decimal points) First, at Second 0, their distance is, of course, 0. For every second afterwards, their distance is the hypotenuse of the lines from their starting position and current positions. Hence, At Second 1, their distance is sqrt(5ft^2 + 1ft^2)=sqrt(26ft)=5.09902 At Second 2, their distance is sqrt(10ft^2 + 2ft^2)=sqrt(104ft)=10.19804 At Second 3, their distance is sqrt(15ft^2 + 3ft^2)=sqrt(234)=15.29706 By second 1, they have a distance of 5.09902ft, or 5.09902ft/second By second 2, they have a distance of 10.19804ft, or 10.19804ft/2 seconds, or 5.09902ft/second By second 3, they have a distance of 15.29706, or 15.29706ft/3 seconds, or 5.09902ft/second Hence, they have a rate of increase in distance at 5.09902ft/second, rounded. Edit: Oh yay I got it correct but for the wrong reasons I guess lol
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.
Part II: calculate an exact expression for the rate at which the boy is moving away from the girl in the girl's frame of reference according to Einstein's theory of General Relativity Part III: repeat Part II under Einsteins theory of Special Relativity assuming the boy and girl are moving on the surface of a sphere
Let S(t) be the distance between the boy and girl at time t > 0 . S(t) = √[ (5t)² + (1t)² ] S(t) = √[ 25t² + 1t² ] S(t) = √[ 26t² ] S(t) = t*√26 [feet] differentiate: dS/dt = √26 [feet/s] ==> rate of separation (at any time after t = 0) is √26 [feet/s] .
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.
At t=2 the boy is at (0,10) with velocity 5 going N and the girl is at (2,0) with velocity 1 going E. We want the magnitude of the sum of their velocity vector components drawn along the line segment drawn from the girl to the boy. Draw the right triangle and then notice the two angles are A=arctan(2/10) and B=90-A. This makes the two components 5*cos(A) and 1*cos(B), in opposite directions. Adding the magnitudes gives 5.099 ft/sec.
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
Because the English of the problem is poor, there is another interpretation, which is that the boy remains "due north" of the girl at all times. Again, since there's no acceleration in the problem (aside from the tears and the rain, I suppose), the velocity must be a constant value, only this time it'll be sqrt(24) ft/sec.
You can solve this problem by taking the vector sum of the two velocities. By geometry, the distance between them is increasing at the same speed. Since the velocities are constant, so is the speed of separation.
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.
That was a long way to declare the obvious-from-the-beginning: That this question has nothing to do with differential calculus. Could it be that the teacher had just a bad day?
There is hardly any room for calculus. The distance s(t) between them after t seconds from the Pythagorean theorem is: s(t) = sqrt(t^2 + (5t)^2) s(t) = sqrt(26)t which is a linear function. No matter what t is, the speed is always: v(t) = s(t)/t = sqrt(26)
It is a relative velocity question. vel BrelG = velB - velG. Draw the vector diagram, similar to yours but with arrows on to indicate vectors, and V BrelG is rt 26 at a bearing of 348.7 degrees. The time is immaterial, so this is where the question-master broke down, because this really isn't hard enough to be a final anything! On the other hand, if the question is saying that he remains due north of the girl, the amended diagram gives he must travel on a bearing of 11.5 degrees at rt 24 fps.
Comment before watching: Assume the earth is flat. North is perpendicular to east, so their paths form the catheters of a right triangle. d(t) = sqrt((t ft/sec)² + (5t ft/sec)²) = t ft/sec sqrt(1²+5²) = sqrt(26) t ft/sec This gives: d'(t) = sqrt(26) ft/sec And, calculating the square root in my head: sqrt(26) ≈ 5.1 - 0.01/10.2 ≈5.099 So the "after 2 seconds" part is a red herring, as they will always be separating at the same speed of 5.099 ft/sec. By the way, this scenario is actually that I knew from Vsauce, iirc it was his video about odd-numbered sums. I can't say anything about who came first though.
Let t = what second we are looking at. Distance at time t = sqrt((5t)^2 + (1t)^2) = sqrt(26t^2) = t*sqrt(26). This is a linear relationship with respect to t so the speed will be a constant which is simply sqrt(26) at any time.
If not otherwise specified, problems like this assume that it takes place significantly far enough from the poles of Earth, that the compass directions are close enough to being perpendicular to each other for all practical purposes.
Ähm, before watching I'd say the rate of ~5,1 ft\sec does not change. Which answers the question of how fast. When they provide constant speeds and directions, the duration in seconds does simply not matter.
For everyone interested: you should _never_ create tables or do anything more than asked. the table is 100% irrelevant. constant speeds added will always result in constant increases in distance, when calculated per second. you will need to save time in your exams, and unnecessary calculations will eat your results.
You don’t even need calculus for this, because the two velocities are constant, so you can just use the Pythagorean theorem and be done with it. If the velocities were NOT constant, then you’d have to actually express the hypotenuse in terms of t, then take the derivative.
Both protagonists are pedestrians, which rules out taxicabs. (-: However, you are right to question the distance metric. "days" rules out an infinite flat Euclidean plane, so this is either spherical or hyperbolic geometry. But is the distance measured along the surface that they are walking on? Or in a 3D straight line through it? How connected is the space in which they are walking, moreover?
They are moving on a sphere, not in an euclidian space. That has some impact on triangles. At some point in time, the boy is stuck at the north pole, while the girl is circling forever at her starting latitude. At least, from the fact of being "a rainy day", it's unlikely that they met first time at the north pole.
*after 1 day* They are together because they thought it would be better for both to not break-up , so if the teacher would ask how far are they from each other after 48 hours, the answer would be : boy's house to girl's house if they are 16- 16cm if they are 16+ I can give you a whole story on this if you want to know , why they break-up?, what will happen to them if they won't be together again? Or would be together? But it'll be imaginary and it'll take time also, i'm not a storywriter but can write basic stories(so you can!)
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 😂)
It would appear he was the jilted one and is "fake running", hiding his tears, in the hope his ex changes her mind. We are not given the conditions that exist at the 3rd second. What a cliffhanger!
Trying before watching the video. Distance from each other(y) = sqrt((5t)^2+t^2)=sqrt(26t^2) dy/dt|t=2 =(52*2)/(2sqrt(26*4))=104/2sqrt(104) = sqrt(104)/2 = sqrt(26)fts-1
I would love to see all of the students answers 😂 If for example you were to asume that the reason for breaking up was his father standing east while also being her new lover and north is a gun store, the given speeds might change a bit during those two seconds, which would make the calculation a bit harder.
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
i am more interested in the passive aggressive backstory of this exam writer.
"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.
Clearly the answer to the question is "Not fast enough."
Valuable lesson on real life scenarios: to be given unnecessary information and be obliged to use inappropriate method.
Bro got a lit too personal 💀
That was not the bro, bro.
@@BKNeifert Being 'bro' is not about the gender, bro.
@@u2bear377 Ah, okay.
Question ❌
Break-up story ✅
Story ❌
Strory ✅
Teacher: Math is everywhere.
Student: It can't possibly be everywh-
Teacher: *Math is everywhere.*
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.
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.
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.
"How do we solve this problem?" Well, when did the relationship dissolve? Was there yelling involved?
Oh! Not *that* problem.
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
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.
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.
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 💔.
In my experience, neither boys nor girls who have just broken up with each other run or walk in an exactly straight line.
Sadly, they had been separating from each other at a constant rate for months prior to the start of the word problem…😢
2:13 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.
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.
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.
The couple should not get disheartened because the boy is longing for the girl by following a line of longitude and the girl is reconsidering by following a line of latitude, and it's possible that at some point in the future their paths will cross and they will meet again as the they initially move apart from each other, but due to the spherical geometry periodically move closer to one another.
It's not only possible, it is required by the "non-parallel" postulate of spherical geometry (as long as we hand-wave the radius of the planet a bit and don't assume point particle boys and girls with zero extents). (-:
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.
For the record I am still traveling at 5ft/s and still crying.
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.
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?
It's awesome how he can explain without bursting into a laughter 👌👌
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?
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.
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.
Mitochondria is the powerhouse of the cell
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. (-:
Wrong! They would stop separating and start to come together, so the distance would be getting lower
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.
A basic, elementary but also interesting differentiation/1st derivative question.
No differentiation needed, since the velocities are constant.
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.
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
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.
Are they meeting at the south pole?
If so moving away East/West is stationary. And the answer is 5'/s without account for the curvature of the Earth, rotation of yhe Earth orbit of the Earth or orbit of the Sun around the Milky Way or travel of the Milky Way from the "centre" of the universe. The existential crisis of "breaking up" is somewhat lost if after 8 years they are at the South pole of an utterly insignificant planet in a vast universe.
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! 😂
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
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
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
A student I was tutoring asked me a very similar question recently, but I had to look up the answer.
Thanks!
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.
before listening to the complete question who thought there would be a role of those 8 years for evaluating the answer
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.
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.
Calculating distance on a spheroid would have not resulted in a constant speed, but flat earthers do have a simpler life.
The writer didnt have to put the "eight long years"❌❌❌
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 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
Let me try to solve it before watching the video (with accuracy of 5 digits after decimal points)
First, at Second 0, their distance is, of course, 0.
For every second afterwards, their distance is the hypotenuse of the lines from their starting position and current positions. Hence,
At Second 1, their distance is sqrt(5ft^2 + 1ft^2)=sqrt(26ft)=5.09902
At Second 2, their distance is sqrt(10ft^2 + 2ft^2)=sqrt(104ft)=10.19804
At Second 3, their distance is sqrt(15ft^2 + 3ft^2)=sqrt(234)=15.29706
By second 1, they have a distance of 5.09902ft, or 5.09902ft/second
By second 2, they have a distance of 10.19804ft, or 10.19804ft/2 seconds, or 5.09902ft/second
By second 3, they have a distance of 15.29706, or 15.29706ft/3 seconds, or 5.09902ft/second
Hence, they have a rate of increase in distance at 5.09902ft/second, rounded.
Edit: Oh yay I got it correct but for the wrong reasons I guess lol
Just put them both at the North Pole so they won't be separated ❤
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.
Shouldn't we take the Earth curvature into account?
Over a distance of 10 feet???
I know the earth is a ball, not flat, but are you aware just how BIG a ball it is?
Seriously, you are doing the math?
Part II: calculate an exact expression for the rate at which the boy is moving away from the girl in the girl's frame of reference according to Einstein's theory of General Relativity
Part III: repeat Part II under Einsteins theory of Special Relativity assuming the boy and girl are moving on the surface of a sphere
Let S(t) be the distance between the boy and girl at time t > 0 .
S(t) = √[ (5t)² + (1t)² ]
S(t) = √[ 25t² + 1t² ]
S(t) = √[ 26t² ]
S(t) = t*√26 [feet]
differentiate:
dS/dt = √26 [feet/s]
==> rate of separation (at any time after t = 0) is √26 [feet/s] .
Literally "Mind Your Decision" question
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.
That's rough buddy.
One problem is there, why the girl is not crying😅
At t=2 the boy is at (0,10) with velocity 5 going N and the girl is at (2,0) with velocity 1 going E. We want the magnitude of the sum of their velocity vector components drawn along the line segment drawn from the girl to the boy. Draw the right triangle and then notice the two angles are A=arctan(2/10) and B=90-A. This makes the two components 5*cos(A) and 1*cos(B), in opposite directions. Adding the magnitudes gives 5.099 ft/sec.
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
5ft/sec? This guy’s running as fast as most people walk.
Because the English of the problem is poor, there is another interpretation, which is that the boy remains "due north" of the girl at all times. Again, since there's no acceleration in the problem (aside from the tears and the rain, I suppose), the velocity must be a constant value, only this time it'll be sqrt(24) ft/sec.
in the end they will meet each other at the exact same point after 4.167464 years
@@kotofyt Actually sooner than that: namely after 2 years and 30 days, at the so-called _antipode_ of their break-up location.
I was just pissed at the feet per second
Ughh... how hard is it really, to understand that a foot is roughly 30 centimeters? It's about the size of a person's foot, for god sakes.
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
Looks like malicious compliance, when you actually solve the problem mathematically and completely ignoring the backstory xD
Story of my life...
You can solve this problem by taking the vector sum of the two velocities. By geometry, the distance between them is increasing at the same speed. Since the velocities are constant, so is the speed of separation.
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.
That was a long way to declare the obvious-from-the-beginning: That this question has nothing to do with differential calculus. Could it be that the teacher had just a bad day?
There is hardly any room for calculus. The distance s(t) between them after t seconds from the Pythagorean theorem is:
s(t) = sqrt(t^2 + (5t)^2)
s(t) = sqrt(26)t
which is a linear function. No matter what t is, the speed is always:
v(t) = s(t)/t = sqrt(26)
It is a relative velocity question. vel BrelG = velB - velG. Draw the vector diagram, similar to yours but with arrows on to indicate vectors, and V BrelG is rt 26 at a bearing of 348.7 degrees. The time is immaterial, so this is where the question-master broke down, because this really isn't hard enough to be a final anything!
On the other hand, if the question is saying that he remains due north of the girl, the amended diagram gives he must travel on a bearing of 11.5 degrees at rt 24 fps.
The question is impossible to answer. You don't know where they started from. At the equator, near the south pole? These give very different answers.
Quite a smokescreen in that text.
Comment before watching:
Assume the earth is flat. North is perpendicular to east, so their paths form the catheters of a right triangle.
d(t) = sqrt((t ft/sec)² + (5t ft/sec)²)
= t ft/sec sqrt(1²+5²)
= sqrt(26) t ft/sec
This gives:
d'(t) = sqrt(26) ft/sec
And, calculating the square root in my head:
sqrt(26) ≈ 5.1 - 0.01/10.2
≈5.099
So the "after 2 seconds" part is a red herring, as they will always be separating at the same speed of 5.099 ft/sec.
By the way, this scenario is actually that I knew from Vsauce, iirc it was his video about odd-numbered sums. I can't say anything about who came first though.
On the real Earth, the two would no longer diverge after 145 days and would move closer together again.
@OrbitTheSun So there is still hope
The question assumes they both can instantly accelerate to their respecftive speeds, which is impossible.
Let t = what second we are looking at.
Distance at time t = sqrt((5t)^2 + (1t)^2) = sqrt(26t^2) = t*sqrt(26).
This is a linear relationship with respect to t so the speed will be a constant which is simply sqrt(26) at any time.
where did they seperate? how close to the poles? how large is the planet they are on?
we live in a curved space, not on a flat earth
If not otherwise specified, problems like this assume that it takes place significantly far enough from the poles of Earth, that the compass directions are close enough to being perpendicular to each other for all practical purposes.
Ähm, before watching I'd say the rate of ~5,1 ft\sec does not change. Which answers the question of how fast. When they provide constant speeds and directions, the duration in seconds does simply not matter.
For everyone interested: you should _never_ create tables or do anything more than asked. the table is 100% irrelevant. constant speeds added will always result in constant increases in distance, when calculated per second. you will need to save time in your exams, and unnecessary calculations will eat your results.
You don’t even need calculus for this, because the two velocities are constant, so you can just use the Pythagorean theorem and be done with it. If the velocities were NOT constant, then you’d have to actually express the hypotenuse in terms of t, then take the derivative.
I wonder if you would get bonus points for writing "Are you okay?" after your answer
Problem was, the exam question wasn't even that hard
the free-est 10 marks ive ever seen in my life
Actually, at t=2 seconds their positions are fixed and they are not moving with respect to each other at all.
That teacher needs a therapist to review her questions
No one specified the Euclidean metric. Far easier in Taxicab geometry.
Both protagonists are pedestrians, which rules out taxicabs. (-: However, you are right to question the distance metric. "days" rules out an infinite flat Euclidean plane, so this is either spherical or hyperbolic geometry. But is the distance measured along the surface that they are walking on? Or in a 3D straight line through it? How connected is the space in which they are walking, moreover?
They are moving on a sphere, not in an euclidian space. That has some impact on triangles. At some point in time, the boy is stuck at the north pole, while the girl is circling forever at her starting latitude. At least, from the fact of being "a rainy day", it's unlikely that they met first time at the north pole.
*after 1 day*
They are together because they thought it would be better for both to not break-up , so if the teacher would ask how far are they from each other after 48 hours, the answer would be : boy's house to girl's house if they are 16-
16cm if they are 16+
I can give you a whole story on this if you want to know , why they break-up?, what will happen to them if they won't be together again? Or would be together?
But it'll be imaginary and it'll take time also, i'm not a storywriter but can write basic stories(so you can!)
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 😂)
"Running" at 5 ft/s? I.e., a bit less than 5.5 km/h. More like a comfortable stroll...
It would appear he was the jilted one and is "fake running", hiding his tears, in the hope his ex changes her mind. We are not given the conditions that exist at the 3rd second. What a cliffhanger!
Trying before watching the video.
Distance from each other(y) = sqrt((5t)^2+t^2)=sqrt(26t^2)
dy/dt|t=2 =(52*2)/(2sqrt(26*4))=104/2sqrt(104) = sqrt(104)/2 = sqrt(26)fts-1
Bro got into a harsh Breakup before writing that question 😢
I would love to see all of the students answers 😂
If for example you were to asume that the reason for breaking up was his father standing east while also being her new lover and north is a gun store, the given speeds might change a bit during those two seconds, which would make the calculation a bit harder.
Love how the question is literally: what is the square root of 26?