Integrating a fall through the Earth
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- เผยแพร่เมื่อ 2 ส.ค. 2016
- This is the integral part of my main "How to mathematically calculate a fall through the Earth" video. • How to mathematically ...
I slipped between x^2 - R^2 and R^2 - x^2. Sorry about that. Do work through it and make sure it all holds together.
Main channel here: / standupmaths
Matt you silly limey of course we want to see the full integration
He's not from England mate
+Nashyj495 he's an Aussie, right?
yes
That would make him an convicted limey, so a lemony?
No Cuba!
I'm here for the hot integration action.
So hot right now.
Me too, I can't say how disappointed I was when I found that the essential calculus was ripped off. No, I'm not joking.
all over his face too
Same I love calculus stuff, sad that Numberphile doesn't cover that stuff
I came watching this.
3:23 Matt Parker: the man who gave spoiler alerts on mathematics.
"0 on top of... WHO CARES?!"
I love maths.
Unless it's 0.
@@DevoutSkeptic
I actually once read some teacher notes on introductory Wave Mechanics where they seriously _had written_ the expression "0/0" while they justified the whole interference/diffraction equation.
I think they did that because they wanted to show what happened when the denominator approached 0, or something like that.
As someone with a background in physics, all these bending of math rules with derivatives and infinities gives me a warm feeling inside.
Sum of all positive integers?
Abiram N
-1/12
Same
That division by zero thing was actually mostly good from a mathematician's POV. The integral is weird with the larger number being the lower limit, so it'd first need to be negated to get the following logic.
The R is now the upper bound of the integral, so it's being approached from the negative direction. sqrt(R^2 - x^2) (the correct order as shown in the footnotes) is therefore always positive and approaches zero from the positive side, so the value of the input to the arctangent approaches positive infinity. And then arctangent's principal branch approaches pi/2 as the input approaches positive infinity.
The only thing I'd change for an audience of mathematicians about that part would be that I'd negate it and take the limit as the upper limit approaches R from the negative direction.
And then of course I'd also actually do the arctangent integral that he blows off.
@@snbeast9545 The mucking around with the differentials and missing limits of integration (v dv from 0 to v is never stated but implicitely assumed)actually got to me, and I'm usually not too big a stickler on formalism (nor a mathematician); though I did once argue with a physical chemistry professor because his solution to a problem had an incorrectly normalized probability distribution and inappropriately treated a vector quantities as scalars (including assuming that the value of a vector's norm could be a negative value), but for the forgiveness of symmetry it would not have yielded a correct answer. I can take the arctan limiting argument without much formalism here. But this is a really roundabout way of dealing with a really simple second order differential equation (it's always worth trying a complex exponential guess before going crazy with these things, once it comes out sinusoidal most of your work is done)
comes for some hot integration action, best part of integration was done by wolfram alpha...
Trigonometric substitution is hot integration action?
David Newman yes. In my eyes integration is a sum of all its parts.
FullTimeSlacker I really do. :)
martinshoosterman It really was xD 👍😃
You mean he Parker integrated it?
this using low dot for multiplication and middle dot for decimal point is killing me
it really confuses me too
But the decimal point should be in the middle unless you're writing it in a cronky old ascii-only computer, or you're German. :)
One cool thing about Matt Parker is that he shows genuine enthusiasm and a sense of humour when he teaches stuff about maths.
Those qualities are definitely good for a maths teacher, and instantly makes that maths teacher more likable and interesting.
He used to be a maths teacher
It's after midnight and the thing is, I haven't even started high school trig. Still very interesting to me though. I love watching this.
You can actually avoid the divide by zero, by replacing the inverse tan with an inverse cotan, where cotan theta = adjacent/opposite of a right angled triangle. Inverse cotan of 0 means the adjacent side must be zero, which happens when the angle is pi/2, therefore inverse cotan of 0 = pi/2.
The negative square root of Blah. Someone give this guy a Fields Medal.
For the tan graph I often describe it like pacman going off the screen on one side, instantly reappearing on the other side.
Summarizing an unwieldy expression as "Blah" and "If you want units on big G...look it up!"
Best channel on TH-cam!
I'm more annoyed about you letting Wolfram Alpha do the most fun part of the integration than I am about you saying tan^-1(inf)=pi/2
something strange defo happened. integral of dx/sqrt(x^2 +- a^2) is ln(|x + sqrt(x^2 +- a^2)|) + C. would love to see some proof that these are the same, no irony
@@arrtemfly Actually, it depends on the sign.
If you take x^2 + a^2 , you get atan
If you take x^2 - a^2 you get ln
@@nadiyayasmeen3928 oh, yeah, of course. thanks
That cheaty tan thing ...
It's not 'cheating', it's an application of understanding what the underlying numbers are doing and what the limits of your system are. If everyone followed mathematical rules completely, never diverging and never thinking below the surface... we wouldn't have most of maths.
- A physicist.
Not cheaty at all! An important identity to have down.
Limits, bitchez.
You can actually artificially find limits by plugging in extremely large or small numbers into a scientific calculator.
For example, if you try calculating arctan(1,000,000,000,000,000) on a calculator like that then it will give the answer π/2 - in fact, if it is a reasonably advanced calculator (like Casio CG-20, for example), then it will give you that answer in fraction form.
Not sure of the full derivation, but i think if you use the fact that tanx = sinx / cosx, i'm sure at some point the dividing by zero can be seperated to just be that cosx = 0 and doing an inverse cos that'll get you pi/2
Ya know, there is something you can use when approaching an unapproachable integer, THE LIMIT
Vincent Loparo III thats what he's doing. He's integrating, which is the limit of adding together infinitely small rectangles under a curve to find area
Watching this as a German student is always funny. It seems like you put your multiplication dots right at the bottom of the lines where as we put them in the middle and on the other hand you put your decimal dot in the middle of the line and we put ours (ok we actually use commas) at the bottom.
Americans do the same thing. Who would have thought we'd be more like Germany than Britain for this?
It looks like hes writing code
+ZipplyZane Matt is actually Australian, so he probably does everything backwards and upside down.
+Scott Goodson he fell through the earth
+ZipplyZane Can actually confirm that Matt is just weird. As far as I know most people here in England do it like our German pals.
12:52 "Oh if you want units on big G........look it up!" LOL
It's amazing and beautiful how something like this can simplify into only a few terms
6:50 LMAO, I've had way to many of those moments
7:14 Hahahaha, exactly what I do
lol 7:22 "...love it.." complete faith in wolfram alpha.
That was something that I always loved about my more advanced math courses;
the teachers didn't give a crap about how I solved single-variable integrals at that point, because that wasn't actually the interesting part about those courses, so they were just like "eh, solve them however you want", so I definitely used Wolfram-Alpha a lot, lol.
I love your enthusiasm as you work this out and the humor that you put in there
Zero on top of WHO CARES!? :D
Couldn't stop laughing :)
I laughed too, because WHO CARES was actually complex because sign error. Are you real?!
3:10 - "It equals blih some blah ..." Love that part xD
13:50 It turns out B equals a-mon-key. You never know what to expect from such a jungle of calculations :)
When I search for this video, I always search for "Matt Parker falls through the Earth".
I'm so glad I'm not the only writing scribbles as variables/constants for the things that just go to a calculator later.
- _Who Cares_
Matt
wolfram alpha told you that the integral of 1/(R^2-x^2)^(1/2) is arctan(x/(R^2-x^2)^(1/2)) instead of arcsin(x/R) which just seems... Needlessly complicated? I think there were a few sign errors here and there under the radical but they all seemed to cancel out somehow
Hmmm I thought that looked like it would be an arcsin...
I feel like he should have R^2-x^2 because all x
+will newman that's what I was thinking too, I think all the imaginary components cancelled out fortunately
***** I know that, I was mainly concerned about how unnecessarily complicated it looked when he was writing it down lmao
You got that using trig sub, right? I let x=Rsin(theta), dx is then Rcos(theta)d*theta, making sqrt(r^2-x^2) equivalent to R*cos(theta), making it the integral of 1 d*theta, which is just theta. Using the first substitution, we find that it is the same as arcsin(x/R). You can check the boundaries, and they all work out. Insert the two values, and you get the same answer.
This was the most brilliant part of my evening, Matt :)
This YT channel and numberphile made me have a huge interest in mathematics, thank you
Excellent fun :) Couple of bits I couldn't quite follow (due to my knowledge of only the most basic integration, not your teaching style!), but I do love it when a nice surprise drops out at the end like that!
if only we could bottle this guys enthusiasm for maths and drip feed it to the majority of school teachers, no child would leave school with a fear of maths.
I have never enjoyed math this much before in my life, and had never liked a video more than this!! You are awesome!!
Using some trig substitution, you get the arcsin(x/R), which ends up getting you the values of 0 and pi/2 as well.
this was an awesome proof and well deliverd, watched the whole thing! 2 questions: is it possible to prove arctan(-inf)=pi/2 using limits? and also can this be modelled a damped oscillition 2nd order ODE? Liked and faved :D
Thank you so much for the reference of Wolfram Alpha. Needed that.
This is why I love you Matt Parker.
Brilliant, Matt! This is why I enjoy maths so much (And physics too, to a lesser extent).
1:33 my analysis prof would burn me alive haha
It delivered more than I expected which is more than it advertised 👍🏼
I enjoyed this video more than the original one on the standupmaths channel that brought me here. The best part is when Matt, tail between his legs, forces himself to admit that the inverse tan of R/0 = pi/2. The video where he explains that any x/0 is undefined is also one of my favorite Numberphile videos.
Being really formal you would have to calculate the limit of that inverse tan, but it would give the same answer.
Nícolas this is not a tutorial. Hes deriving something so it is assumee the.viewers understand the mathematics hes doing.
Explodes to infinity! hahaha I loved that XD. Awesome and funny job you did, btw. Thanks!
The physics methods here hurt
At about 9 minutes, could you integrate with one of the paramaters with respect to the limit approaching infinity? If so, how would you notate this?
Can we get a 10 hour video of 9:55-10:09?
For the evaluation of the integral with the tangent inverse of the infinitely large number, in order to justify that mathematically, would you have to use an improper integral with an infinite limit, then write that part as a the limit as x->R?
This was great, and I thoroughly enjoyed it!
I second your thoughts
+Luis N I guess that means I third them
+Yextus So now I square them?!
¯\_(ツ)_/¯
I'm getting arcsin(x/R) + C for the integral of dx/sqrt(R^2-x^2). Am I doing something wrong?
I miss my high school maths from watching this video. Good o' time. Thanks Matt.
Nice to see you think of the plot of tangent the same way I do.
How did you multiply small t by 2 to get big T. Is it as simple as putting by hand. Should you not do it rigorously.
First thing I thought when your video popped up in my recommendations, was simple harmonic oscillator 🤗
1:20 We don't need to treat derivatives as fractions for that relationship.
We assume that x depends on v (i.e. x is a function of v)
Then by the chain rule,
dx/dt = dx/dv • dv/dt
Rearranging gives us
(dx/dt)/(dv/dt) = dx/dv
As a physicist I must say that I am slightly worried about your lack of attention to the boundary conditions. And signs as well. But nicely done overall.
At least he does not use the silly '+C' in the end of integration.
Well really all of his integrals ought to be definite so there is no constant.
Mikael Jensen is his videos not yours. So he's going to do it however he feels like. Wether you or anyone likes it or NOT.
Your style of explaining this sounds like what Bob Ross would sound like, if he were a mathematician.
As someone who barely understands any of this maths it's nice to see notes like 7:45 because all the way through my maths classes I understood the concepts perfectly but got crappy grades because I couldn't multiply 8x7. Usually I got so excited that I knew the process I didn't pay enough attention to the actual arithmetic.
0/10, not enough brown paper
+Matt Parker, why did you tell the mass of the Earth (M) you used, while in the function the density of the Earth (ρ) is used? I know you can derive it using the function of the spherical volume - V = (4 * π * R³)/3 with R as the radius of the Earth - and the mass of the Earth (M) using ρ = V/M and you get the density of the Earth (ρ).... I don't know if you did it on purpose but I think it's not correct....
I'm starting to love maths. :-) It's a nice new world.
Hard at first glance, but so much rewarding.
Great job!
Hi Matt, a question:
When you're doing the definite integration with limits at 8:10, I understand the 0 on top but if you consider the denominator you would get root(-R^2) which is undefined after you plug in x=0. So it would be 0/something-undefined... does this have no effect?
The denominator should be R^2 - x^2 which is positive so the square root exists.
@@normanstevens4924 I know this is a late comment; you may not read it, but R^2-x^2=(x+R)(x-R), therefore: x^2-R^2=(x-R)(x+R). So they are actually equal because the order of multiplication doesn't matter!
I'm probably missing something obvious, but how is ∫1/(x^2-R^2)^(1/2)dx which he points at on the board at 6:51 the same as ∫1/(R^2-x^2)^(1/2)dx which is what he put into Wolfram Alpha at 7:18? If I'm not mistaken, ∫1/(x^2-R^2)^(1/2)dx = ln((x^2-a^2)^(1/2)+x) while ∫1/(R^2-x^2)^(1/2)dx = arctan(x/(R^2-x^2)^(1/2)), (ignoring the +C). As far as I know, these are not equal. Does this affect the result in any way?
Good old Blah. The most useful mathematical tool.
I manually solved the differential equation and got 41.23961569hrs, which is close enough to the real answer. I solved x(t)=double integral of (-(G(rho)x^3m)/x^2) to get x(t)= Rcos(sqrt(G(rho)m)t)) and then worked out the period of X(t), then divided by 2 to get the time taken to fall from one side to the other, and then plugged in the values, gave it to my calculator and got an answer. "Simple". I love your videos, especially the really mathsy ones, so keep on making them. Also I am 12.
Great calculation.
But how long time will it take for an object with zero velocity, distance x from earth, to hit the earth? I cant do that calculus, tried for hours :(
Can you do an indefinite integral in one side and make it a definite integral in the other? At 4:20.
At 7.20 the shot of wolfram appears to show r^2 - x^2 it Syd be the other way around, but you got the right ans in the end so I suppose that was just a screen shoot
I've somewhat understood some parts of what you were saying. I suck at math, which is a shame, ok. I would give this video a couple of goes. However, i'm interested on how hard it will be to do all of the calculations if we take in a consideration the uneven densety of the earth? Like, at least imagine the earth having a core (lets say half of the radius) 2 times as dense as the rest of it?
Then rho is a function of x and all the integrals are much more complicated since "B" isn't a constant. Probably they have to be solved numerically.
You can solve this problem even if rho is its own function that varies with distance. Unfortunately, the real world isn't so clean. Real rho depends on both distance AND location (theta, phi), and it isn't even consistent. Real rho is neither continuous nor differentiable. It can't be solved analytically.
Density is certainly a smooth function. The problem is just that it is not spherically symmetric, which means that you will get different answers depending on which diameter you choose (e.g. it will take a different time to fall from pole to pole than the intersection of the equator and prime meridian to the intersection of the equator and date line). We could still pick one particular path (say, north pole to south pole) and write down a vector function for the acceleration in terms of depth, but it would not be trivial and indeed probably not elementary. Still, it should be continuous and therefore integrable, and a computer would have no trouble handling the integration.
However, the density of the Earth can be modeled to pretty good accuracy as a function only of depth. In that case, the analysis starts exactly the same way it did here, except ρ is a function of x, not a constant. The right side is still integrable, but the integral is most likely not to be elementary, meaning it cannot be written using only relatively "simple" operations like +, -, *, /, ^, sin, and log. In that case, you would have to solve it numerically.
Calculations similar to this are actually very important for determining quantities like the gravitational binding energy of the Earth, which are relevant to understanding the Earth's primordial heat, among other things.
Wait... Wait.... All rocky bodies no matter what the size will give you same time T, but what if there was a rocky body sufficiently large that velocity approach the speed of light to make that time? Would the one falling still experience time T or would the observer?
Love it but, the density of the earth isn't homogeneous. What if you have a function for the density of the earth in relation to the distance to the center? How would that change things? I would assume that "Blah" would no longer be valid and it would need to be integrated with the rest of it all (but I haven't integrated anything in years so I'm guessing here). And, to spice things up a bit, how much would the density need to vary to achieve a perceptible change in the travel time (within, say, two significant digits)?
Love your videos Matt! Keep up the great work!
If you graph y = tan^-1(x), you'll see the limit as it approaches infinity is pretty clearly π/2. Go graph it.
Well it's alright because y=arctan(x) has range(-pi/2,pi/2) so no problem. And yeah the integral of 1/sqrt(R^2-x^2)can be done by trig sub by x=R*sint and dx=R*cost*dt and then just simplified integration and getting rid of the t by replacing it in terms of x by the help of our previous relation between x and t.
mathematicians.......they always leave out the units
best integral symbols imo
I just realized that Matt draws his Tangent graphs the exact same way he draws his Integration symbols.
Matt, you could simply get the velocity in terms of position using the kinetic and potential energy without any differentials... but i guess it wouldnt be mathematical enough XD
Seriously, this is the first time I see you write the integral sign.
You should have also done the integration for attraction to a thin shelled sphere, and why inside of one, you have zero net force, and outside you have the simple point mass at the center as a simplification. Or why a thick shell you are in, you can ignore, and the thick shell below(a solid sphere) lets you pretend its a point mass.
Why is there no constant from the integral of v.dv? 5:07
I love this
Great vid, but aren't you assuming that at the center of the Earth you are going to be repeled from it, instead of falling past it and then reversing direction until you are stuck at the core?
Cool proof, even cooler is that you get the same result picking any two points on earth to fall through (not just straight through)
This helped me refresh on my high school calculus to prepare for my first year at uni. Hook em!
1:00 Good practice is to bracket individual fractions in fraction columns.
Well, what *is* the reasoning for picking that particular branch of the inverse tan instead of any of the others?
As presented the asymptotes should be any (2n+1)pi/2
It wouldn't matter: if I remember correctly, there was a difference of arctangents (since he evaluates some definite integral), hence even if you chose some other branch, it would cancel out.
He explain it with the tan'x) graph, but since he is actually caclulating arctan(x) for x -> infinity, and the domain of arctan only allow on result
He selects n=0 for simplicity
So cool! Why is this not in the main video?
EDIT: Great... now I had to subscribe to another channel *sigh* ;-)
Well Played
Awesome!
11:05
No problem with that since
sin/cos=1/0 --> sin(x)=1 & cos(x)=0 --> x=pi/2
8:11 is the best bit
The only thing that bugged me a little is that tan^-1. Actually, we use arctan for this since tan^-1 refers to a set, rather than a function. Might just be my rather strict professors
How can we surely say that is is double of time it takes to reach half the distance because you have gained some momentum on reaching the centre which will keep on reducing due to gravity until you get to other side .
YOU MUST HAVE INTEGRATED FROM R to -R and then you would have got the same result
Never been quite so titillated by an integral...
at around 4:30, you evaluated the right side as a definite integral, but the left side was still an indefinite integral. why wasn't there a +C on the left side?
This was fun
The SHM approach is far simpler and more elegant. It solves a straightforward ODE, avoids improper integrals, and easily generalizes to all paths through chords of the Earth, not just diameters. The fact that it takes the same amount of time regardless of the length of the chord is I think a much more surprising result.
I know that you like mathematical challenges, big magic numbers, and weird equations with multiple results, so I have a puzzle for you based on this video. The challenge is this: generalize your equation to work with extreme density. This will require replacing your Newtonian approximation for space and time with relativity. I think you’ll find this challenge interesting for two reasons: 1) The time it takes should asymptote up to infinity at a magic mega number: 6.734e26kg/m (The Schwarzschild density) at which spacetime traps you in a black hole. 2) The equations should be able to give you two separate numbers, that will diverge from each other. At extreme densities, you’ll both hit relativistic speeds and be inside an extreme gravity well, so you’ll need an observer time AND and traveler time due to time dilation and Lorenz stretching.
10:20 - Picky teacher time: your tangents are more like a periodic cubic. The scale could lend itself to showing that picture, but generally speaking, your tangents should have a slope approaching 1 at the x-axis.
I use differential equations to do it.
M is the mass of earth. R is the radius of earth. D=3M/4πR^3 is the density of earth. m is the mass of a person which the distance between earth center is r (r
Really good