Give me a thumbs up if you want to see me film a physics videos with one of our physics professors!

@mit4539

3 жыл бұрын

This will be a great idea.

@rickaryadas9571

3 жыл бұрын

It will be too helpful for us

@sanjibanpaul872

3 жыл бұрын

Yes sir. I am one of the Physics Olympiad aspirants in India. I would really love watching videos on Physics problems on your channel. Thanks!

@debbaranyabiswas1102

3 жыл бұрын

👍

@adi-sngh

3 жыл бұрын

👍🏻

My pure maths prof was so *pure* he forgot how to differentiate x^2 (even though he could easily prove existence and uniqueness for PDEs lol). Good on you :D

@aaronhunt4302

3 жыл бұрын

😆

@electroskylightgaming4085

3 жыл бұрын

Hes too pure for this world....

@grzechu9751

3 жыл бұрын

bruh

@michalbotor

3 жыл бұрын

you know how the joke goes right? math guys and gals, they only show that something (the more general the better) exists and if it is unique, special cases -- such as x^2 -- are of no interest to them. 😁

@MrRenanwill

3 жыл бұрын

Even though this might be the case, he could prove that this is 2x at the point x.

The horizontal friction on the big block should INCREASE the force needed to accelerate the system: F-3mgu = 3ma so F=3mgu + 3ma

more physics videos in the future please

Congratulations on getting verified! You're awesome prof !

13:31 No weekly update today? Nevertheless, applied mathematics problems will probably bring another type of audience on this channel

@atreyamajumdar9836

3 жыл бұрын

I am a physics student, but I am following this channel for almost a year now. Watching this video is very exciting for me.

Liked the video! hoping to see more physics in the coming future

You are one of the coolest math professor

@fearkrypton4565

3 жыл бұрын

Hkr

I would love more videos of this type!

nice. can you overkill it with Lagrangian mechanics?

@santiagoarce5672

3 жыл бұрын

Please this.

@LorencoLisica

3 жыл бұрын

It seems that interpreting the problem with Lagrangian mechanics would be unwieldy because of the external force applied to the system (And you can't say that the force is conservative to define a potential) , if it were just the two blocks interacting it would be pretty simple. It might be necessary to use D'Alambert's principle.

@juliandominikreusch8979

3 жыл бұрын

Friction in the Lagrange formalism is notoriously difficult. As the energy then becomes path dependant, this changes the whole variational approach

@juliandominikreusch8979

3 жыл бұрын

@@LorencoLisicawhat about the potential -m*x the derivative I.e. the force would be a constant term m in some fixed direction

@LorencoLisica

3 жыл бұрын

@@juliandominikreusch8979 I think that approach is more suitable for using D'Alambert's principle. (Which really is a generalization of Euler Lagrange equations) I agree with the method though. But then again as You pointed out, the friction would make the problem practically unsolvable by the varation principle.

I think you have to add your friction force 3m\mu g instead of subtracting, since in addition to what happens "on the slope" for the steady scenario (calculating the acceleration a) you have to apply the additional force of 3m\mu g to compensate for the sliding friction.

at 2:48 by using Newton's second law of motion: F - 3mgu = 3ma ---> F = 3ma + 3mgu

@rcat_

3 жыл бұрын

Yup

@justinrangad4636

3 жыл бұрын

This confused me aswell

@fix5072

3 жыл бұрын

I dont think so: you are talking about the force you have to apply in order to push the blocks with a resulting force of 3ma. Prof. Penn instead is talking about the resulting force when pushen with a force of 3ma. U can missunderstand this easily because of the arrow of F. In a contest you may be allowed to even ask what F is ...

@rezamiau

3 жыл бұрын

@@fix5072 Newton's Second law of motion: "The acceleration of a particle ( or system of particles) is proportional to the resultant force acting on it and is in the direction of this force" Engineering Mechanics: Dynamics 8th edition By: J. L. Meriam & L. G. Kraige Resultant forces = F - (u * N) Acceleration = a Total mass = 3m N = (3m) * g Resultant forces = ( Total mass) * (Acceleration) (F - 3mgu )= ( 3m ) * a

@Fysiker

3 жыл бұрын

@@fix5072 That seems right. I prefer to write F as the force applied, then add forces to get the net force and acceleration, but his way is probably more direct (but harder to follow)

Quite rigorously I must say, as it should be :3 Waiting for more physics!

THANK YOU, Michael.

The F_whole equation is wrong. F - 3umg = 3ma. Acceleration is reduced due to friction on the ground.

@Walczyk

3 жыл бұрын

Right, a is the resultant acceleration, after the horizontal forces are summed/applied. Watching up til 7:37 the sign on the 3m*mu*g term is incorrect. Since the friction of the floor will go against any applied horizontal force, it should be making it more difficult and thus increasing F_min instead of decreasing it.

@foobar5809

3 жыл бұрын

oh, good. I was surprised, but I thought maybe its because I dont know any physics

@user-uv2hk5hs1k

3 жыл бұрын

I literally said "woah woah woah" out loud and paused the video at 2:52 to see if anyone else caught this mistake

@doubleeagle243

3 жыл бұрын

Yes u r absolutely right..i solve it and get confuse in his equation there is plus sign in last eq

Nowadays professor is giving unexpected surprises

But you missed a great trig identity here ... a little algebra shows that the expressions like (sin +/- mu*cos) and (cos -/+ mu*sin) easily give the formula for the tangent of the sum or difference of angles. (One angle is theta, the other is arctan(mu).) This is easy to see if you draw a force diagram: when the static friction reaches its maximum magnitude (at mu*N), there will be a right triangle in the force diagram with sides N and mu*N. Because of the angle of N (=theta measured wrt vertical), the adjacent angle of that right triangle (which of course is arctan(mu)) is going to add to or subtract from theta. These problems should always start with a force diagram: I would very much insist on this. It will help deal with issues like the adding/subtracting of 3mu*mg noted elsewhere in these comments.

@Walczyk

3 жыл бұрын

there is a sign error which is significant

@jimcarrubba2419

3 жыл бұрын

@@Walczyk You are correct: it is significant. But it is just one of many problems with this video that could be addressed by a more careful solution, including (most crucially) the use of free-body diagrams. Here's another problem which---although it does not affect the answer---really needs to be addressed: why is the normal force on the lower triangle equal to 3mg? You *cannot* say it's simply because the surface supports a total mass of 3m. It follows from that, *and* from the fact that the upper triangle is not sliding; for if the upper triangle were sliding, then that would affect the forces between the triangles. (Alternatively, you can say that the vertical acceleration of the system center of mass is related to the net external force, of which the normal force exerted by the surface is a part.) But it seems to me many of Michael Penn's math videos are quite careful---my PhD is in physics, not math, so I'm no expert---and I think if he gets a bit of guidance from his colleagues, then his future physics videos might be worthwhile.

@Walczyk

3 жыл бұрын

@@jimcarrubba2419 aye, i'm on a leave of absence but i'm in the middle of my physics phd. its bold that he tried and i love his math videos. that said, i'd imagine this is one of the easier olympiad problems.

Maybe you could also try out Informatics Olympiads, there are many interesting problems in Olympiads like IOI, APIO, CEOI, BOI... Edit: A more specific suggestion of a problem would be time is money from Balkan OI 2011, it deals with graphs and Minimum spanning trees and also exploring the convexity of the function to find the optimal answer.

3 жыл бұрын

I'd love to see more combinatorics and concrete / discrete math, too.

Man equating the negative of the normal and friction to the acceleration and gravity threw me for a loop, instead of saying "oh, we need the vertical component to cancel" and adding all 4 forces together

I think that the usual place is to note that you can replace the small block by a point particle w.l.o.g. then you put the forces on the diagram. Then it's just a matter of resolving vectors parallel and perpendicular to the inclined plane and look when there is no force, then you will get a relation for the force.

minute 8:00, µ

I am very confused at 3:49 he calls the angle of the small triangle theta when it should be 90 minus theta. I will give him his assumptions that the triangles are right triangles and similar despite the diagram not specifically calling it out in the diagram but now he just assumed they are 45 degree right triangles

@t39an8r

3 жыл бұрын

I was thinking the same, the other angle would be theta making the one he labeled to be 90-theta, but it's alright, the method is perfect, small mistakes happen

@Chris-eh8mi

3 жыл бұрын

I just got to this point in the video and it stopped me dead in my tracks too, so I went to check the comments. I'm glad someone else noticed and posted so I could have a sanity check about basic geometry,

@ohiosucks2010

3 жыл бұрын

I am not going to mess with trying to track the implications other than to say all of his sines and cosines are flipped because of this.

@jagmarz

3 жыл бұрын

He points to the wrong corner of the small block, calling it θ, but when you calculate the x component of the normal force N, it does actually turn out to be N sin θ, and the x component of the friction force is N μ cos θ. It's a lot confusing because θ is nearly 45 degrees, but if you draw the setup with a much smaller θ, you can see he's got it right. Just think about it a bit. As θ approaches zero, the X component of the normal force also approaches 0, which shows that N sin θ is the correct expression. Also, the triangle he's really thinking about is shown at 10:23.

HOLY CRAP!! this question was on my Physics Exam in 2018!! It was such an interesting question; I wondered where the prof got it from!

@dioptre

Жыл бұрын

USPhO 2017

@michael penn. Normally for newton's 2nd law it is written as ΣF=ma. On the left hand side you have all the forces acting on the object and on the right hand side you have the effective force (aka effective acceleration on the object) which changes the sign to + for the expression for F. also at 8:13 you mentioned that you have a negative acceleration when (mu tan theta). It physically wouldnt make sense either for strength of friction to become less (with the minimum of 0) to avoid the object from sliding down. As you mentioned if there is velcro then it wouldnt slide aka mu is getting larger). I hope I didn't make any mistakes in my calculations.

@chaoticlife5569

3 жыл бұрын

I also got Fmin >= 3mg*((1+mu^2)*sin(theta))/(mu*sin(theta)+cos(theta)) and Fmax

:) This is a very formal mathematical solution for the general case with all possible boundary conditions. That there should be tan theta is kind of obvious and is done in one line if you take the specific condition. :) Sin theta / sin (90 - theta) since only one axis needs to be calculated! That's a physicist's solution.

Weird, I'd swear my physics teacher put this question on one of our exams

@brenofilho3320

3 жыл бұрын

that's a famous question

Why does the direction of static friction change for different values of F?

I'm not sure about the part when he says F(small)= . I feel like the mg is incorrect as its not mass times gravity but its just the vertical component. It doesn't really affect the calculation in this case but I'm just confused.

Wow! its been a while since I was here, 125k already!

"F_whole = 3 m a + 3 u m g", not "F_whole = 3 m a - 3 u m g" because the more the friction with the floor, the more the force you will need to apply, while your equation produces the opposite effect. I computed: F_min = 3 m g (1 + u^2) / (u + cot(theta)) also, if "u > tan(theta)" then there is no positive minimum force needed, not if "u < tan(theta)" as you say in video

Edit: I realized that my explanation of two separate ranges for negative accelerations was incorrect, therefore I deleted from below to not confuse who reads this. I remember that the Russian professor on our PHYS 101 course decided to solve something similar to this and I wasn't able to understand the solution at the time. So I challenged myself to solve this before watching the video. Eventually, I was able to solve it for the negative acceleration cases too. The sign of the friction flips for these negative accelerations, as the lower wedge travels backwards. It was a very nice and very inspiring challenge, which also brought back some memories from my freshman year. Thank you very much! PS- 8:10 It should be μ > tan(θ)

I didn't go through the video, but checking the end I think there is a sign mistake : if µ increases, obviously F should decrease ; take θ=45° M.Penn equation becomes F=3mg (1+µ²)/(1-µ), F increases with µ and that can't happen. For my part I found F=3mg.sinθ.(1+µ²)/(µsinθ+cosθ)

@christianpalumbo8278

3 жыл бұрын

Finally someone who got the same answer as me! I was getting very confused with all those variants, signs, trig functions in the comments. My answer is the following: F = 3mg(1+µ²)/(cotθ+μ), which is pretty much the same thing you wrote.

@theotexeditions2658

3 жыл бұрын

@@christianpalumbo8278 Thanks. Now a real physicist wouldn't have said at the end of the video : "this is a good place to stop", because for him maths are just a tool, and he won't stop until he can conciliate his intuition and the formula (which is not always possible, but at least he tries). More over a physicist wouldn't have set the problem in those terms, because the small block doesn't move relatively to the big one, so the friction coefficient to be taken into account is static. But the big block is moving on the table, the friction coefficient then is dynamic, and doesn't have the same value...

What has he done to decompose the Forces at 4:50?

what camera do you use for your videos, Michael?

Ha! Spoken like a true mathematician. I think you will prove to be most useful Michael....

Wow that's soooooooooo amazing :)

Correct math but bad reasoning for physics. For F_whole it should be "moving horizontally, not moving vertically", which (by mutliplying accelecations on mass) give = . For F_small (by same justification) it should be = .

@Fysiker

3 жыл бұрын

I prefer your way

@Walczyk

3 жыл бұрын

no the math is wrong too as a result lol.

Why is the bottom angle of the green block theta

When will the next physics video be?

Summation n=0 to n=88 of 1/cosn×cos(n+l)=coti.coseci then find the value of i Jee Exam 2009 Question

3:46 isn’t that angle 90-theta instead of theta?

This was a lot of fun.

This is what i do when i procrastinate instead of doing my math homework

Great professor 👍😀

We can simple this frication with math formula for tan(ALFA+-THETA). That way the first expreesion becomes mg tan(THETA-ALFA) so that saying us that ALFA must be smaller then THETA. ALFA is equal to arctan(MI) and it will be more understandig for people how dont know physic. We can draw ALFA with force diagram for normal and friction forces. Its simlar for second (maximum) solution

4:30 "..and we've got an acceleration due to gravity down.." He has forgotten the reaction force from the big block.

@icystrangers5482

3 жыл бұрын

oh...ok...he's taking gravity as an acceleration..

3:43 isn't that angle 90-theta, not theta?

hi how can I get the parametric equations of the diocles cissoid?

@tomatrix7525

3 жыл бұрын

I usually scroll past these comments but today I”d thought not to. Why do you just randomely post this in an unrelated video? You are not really helping yourself because it’s unlikely anyone will reply.

My training in physics is reduced, so I could certainly be wrong, but I would like to understand two points : - at 2:40, why not a PLUS ? The needed force = the force who moves the system + the force reacting to the friction. Isn't it ? - I'm confuse by the use of the same μ in the two contacts, because the first must be a dynamic friction and the second a static friction (if the goal is achieved). Thank you for an explanation if you have the time and continue this great channel. And yes for more physics videos.

I'm thinking from a physics perspective you inspect the problem. There is one constraint - M2 does not move with respect to M1 in the solution, therefore F=3ma+3um. Then how to find a? The forces up and down the ramp must add to zero. It is easy to see that this means Framp = m(uNsin + asin - gcos)=0 and therefore a = (gcos -uNsin)/sin or sometime like that. I don't have any paper here.

2:39 That doesn't make any sense. The net force to the left is equal to F - 3mμg, since F is positive valued when acting to the left, and 3mμg acts to the right. And by Newton's second law the net force is equal to M*a, therefore F - 3mμg = 3ma, and F = 3ma + 3mμg, not minus. You can think that if F were 3ma - 3mμg, then the force you have to put in your system to the right is LESS than the force needed to accelerate 3m with an acceleration of a, which would mean that friction is HELPING you to move the blocks!

This is a nice video. In my opinion it could be even better if you focused more on the physics part of the problem. The equation should come from the system modeling part but the video does not tell what is the point (part of the system) that you consider forces in (what are those forces and what is the reference point for considering them). What are the equilibrium constraints for the upper block to not slide up or down. Also, I'm not sure what is the purpose of introducing acceleration `a` in the equations. The math part is great as usual :).

Perhaps you could do some videos on econometrics?

i dont understand why the normal vector and the friction vector have the same labeling ( N ). Dont we have to label the friction some other Letter ? because they are not the same ?

@jagmarz

3 жыл бұрын

Friction is a fraction of the normal force N; this fraction is μ the coefficient of friction. So the Friction vector is μN, which is, generally, different from N.

can someone please explain how the whole negative thing works? thanks

@anugyaseksaria7796

2 жыл бұрын

I mean...it's like easiest physics Olympiad question and u don't even know vectors

it's +3μmg though is n't it ?

about the sign error of the friction force: i think that it is worth reminding ourselves that newton's equation F = ma has the form "cause" = "effect". imho michael made a mistake because he mixed his causes, i.e. +F and -3mug (left = "+", right = "-"), with his effect, i.e. +3ma. had he started his problem from newton's equation and had he plugged in his causes and his effect into it, he would have made no mistake. i.e.: F - 3mug = 3ma, which transforms into F = 3ma + 3mug. the last equation has a nice interpertation btw if we multiply everything by distance d: F*d = 3ma*d + 3mug*d "applied work" = "work used to speed up an object" + "work against friction"

I think there is an almost similar problem in irodov but there, they have asked what is given in this problem

@l1mbo69

3 жыл бұрын

Yeah you're right pretty weird that USPhO would put a question from such a famous book that too in a recent year

@divyanshaggarwal6243

3 жыл бұрын

These types of problems are common in India class 11th syllabus if you are preparing for competetive exams. Drawing a free body diagram makes the problem a lot easier.

@prabkiratsingh4846

3 жыл бұрын

@@divyanshaggarwal6243 me 2 preparing for same

Has anyone here read the book "Sear and Zamansky's University Physics with modern physics"? Its authored by Hugh D Young and Freedman

Could you explain why the equation for F_whole is 3ma - 3mmug? It makes more conceptual sense for it to be +?

@tomatrix7525

3 жыл бұрын

You are correct. He is weong to my knowledge

@joshyman221

3 жыл бұрын

It’s simply a matter of defining what you are talking about. Throughout he clearly is calling 3ma the resulting motion of the body from pushing it. In this case, the 3ma is the F drawn in the diagram and the F in the equations is the resultant force on the body (3ma-3mug)

@ArifSolvesIt

3 жыл бұрын

He is not only making a sign mistake here but also a very important conceptual one. This is what we usually see in most of our Physics students as well, especially when the concepts are not understood thoroughly. He is definitely WRONG! If F in his solution referred to the net force, then the equation would read F=3ma. But since he keeps saying that F is the force applied from the right, he should have written for the net force as F(net)=F-3\mu mg. Then, this net force is equated to 3ma, yielding F-3\mu mg=3ma. So, either way the equation he writes is WRONG, WRONG and WRONG!

@maxmustermann3876

3 жыл бұрын

@@joshyman221 The result he ends up with is wrong though, because he later is using F as the applied force and a as the resulting acceleration.

Professor ur are multi talented

How is the angle at 3:48 also theta?

@djanlukarosi4731

3 жыл бұрын

It is not upper angle is theta

So this is how to use the friction coefficient!!

How the hell could the small block shift upward on the diagonal simply by a horizontal force on the larger block? Does not make sense!

@divyanshaggarwal6243

3 жыл бұрын

A force perpendicular to the contact between the two blocks is generated. This is called the Normal Reaction force. This force will have a component in the upwards direction hence pushes the small block upwards.

In India we have a similar question but there the wedge is PARABOLICAL and a sphere of mass m

Wow sir

At 3:14 something goes wrong when transfering the solution to the chalkboard. It's not "what happens when F is small" but "what happens"with the forces on the small block.

@maxmustermann3876

3 жыл бұрын

Hmm, I think he is correct at this situation. Since friction always acts opposite to the direction of motion, the equations for the forces on the small block change, depending on the scenario (F too small, F too big).

This was one of my physics 101 mid term exam

Thanks you for this beautiful videos .but l holp to explained in your channel computer science and programming and some problem in ioi and in icpc

@goodplacetostop2973

3 жыл бұрын

Oh yeah, I would love to see Micheal videos about computer science : graph theory, algorithm analysis, coding theory, automata...

@goodplacetostop2973

3 жыл бұрын

I actually graduated in computer science so this is my dream lol

@coolmangame4141

3 жыл бұрын

that would be real dope

@noumaneelgaou1624

3 жыл бұрын

I find a lot of video of theoretical computer science like language theory and graphs and automate finit ... ,but l prefer see this video in channel Michel penn. In my opinion this channel the best for understand sience math

4:15 that's not true at all! The small block isn't accelerating in the downwards direction with an acceleration of g. If it were, it would be in freefall, which it isn't. To find the downwards acceleration you'd have to first find the net force in the downwards direction, which is composed of the vertical component of the normal force, and the vertical component of the friction force

that's new and great

I didn't know Physics Olympiads were even a thing that existed.

3 жыл бұрын

There's also Math and Chemistry and even Biology Olympiads. And probably a few more.

Umm,,i think F should be 3ma+3mgu ...which is 3mg[{(sinθ-ucosθ)/(μsinθ+cosθ)}+μ)

2:00 There is another mistake. Even though the block isn't moving in the vertical direction, there is still a force acting on the system. F_vertical = N - 3*m*g would be correct. Since N is exactly 3*m*g, it cancels out. But you still need to keep it in mind, because this vertical force is the reason for the horizontal friction, which needs to be considered. If you write F as a 2D vector (even though there is only one dimension needed to solve the task), then you might as well do this correct.

This is my middle school exam problem ……

3:48 this is NOT theta

@impk100

3 жыл бұрын

Yes, you are right. It should be 90°-θ

@maxmustermann3876

3 жыл бұрын

True, but the following formula (until 5:00) is still correct.

@arifkg9673

3 жыл бұрын

That drove me crazy man

@JCrashB

3 жыл бұрын

Hahaha, I looked for this comment immediately after watching 3:48 :)

Make more videos on physics.

The problem is somewhat incorrectly stated. The coefficient of friction is not mu on every surface. For the bigger block, the friction coefficient with the floor must be at mu if the block is moving (which it must be since we are trying to find the maximum and minimum values for F). But, the friction coefficient between the smaller mass and the big block is, in general, mu_1 F = 3ma + 3mu*mg. There are no two ways about that.

Brave effort :). The problem would be nicer if stated in terms of acceleration (with g as the unit) rather than force. And while you did do some discussion of the results, a physicist would do more thorough discussion in terms of a and mu, in order to get a feel for the whole system. E.g. using some realistic mu like 0.8 and see what how theta and a would mutually relate then.

@Walczyk

3 жыл бұрын

the physical interpretation would point out its all incorrect.

There is no way that the statement @ 8:10 could be true in our world. The small block sits still without a force, when mu is _bigger_ than tan(\theta).

@Walczyk

3 жыл бұрын

Yes, he has the wrong sign on 3*m*mu*g, I think that's the only error. I should really solve the problem on paper.

@garkechify

3 жыл бұрын

Of course signs can be messed up if not treated carefully. Therefore it always a good idea to check end results for plausibility to not disgrace oneself...

@Walczyk

3 жыл бұрын

@@garkechify hes a math guy not a physics guy haha, id be embarrassed though

@maxmustermann3876

3 жыл бұрын

@@Walczyk This sign has nothing to do with the mu > tan(t) though. In case there is no acceleration necessary, mu > tan(t) because sin(t) - mu*cos(t) In this case the force can still vary between a neg and a pos value X, because we need to apply a force greater 3*m*mu*g to overcome friction and get the block moving in any direction. "no acceleration necessary is actually equivalent to "no force necessary" because friction will never accelerate the block on its own, but always act in the opposite direction of the movement.

@Walczyk

3 жыл бұрын

@@maxmustermann3876 Isn't the criterion for tan(pi/4) = 1 < mu, a mu greater than 1 is unphysical. I think you're right, I don't think I listened to the interpretation of the criterion. Gotta look again. If mu was very small then it would slide down under the force of gravity. If mu was that high then gravity still be a downward force on that block. It cannot be symmetrical in both directions. It will take more magnitude of force to move the block up the ramp than it does to bring it down

Thanks,

2:45 I guess there is a mistake The sign of 3*m*g*µ must be a "+". F - 3*m*g*µ = 3*m*a should be the correct equation.

@zombiekiller7101

3 жыл бұрын

Same

uuyy , a qué hora se levantó !!? jajaja

F min -> Body will have tendency to slide down F large -> Body will have tendency to move up

But I've learnt this exactly problem in my IIT-JEE exam but it's something easiest problem for a jee aspirants

Premise: Physics is just applied math. Conclusion: Professor Penn should ace this.

Find last three terms of 63^25×25^63 ans is 375

how could usapho could be so ez,it's like a 30 second problem,try Indian physics Olympiad

This problem is present in IE irodov too

Please try iit jee problems!!

4:07 This should be because there is no acceleration in the horizontal direction. The sum of the forces should be 0, he just sets the forces equal to each other, which does have the same result, but is conceptually wrong.

@icystrangers5482

3 жыл бұрын

Agreed. He takes gravity as an acceleration.

Well....I got something different for the minimum value

Hi, For fun: 1 "so let's may be go ahead and do that", 1 "ok, great", 1 "so, let's go ahead and".

I have a document derivative of the function f(x)=x^x by first principle

@arshaniksa589

3 жыл бұрын

I'd appreciate it if you could possibly provide a link to the doc.

I think this is very easy question for a physics professor....... cause in india any 11 standard students can solve this type of questions who are preparing for jee

@debbaranyabiswas1102

3 жыл бұрын

Absolutely true....I solved it....but got only the minimum value

Have solved very similar question 13 years ago from HC Verma...

Honestly, not quite "olympiad"-level problem. We solved such problems almost routinely in the school (although I was in the "phys-math" specialised class). Olympiad problems seemed to be harder.

At 2:48 F = 3*m*a + 3*n*m*g

This video makes me insecure with my maths

This was at the Olympiad? It's the most basic thing I studied