Just making an observation. The 'Net Work-Kinetic Energy Theorem' does indeed have one underlying assumption. That being the mass/inertia being constant. An example where the 'Net Work-Kinetic Energy Theorem' fails (due to this assumption) is in rocket mechanics because such systems' masses change with time.
Thanks! It was super helpful, I was given to prove that work equals change in kinetic energy in last year's exam and I somehow managed to do it without Calculus.
+Shoaib Rashid If you assumed the net force is constant, then the acceleration is constant, then you can use a Uniformly Accelerated Motion equation in your derivation. I prefer this derivation; it's more fun.
what a PERFECT video from Beginning to End!!... now give the kid his Two Bucks!! lol.. funny!!!!.. You are an Excellent Presenter AND TEACHER!!.. you're one of the RARE ones in this World of CZcams Videos.. THANK YOU!!.. you make me want to work this Derivation again .. but this time I'll substitute Force = dP/dt... change in Momentum... that should work too.. :) ...
Watching from the future, on the first line at 2:25, you can notice that (dv/dt)dx can be rewritten as (dx/dt)dv, where v = (dx/dt). Therefore you have (v)dv. I think that has a more natural flow to solving the integral wrt velocity than dv/dt = (dv/dx)(dx/dt). (And it prevents "canceling" dx which is generally not viewed favorably by mathematicians.) Just a simpler way to solve it.
If we are going to play the mathematicians formal game, then we shouldn't handle derivatives as fractions either. It's an operator that looks like a fraction.
Hey, great video! I saw one of your other videos where you go over how long it takes to make one of these type of videos. It's a lot of work! Maybe you should try making some videos without all the edits so it would save you a lot of time and you could add more content. Also if you made physics caluclus based videos, it would target a lot of college students like myself. And there is no real established physics person on youtube like there is for math. Patrickjmt and ProfessorLeonard are my favorites for math. Thanks for the great teaching you have provided me!
+kevin how I am not currently planning to do this. I already have all of my in-class lecture videos on my website. This includes ~40 hours of calculus based physics videos. flippingphysics.com/calculus.html I know they are not as clear as my Flipping Physics videos, however, the lecture notes are there as well, which helps.
If you're like me, you did not understand how he cut the dx in int(dv/dx * dx/dt) dx. I found a way to prove that this is possible for every function f(x)!!! 1. int(df/dx * f) dx= f * f - int(df/dx * f) dx
6:58 when you derived the work-energy theorem, didn't you technically make the assumption that the mass was constant? So the theorem would not apply in the form you stated?
but why did you treat mass as a constant that you can pull out of the integral vice acceleration which you left in? What compelled you to leave a in and take out m and not the other way around!
Dear sir, you wrote dv/dt=(dv/dx)*(dx/dt) so, does velocity implicitly depends on position here? As I can see (dv/dx ) term, but v is only time dependent. How were you able to write this expression? Could you please make it clear! TiA.
If v were completely independent of x, then dv/dx would equal zero, and you would get nowhere by doing that. But because v, x, and t are all interdependent on each other as kinematics quantities, you will get a derivative dv/dx and dx/dt.
6:50 sir, would it be wrong to say that we assumed mass to be constant and hence thus equation is not applicable for variable mas systems such as rockets??
Yes. You are correct. We used the form of Newton's Second Law which is only applicable for constant mass objects. So this would not be applicable for variable mass systems such as rockets. Thanks!
I have a very important question for me:) In space, there is a space shuttle that does't move. Suppose that its mass is always constant. This space shuttle has a rocket engine with a constant force of 100N. When the engine is started and the ship has a constant thrust of 100N, when the ship starts to accelerate, will the acceleration be constant and remain constant (let's assume 10m / s), or maybe the acceleration will start to decrease as the speed increases? It is related to the rule Ek = 1 / 2mv ^ 2 ??
The acceleration of an object experiencing constant force will be constant. The acceleration comes directly from Newton's Laws (a=F/m); for most mechanics problems, you consider conservation of momentum above that of energy. That isn't to say that energy isn't conserved, just that there are more ways that it can be dissipated, so is generally more difficult to work with. Now in your example: if thrust is constant then the energy being produced by [chemical reactions in] the engines is constant. As you appear to have noticed however, the rate of change in kinetic energy (or "power," dE/dt) of the rocket starts low, and increases over time. To ensure conservation of energy, we have to assume that there is somewhere else that the kinetic energy is going, and that this energy loss changes in the opposite direction, i.e. the loss decreases over time. The answer is to consider the fuel being flung out of the exhaust. As the rocket starts off, it's change in speed is small, but the exhaust gasses are accelerated in the opposite direction to massive speed, so gain a large amount of kinetic energy. Later on however, because the exhaust gasses are flung in the opposite direction to the motion of the rocket, they are having to "slow themselves down, and then accelerate again," so gain less energy. Example: The rocket engines are designed so they have a certain "exhaust velocity", say, 1000m/s. When the rocket is stationary, the Kinetic energy gained by a particle emitted is 1/2*m*1000^2. Later on the rocket is in motion, say at a speed of 200m/s. For a stationary observer, a particle will be travelling at 200-1000=-800 m/s (-800 m/s, means 800 m/s in the opposite direction). The *change in* KE of the particle will be KEfinal-KEinitial=1/2*m*(-800)^2 - 1/2*m*(200)^2 = 1/2*m*632^2. (Much lower).
A rocket is a system of varying mass. You need to integrate the total impulse of the thrust force from the fuel, and then apply it as the change in momentum of the remaining payload. The total mass will decrease over time, so its acceleration will consistently increase if its propellant is burned at a constant rate.
Each position will correspond to a particular velocity though. Once the integral has been expressed in terms of velocity, we only need the starting and ending velocities to evaluate it.
In this context, is it allowed to treat the differentials as fractions? Like when you canceled some out or switched things around to integrate depended on velocity instead of position? It feels so wrong to do so, but how else can you get the same result?
Hi proffesor thank you so much for your great explanations How did we get into the integral of velocity with respect to velocity from the integral of force times distance ? I can understand it in numbers and calculuse but i can't get it in real world when it happens
Any question where you use the concept of kinetic energy as a shortcut to solve, is where you would use this information. You don't need to derive it from first principles every time, but this is to give you an idea of what is behind the equation of KE=1/2*m*v^2.
Actually, this particular derivation is for AP Physics C which is a calculus based physics course and most high school students do not take this class.
Always true? How do you explain to students that that kinetic energy can be converted into other types of energy (like gravitational potential energy) over the duration of the scenario? In those cases the Work is equal to the change in mechanical energy, not kinetic energy.
If I may offer some kind of advice. Do you think it might be a little confusing to refer to both the pre and post calculus velocities as "velocity"? I think it would be enlightening to refer to "pre-calculus" velocity as "average velocity" because that's what it really is. Students would understand that this is only an average and they might also see why it would be desirable to somehow obtain a function that tells us the velocity at any time. In that way, there's clearly some more substantial knowledge to be gained from calculus. If I were a student, I would spend my free time wondering, "well, why do we need two different equations for velocity? That probably means those equations are equal, right?"
+Zuzu Superfly Thanks, that is a valid point. I do describe the calculus equations as "instantaneous velocity" and "instantaneous acceleration", however, it would have been more clear to also refer to the algebra equations as "average velocity"' and "average acceleration". It's not worth redoing the videos over, however, I will keep it in mind for the future and I have updated the lecture notes on my webpage to make it more clear. Thanks. FYI: If you are interested in seeing my video on Newton's First Law, you can see it here: www.flippingphysics.com/first-law.html
Hello! This explanation was easy to understand, however Im a bit confused at part 3:12 as to why the dx in the denominator of velocity dv , did not cancel out with with the dx in the numerator with respect to time and would then yield the integral of (dv/dt) dx?
Seems kinda weird, because if this would be derived for the first time, someone would have intuitively, cancelled out the dx's in the bracket and gotten (dv/dt) dx, however it would lead to deadline (since you just said it would not progress forward in solving the integral). Or mabye this needs some deeper math understanding that I dont have. But thanks :D
lol lolzi there is a formal way of deriving this wich doesnt involve tricks like this. It only requires a definition of the rieman integral and and the mean value theorem
+Abdulkadir Akti 1) If the force is not the net force then you are not able to substitute in "mass times acceleration" for "net force" and the derivation does not work. So yes, it has to be the net force. 2) If you are referring to gravitational potential energy then the net work will include the work done by the force of gravity. If you are referring to elastic potential energy then the net work will include the work done by the force of the spring.
+Abdulkadir Akti Net Work equals Change in Kinetic Energy is valid if acceleration is not constant as well. I'm not sure what you are asking with "what whole equations looks like?"
+Keshav Karat Watch carefully. For some reason I accidentally to wrote dx with the x as a subscript, which doesn't make any sense. So I moved the x up to be in line with the d, however, when my head gets in the way, it doesn't work.
I did this today by algebric method which is more easy than you did here. If you have face book page I will send you the image of those my note book. But again thanks. If you have no facebook page so made a facebook account today
+Swati Jain Nope. Not really a gamer. In my youth I played some computer games, now I find other ways to fill my time. I make these videos, for example.
you cannot cancel dx unless you give a proof why it works. you should state what calculus you do! if you work with standard analysis, dx is just a symbol, it doesn't mean anything! if you work with non-standard analysis, you should say that! most of people who watch this don't know what nonstandard analysis is ,so they don't understand why it works (me too). almost everyone does the same trick. why???? this is the question. i am not looking for trick that no one explain! i think this video is useless bc there is still question, whyyyyy????? trying to explain why it works is easier than endeavoring not to explain it!
Think about what dx means. It means infinitesimal change in the value of x. If we are talking about a change between the same two instants in time, you can cancel the dx, because it is the same change of position during this time.
i want to create a mathematical theorm for energy shortage impact in Pakistan, anyone help me for creating this theorm? please contact me as soon as possible
Watching from the future, on the first line at 2:25, you can notice that (dv/dt)dx can be rewritten as (dx/dt)dv, where v = (dx/dt). Therefore you have (v)dv. I think that has a more natural flow to solving the integral wrt velocity than dv/dt = (dv/dx)(dx/dt). (And it prevents "canceling" dx which is generally not viewed favorably by mathematicians.) Just a simpler way to solve it.
Just making an observation. The 'Net Work-Kinetic Energy Theorem' does indeed have one underlying assumption. That being the mass/inertia being constant. An example where the 'Net Work-Kinetic Energy Theorem' fails (due to this assumption) is in rocket mechanics because such systems' masses change with time.
This was fantastic.
You're a great teacher and a great actor. Thank you for making these videos!
You are welcome!
wow. that was beautifully explained. thanks a lot
Thank you for your lovely comment.
Great explaination...Thanks, admire your work a lot !
+Rahul Jha Thanks. Glad to know my work is appreciated.
I am also from the same country from which you are.
I have the same proof in my physics textbook for class but seeing you explain it so eloquently helped me easily understand it.
GREAT presentation!
Thank you!
you have no idea how much clicked in my head whilst watching this 8min video thank you so much man
I love the word _whilst_
Sir
Just Loved the Way you Teach❤️
Sir you helped thanx alot
Sir you are great
You explained it very well
You are my Newton
Thanks! It was super helpful, I was given to prove that work equals change in kinetic energy in last year's exam and I somehow managed to do it without Calculus.
+Shoaib Rashid If you assumed the net force is constant, then the acceleration is constant, then you can use a Uniformly Accelerated Motion equation in your derivation. I prefer this derivation; it's more fun.
+Flipping Physics While deriving Bernoullis Equation, don't we take work to be equal to change in kinetic energy as well?
+Shoaib Rashid I think it is derived using conservation of energy.
How does this person not have aa million susbcribers
what a PERFECT video from Beginning to End!!... now give the kid his Two Bucks!! lol.. funny!!!!.. You are an Excellent Presenter AND TEACHER!!.. you're one of the RARE ones in this World of CZcams Videos.. THANK YOU!!.. you make me want to work this Derivation again .. but this time I'll substitute Force = dP/dt... change in Momentum... that should work too.. :) ...
wow, it worked!!.. using dp/dt for F in the W =Fx :D . thanks for the Inspiration!!
That is cool!
Dude, u guys are awesome
Watching from the future, on the first line at 2:25, you can notice that (dv/dt)dx can be rewritten as (dx/dt)dv, where v = (dx/dt). Therefore you have (v)dv. I think that has a more natural flow to solving the integral wrt velocity than dv/dt = (dv/dx)(dx/dt). (And it prevents "canceling" dx which is generally not viewed favorably by mathematicians.) Just a simpler way to solve it.
If we are going to play the mathematicians formal game, then we shouldn't handle derivatives as fractions either. It's an operator that looks like a fraction.
i am really looking forward to your answer
I wish my maths and chemistry teacher was as cheerful as this guy!!!
Capital "F" stands for the Faraday constant.
Lowercase "f" is traditionally used for force.
Hey, great video! I saw one of your other videos where you go over how long it takes to make one of these type of videos. It's a lot of work! Maybe you should try making some videos without all the edits so it would save you a lot of time and you could add more content. Also if you made physics caluclus based videos, it would target a lot of college students like myself. And there is no real established physics person on youtube like there is for math. Patrickjmt and ProfessorLeonard are my favorites for math. Thanks for the great teaching you have provided me!
+kevin how I am not currently planning to do this. I already have all of my in-class lecture videos on my website. This includes ~40 hours of calculus based physics videos. flippingphysics.com/calculus.html I know they are not as clear as my Flipping Physics videos, however, the lecture notes are there as well, which helps.
I gotta watch all of these and take notes by Wednesday
I hope it went well.
amazing!! i've been wondering where KE comes from!
this is better very good, am in uni rn and this is helpful
Glad it helped!
Bro this was so helpful you are seriously awesome Thanks!!!
You are welcome!
Great explanation, very clear!
thanks a lot bro i studied entire work energy chapter but our teacher didn't derive this theorem. thanks for helping
Great... had forgotten how to calculate it and here it is perfect
Glad to able to help you out.
this video is amaaaaizing , made me understand what my stupid physics book couldn't make me understand!!!!!!!!!
Thank you so much bro, cheers!
You are welcome!
😊wow Nice explain. Much obliged!! from Sri Lanka ❤
Excellent video.
im in 10th and we did algebra based, but honestly i find this much easier
If you're like me, you did not understand how he cut the dx in int(dv/dx * dx/dt) dx.
I found a way to prove that this is possible for every function f(x)!!!
1. int(df/dx * f) dx= f * f - int(df/dx * f) dx
6:58 when you derived the work-energy theorem, didn't you technically make the assumption that the mass was constant?
So the theorem would not apply in the form you stated?
but why did you treat mass as a constant that you can pull out of the integral vice acceleration which you left in? What compelled you to leave a in and take out m and not the other way around!
Subscribed at the end!!
Thanks btw
Dear sir, you wrote dv/dt=(dv/dx)*(dx/dt) so, does velocity implicitly depends on position here? As I can see (dv/dx ) term, but v is only time dependent. How were you able to write this expression? Could you please make it clear! TiA.
If v were completely independent of x, then dv/dx would equal zero, and you would get nowhere by doing that. But because v, x, and t are all interdependent on each other as kinematics quantities, you will get a derivative dv/dx and dx/dt.
Thank you for the great video!
You are welcome.
*Always True*
Einstein: Hold my Relativity
the best hippie physics professor
6:00 why did the x above your head move downwards
Hes a wizard
6:50 sir, would it be wrong to say that we assumed mass to be constant and hence thus equation is not applicable for variable mas systems such as rockets??
Yes. You are correct. We used the form of Newton's Second Law which is only applicable for constant mass objects. So this would not be applicable for variable mass systems such as rockets. Thanks!
Flipping Physics Glad to know that! Btw the vid is simply superb!!
Thanks. Glad to be able to help you learn!
Nicely elaborated..
I have a very important question for me:) In space, there is a space shuttle that does't move. Suppose that its mass is always constant. This space shuttle has a rocket engine with a constant force of 100N. When the engine is started and the ship has a constant thrust of 100N, when the ship starts to accelerate, will the acceleration be constant and remain constant (let's assume 10m / s), or maybe the acceleration will start to decrease as the speed increases? It is related to the rule Ek = 1 / 2mv ^ 2 ??
The acceleration of an object experiencing constant force will be constant.
The acceleration comes directly from Newton's Laws (a=F/m); for most mechanics problems, you consider conservation of momentum above that of energy. That isn't to say that energy isn't conserved, just that there are more ways that it can be dissipated, so is generally more difficult to work with.
Now in your example: if thrust is constant then the energy being produced by [chemical reactions in] the engines is constant. As you appear to have noticed however, the rate of change in kinetic energy (or "power," dE/dt) of the rocket starts low, and increases over time. To ensure conservation of energy, we have to assume that there is somewhere else that the kinetic energy is going, and that this energy loss changes in the opposite direction, i.e. the loss decreases over time. The answer is to consider the fuel being flung out of the exhaust.
As the rocket starts off, it's change in speed is small, but the exhaust gasses are accelerated in the opposite direction to massive speed, so gain a large amount of kinetic energy. Later on however, because the exhaust gasses are flung in the opposite direction to the motion of the rocket, they are having to "slow themselves down, and then accelerate again," so gain less energy.
Example: The rocket engines are designed so they have a certain "exhaust velocity", say, 1000m/s. When the rocket is stationary, the Kinetic energy gained by a particle emitted is 1/2*m*1000^2. Later on the rocket is in motion, say at a speed of 200m/s. For a stationary observer, a particle will be travelling at 200-1000=-800 m/s (-800 m/s, means 800 m/s in the opposite direction). The *change in* KE of the particle will be KEfinal-KEinitial=1/2*m*(-800)^2 - 1/2*m*(200)^2 = 1/2*m*632^2. (Much lower).
A rocket is a system of varying mass. You need to integrate the total impulse of the thrust force from the fuel, and then apply it as the change in momentum of the remaining payload. The total mass will decrease over time, so its acceleration will consistently increase if its propellant is burned at a constant rate.
woah there, why are we allowed to change the limits of the integral at 3:27. Last i checked, position doesn't equal velocity
Each position will correspond to a particular velocity though.
Once the integral has been expressed in terms of velocity, we only need the starting and ending velocities to evaluate it.
What mic are you using? Sounds amazin in headphones 😍
In this context, is it allowed to treat the differentials as fractions? Like when you canceled some out or switched things around to integrate depended on velocity instead of position? It feels so wrong to do so, but how else can you get the same result?
Hi proffesor thank you so much for your great explanations
How did we get into the integral of velocity with respect to velocity from the integral of force times distance ? I can understand it in numbers and calculuse but i can't get it in real world when it happens
Perfection baby
i love this channel😇
❤️
Work is a scalar quantity and you have it equated to (a change in) velocity (a vector quantity)…
It's a dot product of velocity and change in velocity. That's how it becomes a scalar quantity.
thank you!!
This is great explanation, but what kind of questions do I have to face to use this information?
Any question where you use the concept of kinetic energy as a shortcut to solve, is where you would use this information. You don't need to derive it from first principles every time, but this is to give you an idea of what is behind the equation of KE=1/2*m*v^2.
Whoa I learned this in college . Seems like high-school has included it in syllabus
Actually, this particular derivation is for AP Physics C which is a calculus based physics course and most high school students do not take this class.
is there a video demonstrating the background of this part: DV/DT =DV/DX * DX/DT ? i forget it . its been 20 years
I don't have one, sorry.
ok... i guess i need to take the course again. lol
This is the chain rule for finding derivatives.
thank u for explaining this!!! tomorrow is my exam...
+decorative ideas Good luck!
funny and informative was of teaching.
loved it
Always true? How do you explain to students that that kinetic energy can be converted into other types of energy (like gravitational potential energy) over the duration of the scenario? In those cases the Work is equal to the change in mechanical energy, not kinetic energy.
Please watch this: www.flippingphysics.com/energy-transfer-system.html
Extremely helpful
That's great. Glad to help.
can it be done without using the chain rule at all? cause after the chain rule, when we simplify the dx's it is exactly the same as it was before.
Yes
If I may offer some kind of advice. Do you think it might be a little confusing to refer to both the pre and post calculus velocities as "velocity"? I think it would be enlightening to refer to "pre-calculus" velocity as "average velocity" because that's what it really is. Students would understand that this is only an average and they might also see why it would be desirable to somehow obtain a function that tells us the velocity at any time. In that way, there's clearly some more substantial knowledge to be gained from calculus. If I were a student, I would spend my free time wondering, "well, why do we need two different equations for velocity? That probably means those equations are equal, right?"
+Zuzu Superfly Thanks, that is a valid point. I do describe the calculus equations as "instantaneous velocity" and "instantaneous acceleration", however, it would have been more clear to also refer to the algebra equations as "average velocity"' and "average acceleration". It's not worth redoing the videos over, however, I will keep it in mind for the future and I have updated the lecture notes on my webpage to make it more clear. Thanks.
FYI: If you are interested in seeing my video on Newton's First Law, you can see it here: www.flippingphysics.com/first-law.html
2:40 it makes sense now
Hello!
This explanation was easy to understand, however Im a bit confused at part 3:12 as to why the dx in the denominator of velocity dv , did not cancel out with with the dx in the numerator with respect to time and would then yield the integral of (dv/dt) dx?
dv (dx/dt) equals (dv/dt) dx, however, (dv/dt) dx does not progress forward in solving the problem.
Seems kinda weird, because if this would be derived for the first time, someone would have intuitively, cancelled out the dx's in the bracket and gotten (dv/dt) dx, however it would lead to deadline (since you just said it would not progress forward in solving the integral). Or mabye this needs some deeper math understanding that I dont have. But thanks :D
lol lolzi there is a formal way of deriving this wich doesnt involve tricks like this. It only requires a definition of the rieman integral and and the mean value theorem
PD. en.wikipedia.org/wiki/Line_integral i recommend you seeing the actual definition of line integral
you are best sir this is so good
Thanks!
Ag you the best
You are perfect. Thxxxx
Sentiment enjoyed. Though I am certainly not perfect...
3:13 - mathematicians start cringing....
Is there a more rigorous way to change the integral so that it's a dv integration?
what if force is not net , it it still valid, if not how would you explain it, also where is potential energy in work energy theorem
+Abdulkadir Akti
1) If the force is not the net force then you are not able to substitute in "mass times acceleration" for "net force" and the derivation does not work. So yes, it has to be the net force.
2) If you are referring to gravitational potential energy then the net work will include the work done by the force of gravity. If you are referring to elastic potential energy then the net work will include the work done by the force of the spring.
Flipping Physics is this valid for kinematic as well, I was wondering if acceleration is not constant , what whole equations looks like?
I would like to translate all your videos to Turkish
+Abdulkadir Akti Net Work equals Change in Kinetic Energy is valid if acceleration is not constant as well. I'm not sure what you are asking with "what whole equations looks like?"
+Abdulkadir Akti Translations to Turkish would be awesome. Instructions are posted at flippingphysics.com/translate.html Thanks!
Ótimo!!! Excelente 👏👏👏👏👏
7:38-7:42........ the x in the top changes its position 2 times.... video editing?
+Keshav Karat Watch carefully. For some reason I accidentally to wrote dx with the x as a subscript, which doesn't make any sense. So I moved the x up to be in line with the d, however, when my head gets in the way, it doesn't work.
i gave u a thumb up
Thank you!
thanks....
welcome...
i need proof of word equals force times displacementt
fantastic mam awesome
Most welcome 😊
at 6:14 beau makes the 'x' above his head move
May the force be with you.
and also with you.
at 5:54 notice the x behind his hair, it pops down and at 5:58 again pops up! What SORCERY is ThIS?!
MAGIC!!
Ansariz Bros
The sorcery of video editing.
Time for a nap!
Ooooh, I love naps!
Not only clever but also funny
Thanks!
great........................... :)
good.
I did this today by algebric method which is more easy than you did here. If you have face book page I will send you the image of those my note book. But again thanks. If you have no facebook page so made a facebook account today
If you post it on my Facebook page, I'll take a look at it. facebook.com/FlippingPhysics
Nuceeee
7:41 🙀🙀
Great -_-||-_-#
Sir do you play CS GO ?
+Swati Jain Nope. Not really a gamer. In my youth I played some computer games, now I find other ways to fill my time. I make these videos, for example.
you cannot cancel dx unless you give a proof why it works. you should state what calculus you do! if you work with standard analysis, dx is just a symbol, it doesn't mean anything! if you work with non-standard analysis, you should say that! most of people who watch this don't know what nonstandard analysis is ,so they don't understand why it works (me too). almost everyone does the same trick. why???? this is the question. i am not looking for trick that no one explain! i think this video is useless bc there is still question, whyyyyy????? trying to explain why it works is easier than endeavoring not to explain it!
Think about what dx means. It means infinitesimal change in the value of x. If we are talking about a change between the same two instants in time, you can cancel the dx, because it is the same change of position during this time.
OOOOOOOO YYYEEEEEEEEEEEEEEEEE!!
find the other way and compare.
There should be a plus constant in integration
The constant is added when doing _indefinite_ integrals. This problem uses a _definite_ integral.
yehesssss
absohlutelyyyyyyyy
ooooo. yeahh
agreed
This video does not even discuss the concept behind the written Calculus.
😲
Might just switch to journalism...
Journalism is a noble career.
@@FlippingPhysics not easy also
N E T W O R K
this reminds me of horrible history or murderers maths.
haha
i want to create a mathematical theorm for energy shortage impact in Pakistan, anyone help me for creating this theorm? please contact me as soon as possible
Watching from the future, on the first line at 2:25, you can notice that (dv/dt)dx can be rewritten as (dx/dt)dv, where v = (dx/dt). Therefore you have (v)dv. I think that has a more natural flow to solving the integral wrt velocity than dv/dt = (dv/dx)(dx/dt). (And it prevents "canceling" dx which is generally not viewed favorably by mathematicians.) Just a simpler way to solve it.
why is it not favored by mathematicians?