Thank you so much sir.. That helped a lottt!! .... At 7:11 it should be F(t,a) rather than F(a,t) since u have taken the function to be F(t,x).. and same for F(b,t)... Thanks again..
Clarification question. You can use the taylor linear approximation for f( t + delta t, x) because delta t goes to zero, in the limit. And you used the approximation, in the limit as Δt -> 0, f( t + Δt , x + Δx ) = f(t,x) + ∂f/∂t Δt + ∂f/∂x Δx Here Δx = 0 , so we get f( t + Δt , x ) = f(t,x) + ∂f/∂t Δt
yes, for a function of x and t, the general expansion is as your write here. If you are holding x fixed and looking at changes in t you get the result above. So yes, I think what you wrote here is correct.
It's not correct!!! By Taylor: f(t+ Δt,x) = f + ∂f/∂t Δt + ½∂²f/∂t² Δt² + ... and not only f + ∂f/∂t Δt. Why we break ½∂²f/∂t² Δt² + ... ??? Because it's easy?
@@leonidosovtsov1013 please correct me if I'm wrong, but I thought if you use taylor's, especially if it's for numerical applications, since you would have \Delta t^2, and you expect it to be small, then ^2 would make it "vanish". Is it not the same here?
@@leonidosovtsov1013 its because the other terms get so small they’re insignificant. If delta(t) = 0.0001 then delta(t)^2 = 0.0000001, and delta(t)^3 = 0.0000000001. Imagine when delta(t) -> 0 … The other terms doesn’t really affect anything, and only make the maths Worse, so we ignore them
At 6:33 you say that the term goes away because in the limit you are integrating from a to a. But at 7:00 the integral has the same boundaries and it becomes a constant? I dont really get that.
The terms that stay around are divided by delta t, so that when I take the limit of dt going to zero, those terms aren't zero. It all has to do with whether there is a dt on the bottom of the expression or not.
Brian Storey so ,because the integral is divided by the very small delta t that converges to zero, we still have a value at that point, right? I think i got it now. Thanks for your answer!
@henk Jekel its just because we Get a really small number divided by another really small number, scaling it up to a normal number. Remember, 0/0 can still be a number
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He assumed f evaluated at f(a,t) is constant since delta(t) is really small. So he move f(a,t) out of the integral, and solve integral. The a’s cancel out and he is left with f(a,t)*da/dt * delta(t) / delta(t) = f(a,t)*da/dt
That red term in the bottom line at 7:39 actually has a lowercase f, i.e. it is -f(a,t)*da/dt. You can also think of it this way: (1/Δt) * integral of f dx from a to [a + da/dt*(Δt)] approaches the area of a rectangle, because we are taking the limit as Δt approaches zero. Hence the expression = (1/Δt) * base * height = (1/Δt) * base * (height when x equals a) = 1/(Δt) * {[a + da/dt*(Δt)] - a} * f(a, t) = 1/(Δt) * [da/dt*(Δt)] * f(a, t) = f(a, t) * (da/dt).
Wonderful explanation, thank you. At the very end, why do we subtract the red region, wasn't that excluded from the integral? That is, the green region integral (a->b) (df/dt)dx only included the area above f(a), didn't it? Seems that we would want to subtract the tiny black region immediately above the red region though you show that it --> in the limit. Thanks.
We subtract the region *because* it got "removed" from the original integral (the one we're differentiating, on the LHS). We are adding the green region and blue region, and subtracting the red, to get the new integral.
I lost interest in learning anything. Now I learn it if it makes me money and useful to others. Otherwise you can read 10 zillion books and still feel incomplete.
Wow, this is undoubtedly the best explanation of Leibnitz integral rule
You really make this intuitive. Nicely done, Brian!
A few months ago I saw Kent wink at me at Armstrong chapel and say we got your back buddy! Thanks for the support kent!
Outstanding teaching!!!!
Best exposition on this topic! Thanks a bunch sir.
The most concrete derivation I have seen so far.
Excellent video sir!!!
I just wish you read this. Thank you for making this awsome vedio
This... IS A GODSEND
best explanation on the internet
Excellent way to explain how Leibniz found the formula
Great demonstration! Thanks.
Thank you so much for your cristalline explanation
Fantastic video, increadibly well explained
Thanks Sir :) I had studied this before but came here to revise after hearing a dialog in Young Sheldon ;)
Excuse Sir. Brilliant!!! Thank you.
Thank you so much sir.. That helped a lottt!! .... At 7:11 it should be F(t,a) rather than F(a,t) since u have taken the function to be F(t,x).. and same for F(b,t)... Thanks again..
No. He missed to explain the part about mean value theorem. Check out here ruclips.net/video/wkh1Y7R1sOw/видео.html
incredibly clear explanation, thx sir
Awesome explanation, super intuitive
Awesome explanation! Love it!
Thanks for such an awesome video. It cleared things up quite well.
Great explanation
Logan I still remember that you're proud of me.
Thanks Brian. This was really helpful!
love the illustrations. Thank you so much!
great explaination bro. Thank you very much.
Clarification question. You can use the taylor linear approximation for f( t + delta t, x) because delta t goes to zero, in the limit.
And you used the approximation, in the limit as Δt -> 0,
f( t + Δt , x + Δx ) = f(t,x) + ∂f/∂t Δt + ∂f/∂x Δx
Here Δx = 0 , so we get f( t + Δt , x ) = f(t,x) + ∂f/∂t Δt
yes, for a function of x and t, the general expansion is as your write here. If you are holding x fixed and looking at changes in t you get the result above. So yes, I think what you wrote here is correct.
It's not correct!!! By Taylor: f(t+ Δt,x) = f + ∂f/∂t Δt + ½∂²f/∂t² Δt² + ... and not only f + ∂f/∂t Δt.
Why we break ½∂²f/∂t² Δt² + ... ??? Because it's easy?
@@leonidosovtsov1013 please correct me if I'm wrong, but I thought if you use taylor's, especially if it's for numerical applications, since you would have \Delta t^2, and you expect it to be small, then ^2 would make it "vanish". Is it not the same here?
@@leonidosovtsov1013 its because the other terms get so small they’re insignificant. If delta(t) = 0.0001 then delta(t)^2 = 0.0000001, and delta(t)^3 = 0.0000000001. Imagine when delta(t) -> 0 … The other terms doesn’t really affect anything, and only make the maths Worse, so we ignore them
wow, such a cool intuition
Thanks
Amazing video!
At 6:33 you say that the term goes away because in the limit you are integrating from a to a. But at 7:00 the integral has the same boundaries and it becomes a constant? I dont really get that.
The terms that stay around are divided by delta t, so that when I take the limit of dt going to zero, those terms aren't zero. It all has to do with whether there is a dt on the bottom of the expression or not.
Brian Storey so ,because the integral is divided by the very small delta t that converges to zero, we still have a value at that point, right? I think i got it now. Thanks for your answer!
@henk Jekel its just because we Get a really small number divided by another really small number, scaling it up to a normal number. Remember, 0/0 can still be a number
It is brilliant, timeless proof. Did it come up by yourself?
Finally I get the essence !!! tysm
superb video
sir can you please tell how you learnt so much math being a mechanical engineer. your really cool.
Sam revolinski thought my dad was a genius but the truth is I can do more complex math than my dad now.
so much sense. Thank you!
Good stuff
Very nice sir
thank you sir.
great video
Amazing! Thanks
Great lecture but can u tell me the name of your blue pen.
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Pigma micron 0.45 mm. They are really nice pens. I tried a few different brands and thickness for recording to see what showed up the best and I liked the way these looked.
That was awesome :). Thanks a lot !
How to transform the third term? F(a,t) da/dt at 7:39
He assumed f evaluated at f(a,t) is constant since delta(t) is really small. So he move f(a,t) out of the integral, and solve integral. The a’s cancel out and he is left with f(a,t)*da/dt * delta(t) / delta(t) = f(a,t)*da/dt
That red term in the bottom line at 7:39 actually has a lowercase f, i.e. it is -f(a,t)*da/dt. You can also think of it this way: (1/Δt) * integral of f dx from a to [a + da/dt*(Δt)] approaches the area of a rectangle, because we are taking the limit as Δt approaches zero. Hence the expression = (1/Δt) * base * height = (1/Δt) * base * (height when x equals a) = 1/(Δt) * {[a + da/dt*(Δt)] - a} * f(a, t) = 1/(Δt) * [da/dt*(Δt)] * f(a, t) = f(a, t) * (da/dt).
Thank you
Wonderful explanation, thank you. At the very end, why do we subtract the red region, wasn't that excluded from the integral? That is, the green region integral (a->b) (df/dt)dx only included the area above f(a), didn't it? Seems that we would want to subtract the tiny black region immediately above the red region though you show that it --> in the limit. Thanks.
We subtract the region *because* it got "removed" from the original integral (the one we're differentiating, on the LHS). We are adding the green region and blue region, and subtracting the red, to get the new integral.
four years later, but thank you for the clarification!@@adityaprasad465
My dad is teaching me that computer science is easy even if you can't do math!
My dad admitted I know more about calculus than he does!
Brilliant!
so good!
Thank you so much
Thanks a ton!!!
Wow,thanks
so useful! thanks dude :)
4:36 It would have been nice if you had explained the process of the expansion. It's not especially straightforward.
Finally. Thanks
I lost interest in learning anything. Now I learn it if it makes me money and useful to others. Otherwise you can read 10 zillion books and still feel incomplete.
thanks
It's pronounced LIEPnits. But thanks for the sweet video
actually its pronounced laipnits. Its a german name
@@eternxl8893 I know, we're just representing the same sound two different ways. For an English speaker, the word "lie" has the right sound.
How did I get here? I feel dumb as hell :(
Love u
TY HOST
Brian stop this madness.
.
bruka