AP C is a must take for engineering students. It is very nice to watch this derive for kinematics and then re-fresh derive the formula for uniform acceleration motion using function graph method. With the two methods, it provide a deep understanding of calculus.
Bravo! I've been looking for something like this for ages. Well done. Its just like how my old maths teacher taught us about parabolic equations and laws of motion
Actually 'a' is a constant so just focus on the 't'. When you take the integral of t dt, you get t^2/2. Then multiply this value by a and you get at^2/2.
7:13 Isn't that just the chain rule? Velocity "v" is a function of displacement "x" which itself is a function of time "t". So we get than v(x(t)), the derivative of this with respect to t would then be: d/dt[ v(x(t)) ] = v'(x(t))*x'(t) or (dv/dx) * (dx/dt) :)
So the velocity function in this case does not take a time in, it takes in a position and returns the corresponding instantaneous velocity at that position?
I'd recommend anyone who's starting out in calculus to watch and rewatch 0:37 to 1:05. It's a very good explanation if you want to develop an intuition for what calculus is all about.
At 5:03 I understand how the anti-deravite of at = (1/2at)^2. Why isn't it the same for the Vi? Is it just because it's t isn't t^2 or does it have somethins to do with the constant that I don't know about?
It comes from the power rule for polynomial term calculus. When you take derivatives, the original power compounds with the original coefficient, and the power is reduced by one. d/dx k*x^n = k*n*x^(n-1) n is zero for constants, which means the term disappears when differentiated. n is one, for terms directly proportional to x When you take integrals, you do exactly the opposite. The power increases by 1, and you divide the coefficient by the new power. integral k*x^n dx = k/(n+1) * x^(n+1) + C n is zero for constants, which means the term simply multiplies by x n is one, for terms directly proportional to x, which means the power increases to squaring, and you accumulate a 1/2 term to compound with the coefficient. You notice this rule always works for derivatives, but for integrals, we run in to a problem when n=-1, because we get a divide by zero error. Calculus has the solution, and that is that the integral of x^(-1) is ln(abs(x)) + C.
a/2t^2 is a messy way of saying 1/2(at)^2, you can take the anti derivative of the term (at) and you would end up with 1/2(at)^2, you can check your logic by taking the derivative of 1/2(at)^2, and 2(1/2)(at) = (at). Hope that helps.
samartha s.r No you can't do that. You have to use the constant rule of derivatives. (1/2)(a) is a constant. You set that aside. Now take the derivative of t^2. It is 2t. Now you can bring back the constant. Hence, derivative of (1/2)(a)(t^2) is (1/2)(a)(2t). Cross out the "2"s on the top and bottom and you get (at).
I'm in AP physics, I was told it was going to be algebra based... I'm barely in my first few units of pre-calculus and my teacher can't teach, he sprung this on us from nowhere.
according to u which gives more depth calculus proof or other geometrical/algabriac proof given in university zemanskys physics? Becasue i am more comfortable with calculus proof.
But why is time boundary is 0 and t, not time initial and time final just like velocity having boundary of velocity initial and velocity final? I'm dumb but pls answer my question Sir, Thank you!
The bounds of integration on time, could really be either time initial and time final, or zero and t. Since it is arbitrary where we define time = 0, you might as well define it to start at t=0. The times when you would have the distinction, is if you have multiple intervals, each with a different acceleration.
Why do you rearrange the acceleration and ve,ocity equations so you have vi and vf, but when you derive the distance equations you just combine them into (delta)x? Why did Newton decide he wanted two velocity veriables but that he didn't want two distance veriables?
thank you so much this video blew my mind. you called "a" the intergrant, is there other names for the other parts of the intregal you could tell me about?
Glad you find the viideos helpful. "a" is the integrand, "dt" is called the differential. To solve the integral you take the antiderviative of the integrand and the differential tells you what the variable is, namely t (in this case.) So the (integral) 5 dt = 5t but the (integral) 5 dx = 5x.
@@lasseviren1 so A , which is a constant like some number such as 5 means that you plug in the integral of acceleration like dt =at and dx= ax using the constant?
@vinnienauta Hmm if i where teaching this I would say to take the anti-derivative -- or integral -- of at with respect to t so then that would give you the over two since when you take d/dt of (a/2)t^2 it = at im not sure if this helps i tried my best (x
If you're not familiar with derivative, all that's really saying is that the acceleration=change in velocity/change in time. Basically, if a car starts off going 10 m/s then it goes up till 15 m/s in 5 sec, then... Acceleration=change in velocity/change in time a=(15-10)m/s /5-0s =5/5 =1 m/s^2 Meaning, the "m/s^2" means the car was going 1 meter FASTER every second. So its the velocity (m/s) after each second. So m/s^2. So, that's the average velocity of the car. But what the derivative means is that.... Acceleration=change in velocity/change in time as the change in time approaches 0. So you're finding the infinitesimally small change in velocity in an infinitesimally small change in time. The reason why this " infinitesimally small change" part is useful is because we can derive and do other things with the equation. Such as, taking its integral.
@MaxxG94 HEY thank you for TRYING!! I mean!! DAYUM IM LEARNING DERIVATIVES! AND IM ACTUALLY GETTING IT! yeah, im taking physics b. but i want to take physics c test.. physics b is boring
@@javierarana2349 Because he took the integral of Vi (initial velocity) with respect to t (time). On the right side there was t^1 and when you take the integral of that it becomes t^2/2. On the left side there's no t so we assume it's Vi x t^0. When you take the integral of that it becomes Vi x t^1/1 which is Vi x t. Hope I've made it clear.
this is more basic differential equations, than general calculus...yes calculus is used, but a separable differential equation is still a differential equation.
That's how you do the antiderivative. It's a little hard to explain in a few sentences, so why don't you check out some other videos about that? If you just want to see why it gives you the correct solution, you can just take the solution and take the derivative.
@vinnienauta cool I am taking this test too but i have no teacher Im self tought by a princenton review ap physicsc review books and any internet resources which are handy. ( and especially the videos of this teacher, he is great) I got a 3 in Ap phyiscs becuase i sucked at everything that WASN'T mechanics so i want my revenge lol
I wish there was a single calculus textbook produced on planet earth that adequately explained integral notation.
AP C is a must take for engineering students. It is very nice to watch this derive for kinematics and then re-fresh derive the formula for uniform acceleration motion using function graph method. With the two methods, it provide a deep understanding of calculus.
dv/dx describes how an object's velocity changes with small changes in its position.
Bravo! I've been looking for something like this for ages. Well done. Its just like how my old maths teacher taught us about parabolic equations and laws of motion
Very insightful, thank you. I hate just plain memorizing these formulas. It helps me much more if I can understand where they came from.
May allah bless you ... this is an ehsan from you sir
After hours searching the internet I FINALLY found this video with the answer to my math IA, thank you very much Sir!!!!!!
+Dan Al OMG SAME are you doing the IB?
+aleksdurowicz yes
Haha I included it in my math exploration as well :)
You graduated in the wrong year hehehe. M20 exams are cancelled! >
@@stephanielue8454 welp i graduate in 2021...
Actually 'a' is a constant so just focus on the 't'. When you take the integral of t dt, you get t^2/2. Then multiply this value by a and you get at^2/2.
understood it better than with my physics teacher, thanks! :)
JFK assassination of my teacher in your country and we are going through my work
I am a class 11th Student and was searching for this because tomorrow is my exam and I understood it completely.
May God bless you sir for this superb explanation
This videos is awesome, I always forget equations so this is really useful for when I forget. If I do forget I can just derive them myself!
I don’t like memorizing so I try to derive everything as much as possible
Holy damn I needed this so much for my physics lab. Thank you!
Thank you so much for explaining to me how to have the kinematic formulas using calculus.
thank you for using the convention of "V-initial" instead of "V-not". Its like a personal tick I have been dealing with since high school.
no
Thanks so much. I'm taking Dynamics and I haven't taken differential equations yet. The book just sort of expects you to know how this is done.
So far I've used a little. Just basic integration to derive these formulas.
no
Wonder how much you progressed since this
wow this is very useful instead of trying to memorize the formulas, thanks!
7:13 Isn't that just the chain rule? Velocity "v" is a function of displacement "x" which itself is a function of time "t". So we get than v(x(t)), the derivative of this with respect to t would then be: d/dt[ v(x(t)) ] = v'(x(t))*x'(t) or (dv/dx) * (dx/dt) :)
So the velocity function in this case does not take a time in, it takes in a position and returns the corresponding instantaneous velocity at that position?
I'd recommend anyone who's starting out in calculus to watch and rewatch 0:37 to 1:05. It's a very good explanation if you want to develop an intuition for what calculus is all about.
At 5:03 I understand how the anti-deravite of at = (1/2at)^2. Why isn't it the same for the Vi?
Is it just because it's t isn't t^2 or does it have somethins to do with the constant that I don't know about?
It comes from the power rule for polynomial term calculus.
When you take derivatives, the original power compounds with the original coefficient, and the power is reduced by one.
d/dx k*x^n = k*n*x^(n-1)
n is zero for constants, which means the term disappears when differentiated.
n is one, for terms directly proportional to x
When you take integrals, you do exactly the opposite. The power increases by 1, and you divide the coefficient by the new power.
integral k*x^n dx = k/(n+1) * x^(n+1) + C
n is zero for constants, which means the term simply multiplies by x
n is one, for terms directly proportional to x, which means the power increases to squaring, and you accumulate a 1/2 term to compound with the coefficient.
You notice this rule always works for derivatives, but for integrals, we run in to a problem when n=-1, because we get a divide by zero error. Calculus has the solution, and that is that the integral of x^(-1) is ln(abs(x)) + C.
thank usir because of u i leaarnt kinematic equations derivations andd hence i learnt what is integration...thank u vry much
At 5.03, I know you said acceleration is a constant but I don't get why it has to be 1/2a, I get why it is t^2. Could you explain please. Cheers
a/2t^2 is a messy way of saying 1/2(at)^2, you can take the anti derivative of the term (at) and you would end up with 1/2(at)^2, you can check your logic by taking the derivative of 1/2(at)^2, and 2(1/2)(at) = (at). Hope that helps.
But then the derivative of 1/2 is 0, and 0 times anything is 0, hence 1/2(at)^2 is 0, but that's not the case?
:-(
samartha s.r No you can't do that. You have to use the constant rule of derivatives. (1/2)(a) is a constant. You set that aside. Now take the derivative of t^2. It is 2t. Now you can bring back the constant. Hence, derivative of (1/2)(a)(t^2) is (1/2)(a)(2t). Cross out the "2"s on the top and bottom and you get (at).
Aaaaaaaaaaaaa
guys, take the analogy of integral of (kx)dx
you are going to find:
k times integral of x wich is k/2*x^2
so integral of "a" times tdt is:
a/2*t^2
You could also leave it as an indefinite integral to get a form for quadratics.
We usually called Vi(initial velocity) and vf(final Velocity)
if only I knew when I was 3, 14 years ago that I would need to watch these videos going into AP Physics C next year
Only now i finally totally understood this! Cheers!
I'm in AP physics, I was told it was going to be algebra based... I'm barely in my first few units of pre-calculus and my teacher can't teach, he sprung this on us from nowhere.
@@beoptimistic5853 THANK YOU SO MUCH.. This would've been so nice to have a year ago :(
Thank you so much. Very helpful and well presented
no
thank you, I have been learning physics for +3 years and I have just discovered how to derive these formulas
it was really cool calligraphic explanation dude!!!!!!!!!thanks tonssss!!!!!!!!!!!!
Thanks, very helpful for my physics class
Are you a hand model?
shutupp.............(o_o).
according to u which gives more depth calculus proof or other geometrical/algabriac proof given in university zemanskys physics? Becasue i am more comfortable with calculus proof.
Amazing video, thank you brother!
Thank you so much... now i understant
Thnks sir...Your way of teaching is quite easy....Got it in the first time 🙏🙏
Lasse Viren, the flying Finn!' I remember watching him in the Olympics when I was a kid. Great video...
very useful video, but how did 2 came below A in the third equation?? please help me with that
Thanks you so much for this video....now I got it....🙂🙂🙂😙😙😙😉😉😀😀😀
Great Video! I have one question, is V final in the last equation supposed to be negative or positive?
Marcus Cowan It depends maybe?
no
Are you all doing big job meow?
This video is amazing, thank you so much
my friend you truly are the second coming of christ thank you so much
But why is time boundary is 0 and t, not time initial and time final just like velocity having boundary of velocity initial and velocity final? I'm dumb but pls answer my question Sir, Thank you!
The bounds of integration on time, could really be either time initial and time final, or zero and t. Since it is arbitrary where we define time = 0, you might as well define it to start at t=0. The times when you would have the distinction, is if you have multiple intervals, each with a different acceleration.
Why do you rearrange the acceleration and ve,ocity equations so you have vi and vf, but when you derive the distance equations you just combine them into (delta)x?
Why did Newton decide he wanted two velocity veriables but that he didn't want two distance veriables?
To spite Leibniz perhaps...
can please explain how you got a/2.. I MEAN I GET IT but I WANT TO LEARN HOW TO TEACH IT
thank you so much this video blew my mind. you called "a" the intergrant, is there other names for the other parts of the intregal you could tell me about?
Glad you find the viideos helpful. "a" is the integrand, "dt" is called the differential. To solve the integral you take the antiderviative of the integrand and the differential tells you what the variable is, namely t (in this case.) So the (integral) 5 dt = 5t but the (integral) 5 dx = 5x.
@@lasseviren1 so A , which is a constant like some number such as 5 means that you plug in the integral of acceleration like dt =at and dx= ax using the constant?
how is the antiderivative of vi vi(t)? Shouldn't it be 1/2 vi ^ 2 ?
Josh Golden
Because it's with respect to t.
You treat v as a constant
This is great material
For the second equation I’m confused for why the integral of (at)dt has t^2 and not t^3
integral of x is simply x^(n+1)/(n+1)
@vinnienauta
Hmm
if i where teaching this I would say to take the anti-derivative -- or integral -- of at with respect to t so then that would give you the over two since when you take d/dt of (a/2)t^2 it = at
im not sure if this helps i tried my best (x
i did not understand first page last step. can u pls explain it again @lasseviren1
Which ap physics is this done in?
why do you need to divide it by 2???? please explain!!
Thanks sirr u r the best😄😄
okay this is a stupid question but where did a=dv/dt come from, at the very start
Acceleration = derivative of velocity with respect to time.
Ahhh okay thank you
If you're not familiar with derivative, all that's really saying is that the acceleration=change in velocity/change in time.
Basically, if a car starts off going 10 m/s then it goes up till 15 m/s in 5 sec, then...
Acceleration=change in velocity/change in time
a=(15-10)m/s /5-0s
=5/5
=1 m/s^2
Meaning, the "m/s^2" means the car was going 1 meter FASTER every second. So its the velocity (m/s) after each second. So m/s^2.
So, that's the average velocity of the car.
But what the derivative means is that....
Acceleration=change in velocity/change in time as the change in time approaches 0. So you're finding the infinitesimally small change in velocity in an infinitesimally small change in time.
The reason why this " infinitesimally small change" part is useful is because we can derive and do other things with the equation. Such as, taking its integral.
Thanks, I got it. I was clearly have an intellectual crisis when I asked this question aha.
We are integrating and now we are doing slope fields
Thank you so much that is great job for me
why did u divide by 'a' by 2???please explain!!!
He integrated it. You can check it by taking its derivative.
Hi I am an Indian student and want to know in foreign this topic is covered in which standard????????
kindergarten
@@9678willy oohh thats right i learnt this topic in kindergarden* from your mom
@@Burner. it’s kindergarten bruh
@MaxxG94 HEY thank you for TRYING!! I mean!! DAYUM IM LEARNING DERIVATIVES! AND IM ACTUALLY GETTING IT! yeah, im taking physics b. but i want to take physics c test.. physics b is boring
He divided by two because he was taking the integral. Therefore, a/2 is equal to the usual 1/2(a).
Tomodachi666 yes but how come he didn’t divide the left side by two when he took the Integral of the left side?
@@javierarana2349 Because he took the integral of Vi (initial velocity) with respect to t (time). On the right side there was t^1 and when you take the integral of that it becomes t^2/2. On the left side there's no t so we assume it's Vi x t^0. When you take the integral of that it becomes Vi x t^1/1 which is Vi x t. Hope I've made it clear.
this is more basic differential equations, than general calculus...yes calculus is used, but a separable differential equation is still a differential equation.
Thank you very much :)
Yey
I wont have to memorize the kinematics equations for the AP exam because I can derive them now haha
lol jk
simply differensome! 😉awesome!!!
thank u
at last the thing i wanted
great video
Thanks sir 👍
@@beoptimistic5853 what's this
Thanks
The calc involved in mechanics is pretty basic--it almost harder to learn it the "easy" way! I dont know how you draw with a sharpie and never smear
What about the constant of integration
Thank you!!.
Nice marble table top
very very heipful...
That's how you do the antiderivative. It's a little hard to explain in a few sentences, so why don't you check out some other videos about that?
If you just want to see why it gives you the correct solution, you can just take the solution and take the derivative.
Anybody here in 2019 ?
I am. I was barely ten years old when this was posted.
thank you !
Dad, is that you??!!! LOL :))
👍🏻👍🏻👍🏻👍🏻 thanks 😘
legend
You da man lasseviren1!
Nope
the hardest if seen it go is u subsitution and then integrating in the form of du/u
and that was in a free response question.
You are awsome
Super cool!
Thnx
Nice
Thank u
mind. blown.
@vinnienauta
cool
I am taking this test too
but i have no teacher
Im self tought by a princenton review ap physicsc review books and any internet resources which are handy. ( and especially the videos of this teacher, he is great)
I got a 3 in Ap phyiscs becuase i sucked at everything that WASN'T mechanics so i want my revenge lol
Umm a bit confusing
@legoindianajones1000 Physics C Mechanics
thanks :)
M A S T E R -------------------------------- P H Y S I C S
U almost ended up confusing me
me toooo
I do been doing this in my head since 8th grade im now in MIT as a 16 year old
Andrew Aguirre sure...
You keep forgetting about the integration constant 😂😂
Ali Al-Musawi With indefinite integrals the constant of integration cancels out so there isn't much benefit writing it out every time
very very very very XD . amazing though
Ill be Darn'd XD
i love you
when i m serching for my physics i found a vdo that is yours..... But i am not impressed with your vdo