It's "little" things like this that make Calculus pointlessly difficult. We're introduced a formula that says W = Fd. Simple, right? Then suddenly we're using integration to solve problems involving Work without throughly being explained why we moved from a simple multiplication to integration. And you're just supposed to nod along. In my book for example it says, if a book of 1.2kg is moved up 0.7m, the Work applied is W = (1.2)(9.8)(0.7). Then the subsequent problems are all done by way of integrating. Why didn't we apply integration when dealing with the book-lifting? Who knows? But apparently an irrelevant detail. Calculus 2 is not hard, it's just taught by people who managed to hack it and are now faced with the challenge of having to _teach_ it.
Dude here we are using integration because the force is varying now for every single unit of displacement so we are finding the work done by adding every single infinitely small piece of work! I hope that helps!!!
He said if it weren’t HORIZONTAL, not in a straight line. 3x-1 is a linear function but is not horizontal, it would be at an angle. So there’s nothing wrong with his choice of example
@@cheshstyles His example is misleading, the function is supposed to represent the amount of force being applied. Therefore the force is not constant, i.e it fluctuates so we need to figure out the force being applied at each point along the distance it is pushed which is where the need to integrate comes from.
In metric, the unit for work is joules, which is the same as Newton-Meters (just like force times distance) which is the same as a kg M^2/sec^2. There are many other units of energy like foot-pounds, calories, BTU's, Kilowatt-hours, and gallon of gasoline equivalents.
The question doesn't make sense the way you've asked it. Centripetal force is an instantaneous concept; it is not an integral. If angular velocity is continuously changing, then this means centripetal acceleration is also changing. There will be both tangential and centripetal acceleration coexisting at the same time, perpendicular to each other, in order for this to happen.
Work is force times distance. (or maybe you can think of it as adding up the forces over a distance) An integral is just a continuous summation. So, if we continuously add up all of the tiny amounts of force done over the distance from a to b, that gives us the work on [a,b] and can be calculated with a definite integral. You can read a more depth (and probably better) answer here if you'd like : en.wikipedia.org/wiki/Work_(physics)
@@BriTheMathGuy Yes as he was saying, we have no problem calculating the work done by a constant force. But if the force is variable, changes its value along a time interval, we can't apply the definition there. The thing is, we analyse; talking about all the distance in wich the variable force has acted: A to B, if we take 2 points between AB, but we say they're separated by an infinitely small distance, they are separated by almost nothing, they are almost the same point --> So we can say on interval MN the variable force has barely changed at all, the time that has passed on that interval it so small that we can say the force has been constant (at least on that infinitely small period). The job is done, we now can calculate normally the work between a small fraction (MN) of the distance: a diferential of the distance (dD). So in order to calculate the total work on AB we have to sum all the "miniworks" resultant for each infinitely small distances until complete all the distance AB. I don't know how they've created this, but the thing is the INTEGRAL operation its defined as the SUM of ALL of the small products (areas): (Y) * (a small Δx), during a certain interval of x. So the Work is the Integral of the variable F with respect to many small Δx of an interval of x because that means the sum of all miniworks until complete all the work, at least this is our method. Obviously F variation needs to be definable in terms of the position or maybe as a function of time, because if we can't define the curve line of F we can't do anything. Note: in this case x is the position, so i would be F*d, but d its Δx(change of position), and in this case we are talking about many Δx -> 0 = dx (really smal Δx). Hope you get it.
Personally I have always used the TI-89 Titanium and found it to be the best. (It takes some getting used to and not all classes will allow it) I haven't used the other so I can't speak for it. Just my opinion.
+BriTheMathGuy I seen the TI Nspire and it looks like a phone xD . I rather go with the TI 89 titanium . Did your teachers make you calculator savvy or did they expect you to know everything by hand ?
Kinda bugs me when no unit is included. "The work you would've done is just ...200". 200 what, should be 200 Joules and it conveys information more clearly...
Then you will need a calculation that will convert or relate force to power. Once you have power with respect to time, you can integrate with respect to time. Much like force x distance is work, power x time is also work. Alternatively, you will need a calculation that will relate thrust to distance. Then after substituting, you will have a function for force and distance. These relations between force and power or thrust and distance can become their own related rates problems. The easiest case is if you find that thrust (force) is constant.
8 years and you're still saving lives with your videos ❤😂
More helpful in 4 minutes than my physics teacher in 1 semester
Glad to help. Have a great day!
सच में
Ouch... I think he's implying our 9-to-5 jobs have no quantitative or measurable value.
Is he wrong tho?
It's "little" things like this that make Calculus pointlessly difficult. We're introduced a formula that says W = Fd. Simple, right? Then suddenly we're using integration to solve problems involving Work without throughly being explained why we moved from a simple multiplication to integration. And you're just supposed to nod along.
In my book for example it says, if a book of 1.2kg is moved up 0.7m, the Work applied is W = (1.2)(9.8)(0.7).
Then the subsequent problems are all done by way of integrating. Why didn't we apply integration when dealing with the book-lifting? Who knows? But apparently an irrelevant detail. Calculus 2 is not hard, it's just taught by people who managed to hack it and are now faced with the challenge of having to _teach_ it.
Dude here we are using integration because the force is varying now for every single unit of displacement so we are finding the work done by adding every single infinitely small piece of work! I hope that helps!!!
th-cam.com/video/vFDMaHQ4kW8/w-d-xo.html 💐..
@@biancaa-ramesh Perfectly explained
We can only use W=Fd if the force is constant, and what if the the force is varying through the distance it travels? Thats why we use integrals.
It’s because it’s a variable force. But yea I agree. Pointlessly difficult 😭
Thaaaaaaank you very much this was so helpful 🤧🤧 🙏🙏😊😊
Thanx man this was really easy and simple
Awesome topic :) it's starting to make sense now .
I'm glad to have helped!
Thanks. Nicely explained in real time of 4 min!!
Wow, and he even makes integrals make sense lol
Glad you thought so!
I don't understand when to integrate a problem. What is the "trigger" in a problem that would indicate that I need to integrate?
When the force isn't a constant function
I am class 9 its new to me and i can't understand. Thanks for the information sir please make more now i understand 🥰🥰🥰🥰
THANK YOU SO MUCH. u did what my physics teacher couldnt with 1 week in 4 minutes
Listen here: 1:02. Then why give an example in a straight line??? Pretty confusing!
Thanks for the feedback. I hoped to make the difference between the two clear, perhaps I didn't for you. Have an awesome day!
He said if it weren’t HORIZONTAL, not in a straight line. 3x-1 is a linear function but is not horizontal, it would be at an angle. So there’s nothing wrong with his choice of example
@@diaabnasir471 not true. He said "straight".
@@cheshstyles His example is misleading, the function is supposed to represent the amount of force being applied. Therefore the force is not constant, i.e it fluctuates so we need to figure out the force being applied at each point along the distance it is pushed which is where the need to integrate comes from.
work is only measured in Joules right?
I'm no physicist, but I think you're correct.
It is also measured in erg
In metric, the unit for work is joules, which is the same as Newton-Meters (just like force times distance) which is the same as a kg M^2/sec^2.
There are many other units of energy like foot-pounds, calories, BTU's, Kilowatt-hours, and gallon of gasoline equivalents.
@@Firebert79TA *hears foot-pounds* I smell scientific blasphemy-
@@ObtainThePain just wait until we put them in groups of 550 per second and then multiply by the time duration in hours! (horsepower-hours)
Thank you so much, explained so muc better than my college professor.
can you prove this formula
What's the relationship between f(x) and dx over the intregal sign🙏
dx is the variable of integration, if the fuction were f(t) the the variable of integration would be written “dt”
Hi, what if it's centripetal force, F= m(wr)^2 /2 from r initial to r final, with a changing omega (w)?
The question doesn't make sense the way you've asked it. Centripetal force is an instantaneous concept; it is not an integral.
If angular velocity is continuously changing, then this means centripetal acceleration is also changing. There will be both tangential and centripetal acceleration coexisting at the same time, perpendicular to each other, in order for this to happen.
so, why is work the integral of force?
Work is force times distance. (or maybe you can think of it as adding up the forces over a distance) An integral is just a continuous summation. So, if we continuously add up all of the tiny amounts of force done over the distance from a to b, that gives us the work on [a,b] and can be calculated with a definite integral. You can read a more depth (and probably better) answer here if you'd like : en.wikipedia.org/wiki/Work_(physics)
@@BriTheMathGuy Yes as he was saying, we have no problem calculating the work done by a constant force. But if the force is variable, changes its value along a time interval, we can't apply the definition there.
The thing is, we analyse; talking about all the distance in wich the variable force has acted: A to B, if we take 2 points between AB, but we say they're separated by an infinitely small distance, they are separated by almost nothing, they are almost the same point --> So we can say on interval MN the variable force has barely changed at all, the time that has passed on that interval it so small that we can say the force has been constant (at least on that infinitely small period).
The job is done, we now can calculate normally the work between a small fraction (MN) of the distance: a diferential of the distance (dD). So in order to calculate the total work on AB we have to sum all the "miniworks" resultant for each infinitely small distances until complete all the distance AB. I don't know how they've created this, but the thing is the INTEGRAL operation its defined as the SUM of ALL of the small products (areas): (Y) * (a small Δx), during a certain interval of x.
So the Work is the Integral of the variable F with respect to many small Δx of an interval of x because that means the sum of all miniworks until complete all the work, at least this is our method. Obviously F variation needs to be definable in terms of the position or maybe as a function of time, because if we can't define the curve line of F we can't do anything.
Note: in this case x is the position, so i would be F*d, but d its Δx(change of position), and in this case we are talking about many Δx -> 0 = dx (really smal Δx).
Hope you get it.
We are calculating force with respect to the distance.
does this change if the lower limit does not equal 1? if it were a greater number or even a negative lower limit?
barrington hebert the process is the same!
Thanks u ❤️
HELLO! Fantastic video. Can you please provide the same video with a F-X graph function for the area ?
And is or better to use a TI-89 or TI -Nspire CX CAS for school
Personally I have always used the TI-89 Titanium and found it to be the best. (It takes some getting used to and not all classes will allow it) I haven't used the other so I can't speak for it. Just my opinion.
+BriTheMathGuy I seen the TI Nspire and it looks like a phone xD . I rather go with the TI 89 titanium . Did your teachers make you calculator savvy or did they expect you to know everything by hand ?
Once I got to college I had to do almost everything by hand. I got to use calculators in statistics and physics though.
Kinda bugs me when no unit is included. "The work you would've done is just ...200". 200 what, should be 200 Joules and it conveys information more clearly...
th-cam.com/video/vFDMaHQ4kW8/w-d-xo.html 💐..
How to compute the force?
In many examples it’s given. But Newton’s 2nd law tells us Force=mass*acceleration
I was with u until the anti thingy💀 I havnt even learnt this in school so im a bit baffled lmao
what if the force is given as a function of time, like in the case of thrust.
Then you will need a calculation that will convert or relate force to power. Once you have power with respect to time, you can integrate with respect to time. Much like force x distance is work, power x time is also work.
Alternatively, you will need a calculation that will relate thrust to distance. Then after substituting, you will have a function for force and distance.
These relations between force and power or thrust and distance can become their own related rates problems. The easiest case is if you find that thrust (force) is constant.
Thanks Sir 🥰 me from India that why namaste🙏🙏
You're most welcome