To find x in terms of y, you could also argue by similar triangles that 2x/y=3/3, and thus 2x=y. Your approach is more general, but this was the way I saw it first.
thank you so much. I just subbed. I'm watching this outside of my classroom right before I'm about to take an exam. big help, very concise, and funny with the mic. thank you
Desde México City: He visto sus vídeos en los que se emplea la integral para calcular la fuerza que ejerce un fluido sobre una pared vertical. También en donde emplea la integral para calcular el trabajo para bombear cierto liquido. Sus vídeos son excelentes, pues explica muy bien los conceptos para ir armando la integral. Muchas gracias profesor por su aporte y saludos.
As im listening to you "yes, yes, that makes sense", 1 minute after the video is done, "wait, what?" Nonetheless this is great info., ... now onto another video... lol
Volume × Density is NOT the weight! That is only the MASS of an object . You have to multiply with ACCERELATION to get the WEIGHT of an object. ELEMENTARY PHYSICS!!!!
He's probably oversimplifying, and assuming that "density" could really be specific weight. If you want to get technical, it isn't acceleration you multiply by to get the weight either. It is gravitational field strength. A body accelerating due to gravity alone has no weight, because the gravity is all "used up" in causing its acceleration, that it doesn't experience any gravity to give it an ability to detect its weight.
He does. That’s the 5 in the term (5-y) at 7:40. It is used as the term for distance the water must move. He did not use 5 back when computing the slope because you can compute slope with any “rise over run,” and he just used the numbers available. There was no “run” number available for the “rise” of 5. I would agree that I wish he had made it clearer why he used 5 and how that changes the value of the integral. It actually has a large effect on the final work done, versus just pumping to the top of the trough.
You don't get enough recognition for the amount of work you put into this. Thank you for helping me get through Calculus!
Patrick Chevis thank you. I am very glad to hear this. Hopefully one day, I will.
was... was that a math pun...
he does now
@@blackpenredpen Now you do🙂
This is problem 23 in Stewart 8th edition!!!! 9 years later and I found an exact explanation. you are the best
Exact problem from same book assigned to my HW, and is probably on the exam. He is the best!
@@brax300 good luck! Hydrostatic force was one of my favorite parts of calc 2
i was studying this for 2 hours, i learned more from you in a few minutes.
I have learned more from you in a few minutes than while in calculus class. What would you use the Work calculated for in engineering?
To find x in terms of y, you could also argue by similar triangles that 2x/y=3/3, and thus 2x=y. Your approach is more general, but this was the way I saw it first.
Yeah you can use similar triangles or slope they'll give you the same answer
Anyone can learn anything... even rocket science...if they have the right teacher.
Best tutorial out there for me personally on this topic. I love how intuitively you broke this all down.
Dude, that's a kick-ass mic you got there. Your presentation is awesome, I can now solve work problems with triangular shapes. Thank you.
This is the best explanation of this type of problem I have found. Patient and descriptive, not too fast. Thank you
This is the best explanation of this problem that I have seen. Thank you!
thx 4 not going out of your way to make the problem more difficult !!
In Physics, Work is the product of force and DISPLACEMENT.
thank you so much. I just subbed. I'm watching this outside of my classroom right before I'm about to take an exam. big help, very concise, and funny with the mic. thank you
Desde México City:
He visto sus vídeos en los que se emplea la integral para calcular la fuerza
que ejerce un fluido sobre una pared vertical. También en donde emplea
la integral para calcular el trabajo para bombear cierto liquido.
Sus vídeos son excelentes, pues explica muy bien los conceptos para ir armando
la integral.
Muchas gracias profesor por su aporte y saludos.
thank you for helping me understand calc 2 in 9 minutes
thank. you. youre the hero of my education.
Rewatching this. It is a classic
You make great help videos. They are spot on. Keep up the great work!!
gracias, te amo, me ayudaste
My professor is teaching us in British units but using si-units is so much easier.
I got the exact answer when using integral 78400(3-x)(2+x)dx from x=0 to x=3.
I love your mic. Also thank you for the help.
I love it, too! It's a Blue Snowball USB. Had mine for years.
everybody gangsta till he pulls out the blue pen
super simply put. well done and thank you!!!
hi there, how do you know when to start your cartesian plane at the tip of the triangle or at the top?
I'm so grateful that I've found this video sir thank you po hihihihi
As im listening to you "yes, yes, that makes sense", 1 minute after the video is done, "wait, what?" Nonetheless this is great info., ... now onto another video... lol
Great work!
jesus christ you’re phenomenal!
Thank you!!!!!!!
Thanks so much man..you are good at explaining
Thank you so much! This has been so much help to me!
Thank You (EN). تشکر(Farsi)
在这里是否可以用X取代Y啊？ 算出来会不会不是一样的结果？
you're a legend.
Thanks!
Excellent!!!
you are awesome
Thank you so much!
Thank Youuuuu!!!!!!!!!!!!!!!!!
Where did the initial 2xcome from? I'm not sure i understood.
Nice
You are awesome.
thank you jackie
Jackie? as of Jackie Chan?
yes you are the jackie chan of calculus
Will using the idea of similar triangles work the same in finding this value of x? 4:50
@@a.koomson Yes! Either way gives the same answer.
it seems that everyone uses the stewart textbook
Holy fucking shit i finally got it. Thanks bro
bless you.
Good
why would you not take into account the complete distance from bottom to top of the spout? from 0 to 5?
he did in (5-y). the limits of integration tells you where to sum the little dy volumes. there is no water at 5m so the max musy be 3
Bcs 3/1.5=2 eh?
Good lad
Volume × Density is NOT the weight! That is only the MASS of an object .
You have to multiply with ACCERELATION to get the WEIGHT of an object. ELEMENTARY PHYSICS!!!!
He's probably oversimplifying, and assuming that "density" could really be specific weight.
If you want to get technical, it isn't acceleration you multiply by to get the weight either. It is gravitational field strength. A body accelerating due to gravity alone has no weight, because the gravity is all "used up" in causing its acceleration, that it doesn't experience any gravity to give it an ability to detect its weight.
SOS FAN DEL DY??
who's this??? DA-DDY, YAAAN-KEEE, DY
im still confused lol
I dont understand how he got 2x as his length. Could anyone explain this a bit more?
if x represents the length from the center line to the hypotnuse of the triangle, then 2x is the length of the whole base.
why would you not take into account the complete distance from bottom to top of the spout? from 0 to 5?
why would you not take into account the complete distance from bottom to top of the spout? from 0 to 5?
He does. That’s the 5 in the term (5-y) at 7:40. It is used as the term for distance the water must move.
He did not use 5 back when computing the slope because you can compute slope with any “rise over run,” and he just used the numbers available. There was no “run” number available for the “rise” of 5.
I would agree that I wish he had made it clearer why he used 5 and how that changes the value of the integral. It actually has a large effect on the final work done, versus just pumping to the top of the trough.