Related rates intro | Applications of derivatives | AP Calculus AB | Khan Academy
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What's the relationship between how fast a circle's radius changes, and how fast its area changes? Created by Sal Khan.
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the derivative with respect to time, the derivative with respect to time, the derivative with respect to time, the derivative with respect to time, the derivative with respect to time, the derivative with respect to time, the derivative with respect to time...
His tutorials are really good but yeah that "derivative with respect to ____" always just makes me a little more confused
@@Odeh"the derivative with respect to _____ " basically just means "the rate of change as _____ changes".
you are sent from heaven. you are a good dude. thank you for these videos. im gonna get my degree probably cause of you. keep up the good work, bro
I love this guy's videos, but sometimes the way he repeats things, repeats things. It gets to be a little distracting, a little distracting.
true, but actually i hear that repetition subconsciously helps with absorbing the material. it annoyed me a bit but I found myself paying attention a little more.
I could see how that could help a bit, but sometimes I'm just like, "get on with it man!!" lol
+Dylan D'Alatri Well I guess you also need to know that his native language is not English. Sometimes it just happens. Same with me.
How is his native language not English, i know his parents are from Bangladesh, but he was born and grew up in the U.S, which makes English his native language. Bengali is his parents native language while English is his. There's literally no mistakes in his speech, plus he has a higher mastery of the language than the average American.
@@HoshinoMirai lmao his native language is english
I think it is important to point out that there is no physical reality to this, and that it is a mere exercise in math. But the exercise is to prepare us for a similar math process that tells us how the area of a sphere changes with time and that DOES have a physical reality.
Most math relationships, and there's an infinite number of them have no meaningful physical application (such as the third derivative of position with time), but those few that have application are profoundly important.
Sal is great
3:45 #rules
Derivative of constant * something
you can take the constant out
4:30 #rules
Chain Rule - well explained
Why are you just able to take a constant out, is it because it derives 0, which makes it disappear, so how is it reasoned that you can just take it out
this was excellent! please have more calculus word problems
The rate which the area is growing is different to each value of radius. you forgot to mention that the number you found is the rate that the area grows when the radius is equals to 3 cm. At 6:29 you said that r is 3 cm but you forgot to said that if we want the rate of any other value of radius, we won't use 3, because it would have changed according to the dr/dt.
I love you
Yes thank god that was freaking me out. Thought maybe I missed something.
Yep, Derivatives is used in finding rates, velocity, any changes in times, size,etc.
I paused the video and tried to solve it before him, what I did was; since A is equal to pi r^2 then just take the derivative of pi r^2 with respect to t, using the chain rule: (the derivative of pi r^2 with respect to r) times (the derivative of r with respect to t), and so it will be equal to (2pi r) times (the derivative of r with respect to t, which equals to 1 according to the question) so (2pi rcm) times 1cm/t is just (2pi r cm^2/t) then replacing r with 3 gets us left with 6pi cm^2/t.
Isn't this way easier? Or is understanding to do it how he did is important?
The chain rule thing doesn't make any sense to me. I can see where you might use it if you had [r(t)]^2, but then you have the bizarre case of the derivative of a function (what's the derivative of f(x)? Well you have to define the function). Whatever he's getting at here it's not at all as straightforward as he makes it seem.
I also find it confusing
Thank you a LOT Sal..
I have just understood. thanks
I actually find it really helpful since math is pretty hard for me. If it bothers you, just fast forward :)
Raise your right eyebrow every time he says "Rock"...
guys a question on this, shouldn't the rate of how the Area increases not depend on the actual Radius of the circle at a specific time? i mean if you see the answer you can clearly see that the rate for a already formed circle with 3cm on radious at that time is going to have a different rate against a circle with 1cm at that time but with the same increase of cm/sec
The rate of change of area does indeed depend on the radius.
I Respect Him for THAT
Great!!
What a good video!
Isn't it strange that if radius is changing by 1 cm, that means area goes from 9pi to 16pi and the change in area is 7pi. I don't see how derivatives are supposed to explain the change in area in a second?
The derivative isn't about finding the average rate of change over a 1-unit interval; we don't need calculus for that. It's about finding the average rate of change over smaller and smaller intervals and taking the limit as the size of the intervals approaches 0.
What are you using in this video to draw?
Facts
@@jaredmorales9763 he isn't using facts
@@prodjvenchy lmfao
So So in in this this case case its its literally literally just just multiply multiply the the Area Area formula formula by by the the rate rate of of growth growth.
That's the circumference formula.
@@isavenewspapers8890 that’s that’s the the circumference circumference formula formula
i feel like my mind is like a machine
If you are unfamiliar with the derivative, there are a lot of prior videos explaining it.
Thx it help on my 7th trade finals
Calculus in 7th grade?
U r genius
omg. at first i was like oh yeah i like this guy. he's explaining it so simple. then... like you kept adding things. i knew exactly what to do for my next step but you kept adding "to make sure." There were times you just confused me but thanks for the video anyway. i've watched many related rates video and got it down finally. just gotta be careful.
Well, there is some unclear moment. Okay, 6pi, but let's find area at 3 cm radius = 9pi, at 4 sec = 16 pi, so if we divide A(4sec)/A(3 sec) we get 4/3(not 6 pi), it's a rate of growing area between 3 and 4 sec. So where is here mistake?
actually, I'm equally confused except that I would say that the area changed by A(4) - A(3) which gives me 7pi, not 6
@@jdlopez131 yeah you're right
That's the wrong way of thinking about it. Your method gives the average rate of change of area over the course of 1 second, but what we want is the rate of change at a single instant. To do this, you can think about finding the average rate of change over smaller and smaller intervals, like 0.1 seconds and 0.001 seconds. The derivative is whatever this rate of change approaches as the size of the interval approaches 0.
ya let me just change that to blue, oh no thats purple, perfect, oh wait thats too dark, whoops
Too much emphasis on the derivative with respect to time and not enough emphasis on the chain rule: dA/dt = dA/dr · dr/dt.
this is so confusing why calculus gotta be so extra
nvm i got it but sal you kinda missed on this one. made it way more complex then it needed to be imo.
its sad cuz u always have bangers but this one was a miss. i had to watch a vid by an organic chemistry tutor to understand this.
so many videos are posted that there are no views for an amount of time
I hear him saying derivative, and I am wondering, is this calculus?
PoweredMinecart McUnite yup, np for 6 year response
who put you in charge?
I believe he has those already, go to his channel uploads and try looking throught the.... 3,411 videos... right...
Oh here you go
watch?v=ANyVpMS3HL4
I searched "derivative khan academy"
Why was it necessary to constantly repeat yourself?
Hehe soup