So we have got a very

interesting scenario here. I have this conical thimble-like

cup that is 4 centimeters high. And also, the diameter

of the top of the cup is also 4 centimeters. And I’m pouring water

into this cup right now. And I’m pouring the water at

a rate of 1 cubic centimeter. 1 cubic centimeter per second. And right at this

moment, there is a height of 2 centimeters of water

in the cup right now. So the height right now from the

bottom of the cup to this point right over here

is 2 centimeters. So my question to you

is, at what rate– we know the rate at which the

water is flowing into the cup, we’re being given

a volume per time. My question to you is

right at this moment, right when we are

filling our cup at 1 cubic centimeter

per second. And we have exactly

2 centimeters of water in the cup,

2 centimeters deep of water in the cup,

what is the rate at which the height of the

water is changing? What is the rate at which

this height right over here is actually changing? We know it’s 2 centimeters,

but how fast is it changing? Well let’s think about

this a little bit. What are we being given? We’re given– we are being

given the rate at which the volume of the water is

changing with respect to time. So let’s write that down. We are being given the rate at

which the volume of the water is changing with

respect to time, and we’re told that this is 1

cubic centimeter per second. And what are we

trying to figure out? Well we’re trying to

figure out how fast the height of the water is

changing with respect to time. We know that the height

right now is 2 centimeters. But what we want to

figure out is the rate at which the height is

changing with respect to time. If we can figure

out this, then we have essentially

answered the question. So one way that we

can do this is we can come up with the

relationship between the volume at any moment in time and the

height at any moment in time. And then maybe

take the derivative of that relationship,

possibly using the chain rule, to come up with a relationship

between the rate at which the volume is changing and the

rate at which the height is changing. So let’s try to do

it step by step. So first of all, can we

come up with a relationship between the volume and the

height at any given moment? Well we have also

been given the formula for the volume of a

cone right over here. The volume of a cone

is 1/3 times the area of the base of the

cone, times the height. And we won’t prove

it here, although we could prove it later on. Especially when we start

doing solids of revolutions within in integral calculus. But we’ll just take

it on faith right now, that this is how we can figure

out the volume of a cone. So given this can we

figure out volume– can we figure out an

expression that relates volume to the height of the cone? Well we could say

that volume– and I’ll do it in this blue color–

the volume of water is what we really care about. The volume of water

is going to be equal to 1/3 times the area

of the surface of the water– area of water surface– times

our height of the water. So times h. So how can we figure out the

area of the water surface, preferably in terms of h? Well we see right over

here, the diameter across the top of the

cone is 4 centimeters. And the height of the

whole cup is 4 centimeters. And so that ratio is

going to be true of any– at any depth of water. It’s always going to have the

same ratio between the diameter across the top and the height. Because these are

lines right over here. So at any given point, the

ratio between this and this is going to be the same. So at any given

point, the diameter across the surface of the

water– if the depth is h, the diameter across the

surface of the water is also going to be h. And so from that

we can figure out what are the radius

is going to be. The radius is going

to be h over 2. And so the area of

the water surface is going to be pi r squared,

pi times the radius squared. h over 2 squared. That’s the area of the

surface of the water. And of course we still

have the 1/3 out here. And we’re still multiplying

by this h over here. So let me see if I

can simplify this. So this gives us 1/3 times pi h

squared over 4 times another h, which is equal to– we have pi,

h to the third power over 12. So that is our volume. Now what we want to do is relate

the volume, how fast the volume is changing with

respect to time, and how fast the height is

changing with respect to time. So we care with respect to

time, since we care so much about what’s happening

with respect to time, let’s take the derivative of

both sides of this equation with respect to time. To do that, and just so I

have enough space to do that, let me move this over. Let me move this over to

the right a little bit. So I just move this

over to the right. And so now we can take the

derivative with respect to time of both sides

of this business. So the derivative with

respect to time of our volume and the derivative with respect

to time of this business. Well the derivative with

respect to time of our volume, we could just rewrite that as dV

dt, this thing right over here. This is dV dt, and

this is going to be equal to– well we could

take the constants out of this– this is going

to be equal to pi over 12 times the derivative

with respect to t of h, of h to the third power. And just so that the

next few things I do will appear a

little bit clearer, we’re assuming that height

is a function of time. In fact, it’s definitely

a function of time. As time goes on, the

height will change. Because we’re pouring

more and more water here. So instead of just writing

h to the third power, which I could write over here,

let me write h of t to the third power. Just to make it clear that

this is a function of t. h of t to the third power. Now what is the derivative

with respect to t, of h of t to the third power. Now, you might be getting

a tingling feeling that the chain rule

might be applicable here. So let’s think about

the chain rule. The chain rule tells us– let

me rewrite everything else. dV with respect to t, is going

to be equal to pi over 12, times the derivative of

this with respect to t. If we want to take the

derivative of this with respect to t, we have something

to the third power. So we want to take the

derivative of something to the third power with

respect to something. So that’s going to

be– let me write this in a different color,

maybe in orange– so that’s going to be 3 times

our something squared, times the derivative of that

something, with respect to t. Times dh– I’ve already used

that pink– times dh dt. Let’s just be very clear. This orange term right

over here– and I’m just using the chain rule–

this is the derivative of h of t to the third power

with respect to h of t. And then we’re going to multiply

that times the derivative of h of t with respect to t. And then that gives

us the derivative of this entire thing, h of t to

the third power, with respect to t. This will give us

the derivative of h of t to the third power with

respect to d with respect to t, which is

exactly what we want to do when we apply

this operator. How fast is this changing? How is this changing

with respect to time? So we can just

rewrite this, just so gets a little bit cleaner. Let me rewrite

everything I’ve done. So we’ve got dV, the rate

at which our volume is changing with respect to time. The rate at which our volume is

changing with respect to time is equal to pi over 12

times 3 h of t squared, or I could just write

that as 3h squared, times the rate at which

the height is changing with respect to

time, times dh dt. And you might be

a little confused. You might have been tempted to

take the derivative over here with respect to h. But remember, we’re

thinking about how things are changing

with respect to time. So we’re assuming–

we did express volume as a function of

height– but we’re saying that height itself

is a function of time. So we’re taking the

derivative of everything with respect to time. So that’s why the chain

rule came into play when we were taking

the derivative of h, or the derivative of

h of t, because we’re assuming that h is

a function of time. Now what does this thing

right over here get us? Well we’re telling us

at the exact moment that we set up this

problem, we know what dV dt is, we know that it’s

1 centimeter cubed per second. We know that this

right over here is 1 centimeter

cubed per second. We know what our height

is right at this moment. We were told it

is 2 centimeters. So the only unknown

we have over here is the rate at

which our height is changing with respect to time. Which is exactly what

we needed to figure out in the first place. So we just have

to solve for that. So we get 1 cubic

centimeter– let me make it clear– we get 1

cubic centimeter per second– I won’t write the units

to save some space– is equal to pi over 2. And I’ll write this

in a neutral color. Actually, let me write

in the same color. Is equal to pi over 2,

times 3, times h squared. h is 2 so you’re going to get

4 squared centimeters if we kept the units. So 3 times 4. All right let me be

careful, that wasn’t pi over 2, that was pi over 12. This is a pi over

12 right over here. So you get pi over 12, times 3

times 2 squared, times dh dt. All of this is equal to 1. So now I’ll switch

to a neutral color. We get 1 is equal to,

well 3 times 4 is 12, cancels out with that 12. We get one is equal

to pi times dh dt. To solve for dh dt

divide both sides by pi. And we get our drum roll now. The rate at which our height

is changing with respect to time as we’re putting 1 cubic

centimeter of water per second in. And right when our height

is 2 centimeters high, the rate at which this height

is changing with respect to time is 1 over pi. And I haven’t done the

dimensional analysis but this is going to be

in centimeters per second. You can work through

the dimensional analysis if you like by putting in the

dimensions right over here. But there you have it. That’s how fast

our height is going to be changing at

exactly that moment.

thanks. you sound like Joe from family guy lol

That diagram is very

khanical.

drawing tutorial link?

👍👍👍

Why were you able to take the constants out at 5:36?

No matter what i do, I can't understand all the random problems they throw at me. Neither one helps with any of the others. Think i'll just stop trying this bullshit.

I think the problem is even simpler if you use the chain rule this way: dV/dt = dV/dh * dh/dt. Once you have the formula for V(h), take the derivative and replace dV/dh with that. You're given dV/dt so plug that in too, then all you have to do is then solve for dh/dt.

Who knew Sal was an artist too, oh dang.

This guy kinda sounds like Ian Riley from Barney's Mind fame.

Ugh… I understand it when he does it but I can't figure it out by myself :'(

why the fuck do you feel the need to repeat every sentence 5 times, super helpful vid tho thanks alot!

Something with respect to something and the derivative of something.

i have been staring at my question for good 3 hrs now, and now my head hurts. IM DONE

where you get v=(πh^3)/12 if you get the derivative of that then you have dv/dh and if you get the inverse of that you get dh/dv and then you can times dh/dv*dv/dt and you get dh/dt and then just sub in 2 for h, i find that less complicated, even though it'll look way more complex in this notation

Cool vid, for anyone confused, when you take the derivative of the rearranged volume formula, you can just tag on the dh/dt next to the new derivative of height you just found instead of elaborating with the h(t) part.

Making diameter and height of the cone the same made simplifying too easy. None of the problems I am assigned use the same dimension like this example. How would you simplify and pull the derivative when you can't just substitute (h/2) for r?

i love the drawing of the cone and faucet. unfortunately im still trying to figure out the problem…

I used quotient rule and got 1/(pi^2) as the answer. why is the answer different with quotient rule

Why does he keep writing d/ot

I can understand when other people do it but when i do it i get wrecked

fuck calculus

I love this, it helped me understand how to take related rates. I couldn't have done it without ya ;D

how did pi/12 become pi/2

let my understanding of related rates=r. Find dr/dt.

dr/dt=0

holy shit drawing skills on point tho

thnx man

god Im so fucked

wtf this guy making up his own shit

Why do I hate ur voice

wtf calculus ./.

Dude. Just use the fucking power rule and multiply by the derivative. All the chain rule stuff is unnecessary for this problem.

when you fill a cone with water, water height and radius increase if its rate of change of its volume is constant.

the rate of change of the speed of increasing of the radius and height is decreasing (logically due to the cone shape) so how can we calculate this acceleration of in this case deceleration.

He wrote pi/2 instead of pi/12……

Yeah, I hate this shit.

Hello derp guy me dwee,please reply to this comment everyone who is reading this please.

I too will use calculus to find how much water I have in my 4 cm tall conical shaped cup at 2 cm of water when water is running at 1 cm^3/sec

"Let me make it clear" yes because writing it in some weird way makes everything so much clearer hahaha

I love you videos. thank you

Fck related rates bc every problem is different so I get stuck at every new problem fuuckk dissss shiiiittt

I've taken calc for 3 years and still have never done one of these correctly on my own

Might as well embrace it kiddies. Got a degree in CS with 4 certifications but still can't find work. I thought I was done with this but here I am reviewing in case I have to become a math teacher. lol

He messed up when he rewrote pi/12 as pi/2. messed with my head for a while.

i am getting analed…… in the brain.

did anyone else notice that he messed up the problem by saying it is pi/2 instead it should be pi/12

my man, why is the radius irrelevant?

Am I the only one to realize its supposed to be pi/12. Better delete this video before the whole youtube community drops out of calculus.

Is this the dog from family guy

why does khan Academy use the same problems in my textbook?

:(((((((

i watched this whole video and absorbed nothing

hahaha test is tmr I'm totally screwed 🙂

Finals RIP

our teacher was writing much faster its hard to slow down my thinking this much

This problem isn't so hard really; just set up your volume equation, take the derivative, and live happily ever after.

Hey Khan, i think it would be easier if you'd just go with the idea of implicit differentiation.

Thanks khan academy has helped so much.

It's pi over 12 not 2 u forgot the 1

Lmk why its so easy when other people do it but then when i get a problem on the test i cant even draw the diagram correctly

I'm good at calculus but this was probably the most confusing thing I've ever watched

why tf does he repeat everything he says 18 times its so distracting

what

h(t), cuz it just wasnt confusing enough already.

Thank you very much!

I'm dropping out

where did you get pi/2?? tffff

That should not of been pi/2 it should of been pi/12 and then when combined turned into (pi*h^2)/4

anyone else internally screaming when he wrote pi/2?

Got an exam tomorrow and this is the one thing I didnt understand. Thanks khan academy, for saving my ass year after year

Rewriting pi/12 to pi/2 is the 100% the reason why i hate math…. i do that 90% of the time.

I could barely focus because I was so busy admiring the drawing

r over h times h. much easier and faster way to do it

Obviously, he used graphics tablet for the drawing

Could you guys do related rates for troughs

bro you made this way more difficult then it had to be…why…

Setting it up always gets me. Reason why I failed the exam twice. Couldn't set it up. 🎃

He fixes pi/2 to pi/12 @10:29, don't trip guys LOL

Why are you able to take the 3.14/12 out, if I may ask.

got my ap calc ab exam in about 9 hours im so dead

the guy talks for 11: 31 minutes and change my whole life…and not only me and for millions and billions of people out there…..

Khan academy freakin blows

I think you are complicating things when making h the function h(t). Its an unnecessary additional step IMO. Just derive H^3 to 3h^2(dh). Good video otherwise.

Dammit the denominator is "dt" not "ot" and quit repeating yourself 50 thousand times! Good stuff though. Lol

Stop caring about the color. It interrupts the flow. Just kept crashing the flow over the stupid color.

You meant pi over 12 at 10:13

all your d's look like o's just so you know

…

The old video on rates of change provides a much shorter and time-saving method to solve this.

w h a t

What about if the height of the cone is not 4cm, what if it's 6cm?

this is alien speak to me … lol

His D’s make me wanna kms

I GOT IT!!!

I think it would be better if you were to explain the mentality behind the concepts you are using in the actual problem, especially because every example is different. I think most people understand things best when they are given a reason as to why they’re doing something, such as just finding the derivative of both sides of the basic area of a cone in respect to time, then plugging in the values that are given; Always dumb things down as much as possible, it will make more sense later

Poorly explained. I am lost on the process. I have no clue what to do if I know what h is.

Here an interesting question is How much is the dH/dT changing each moment. its much more useful.

How would u solve for the deriviative of h with respect to t in general

This is a function of time, that has respect to time, within the derivative of t, and the chain rule with respect to time

what if the diameter is not equal

8:23 why is it pi/2 and not pi/12?? nvm he fixes it at 10:27

What is the given?