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Related rates: water pouring into a cone

In this video, we explore an intriguing scenario where we pour water into a cone-shaped cup at a constant rate. We'll discover how the rate of change in the water's depth connects to the rate of change in volume, all with the help of our new related rates tools. Created by Sal Khan.

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  • mr pink red style avatar for user Raghav Srikanth
    At how are the diameter and height having the same ratios. What does he mean by "these are lines"?
    (50 votes)
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    • piceratops tree style avatar for user TelstarB
      Imo is easier if u notice the ratio between the Diameter and the height in the bigger cone. for every 4 cm of D, there is 4 cm of height, so the ratio D/h = 4/4 = 1/1. Since D = 2r u can say then the ratio 2r/h = 1/1 ---> r= h/2
      (17 votes)
  • blobby green style avatar for user ma832
    how is r h/2? I still don't get it
    (27 votes)
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  • mr pants teal style avatar for user tomi_lerner
    At I don't understand why Sal uses "h" as height when he has real value for that height which is 2.
    (15 votes)
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  • blobby green style avatar for user abplanalpba
    Why can't you make the volume formula h = V/.3(area of base) and then take the derivative of that?

    d/dt [V(.3(pi(h/2)^2)^-1] = d/dt [h]

    then solve? The answer doesn't seem to match, but I can't figure out why. Is it because the h in the Volume formula is not implicitly a function of time?
    (6 votes)
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    • blobby green style avatar for user jw
      Is it as simple as you changed 1/3 in the Volume formula to 0.3? I followed that process, left it as 1/3 and was still able to get 1/pi as dh/dt:

      h = (12/pi)vh^-2. Take derivative of both sides, use product rule:

      dh/dt = (12/pi)(h^-2 dV/dt + V*(-2h^-3) dH/dT). - Substitute h=2, V=pi/3, dV/dt = 1 & simplify:

      dh/dt = (12/pi)(1/4 - pi/6 dh/dt) - I multiplied both sides by pi/12 to move it over:

      pi/12 dh/dt = 1/4 - pi/6 dh/dt - get rid of fractions by *12 both sides:

      pi dh/dt = 3 - 2pi dh/dt - Move the right term over (add 2pi dh/dt):

      3pi dh/dt = 3 - divide both sides by 3pi:

      dh/dt = 1/pi
      (17 votes)
  • marcimus pink style avatar for user Raymond Greenwood
    How would we derive a function to determine the rate at which the increase of the height of the water is decreasing with respect to time?
    (5 votes)
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    • male robot hal style avatar for user CMcIntyre
      The first derivative of the variable h with respect to time (dh/dt, or h' ) shows how the height changes with time. (ie. where is the height at any time). The second derivative of the variable h with respect to time (h'' ) would show how fast the rate the of the height is changing with respect to time. (ie. how fast is the rate of the height changing at any time). Solve for the second derivative.
      (8 votes)
  • mr pants teal style avatar for user Diana Dai
    Why can he take out the constants at ? I'm guessing it won't change the answer, but I don't get how
    (3 votes)
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  • spunky sam green style avatar for user Not Friedrich Gauss
    Sal always surprises me with his deep understanding of math, i would have never thought about relating diameter and height and somehow getting r^2 in terms of height. Anyways, is the reason we want the radius in terms of height is that so when we differentiate the equation we only have one unknown (dh/dt) instead of two (dr/dh and dh/dt)?
    (7 votes)
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  • aqualine ultimate style avatar for user Liang
    at , why can't we substitute radius=1 cm at the exact moment to V=pi*r^2*h/3? so we'll get V=pi*h/3 to proceed? with this method, I got 3/pi as the answer. thanks.
    (3 votes)
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    • male robot donald style avatar for user Venkata
      That's a good question, and there's a reason why we can't do that.

      So, we have V = (1/3)πr^2(h). Now, by substituting r = 1, you're treating r (and subsequently r^2) as a constant. So, when you differentiate w.r.t. t, you get,

      dV/dt = 1/3πr^2 (dh/dt)

      Now, here's the main problem: r is not a constant. See that as the height changes, the radius does too. So, when you differentiate both sides of V = (1/3)πr^2(h), you'll get a dr/dt term as well. However, we do not know what dr/dt is, and hence we can't solve for dh/dt. This is why we express r in terms of h and don't substitute r = 1.
      (8 votes)
  • blobby green style avatar for user NicoleJFriedman
    I have a general question about related rates. I am trying to solve a problem two ways and keep getting two different answers.

    The volume of a cone of radius r and height h is given by V = 1/3 pi r^2 h. If the radius and the height both increase at a constant rate of 1/2 cm per second, at what rate in cubic cm per sec, is the volume increasing when the height is 9 cm and the radius is 6 cm.

    I tried letting r = 2/3 h and doing a substitution. Then when I differentiated with respect to time, I got an equation for dV/dt involving dh/dt. This process didn't give me the correct answer though.

    Better yet, if r/h = 6/9 = 2/3, how could dr/dt = dh/dt = 1/2???? Isn't r = 2/3 h and therefore dr/dt = 2/3???

    Your help is greatly appreciated,
    Nikki
    Calculus instructor
    (2 votes)
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    • piceratops tree style avatar for user Theresa Johnson
      The reason you are having problems is that you are assuming that r and h have a constant ratio. They do not. The height and radius are 9 and 6 at the point in time that you are looking at. However, two seconds before they would have been 8 and 5. The ratio changes as their measures change.

      To solve this problem, you do not need to get rid of one of the variables. You have rates of change for both. Differentiate using the product rule and substitute in both dr/dt and dh/dt.
      (6 votes)
  • mr pink red style avatar for user Arash Singh
    I still don't get why do we do h(t). Is it like that in every situation? When do you know when to do it?

    Thanks
    (3 votes)
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    • leaf green style avatar for user kubleeka
      We do it because we know that height is changing as time passes. So when looking at an equation to differentiate, you need to look at each variable and ask yourself whether that variable changes with time (or with whatever variable you're taking the derivative with respect to).
      (3 votes)

Video transcript

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.