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### Course: AP®︎/College Calculus AB > Unit 4

Lesson 5: Solving related rates problems- Related rates intro
- Related rates (multiple rates)
- Related rates: Approaching cars
- Related rates: Falling ladder
- Related rates (Pythagorean theorem)
- Related rates: water pouring into a cone
- Related rates (advanced)
- Related rates: shadow
- Related rates: balloon

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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.

## Want to join the conversation?

- At3:30how are the diameter and height having the same ratios. What does he mean by "these are lines"?(50 votes)
- 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)

- how is r h/2? I still don't get it(27 votes)
- The proportion of height and diameter is the same in this problem. So h = d. The diameter is 2 times the radius.
`h = d`

h = 2r

h/2 = r

Now we have r in terms of h.(63 votes)

- At3:53I don't understand why Sal uses "h" as height when he has real value for that height which is 2.(15 votes)
- We also need to know the height of the water at times near "now" so that we can answer the question of how quickly the height of the water is rising. Therefore, we do need to define h as a function of t so that we can calculate its derivative.(12 votes)

- 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)- 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)

- 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)
- 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)

- Why can he take out the constants at6:35? I'm guessing it won't change the answer, but I don't get how(3 votes)
- the constant rule states that we can take the constant out because a derivative of a constant is 0(12 votes)

- 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)
- Exactly. Differentiation and integration are very easy to do with one variable but highly complicated with multiple variables.(3 votes)

- at4:12, 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)
- 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)

- 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)- 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)

- 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)- 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.