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### Course: Calculus, all content (2017 edition)>Unit 3

Lesson 13: Related rates

# Related rates intro

Join us as we explore the intriguing relationship between the rate at which a circle's radius expands and the corresponding rate of area growth. Using the tools of calculus, specifically derivatives, we'll tackle this concept, providing a practical application to real-world problems involving rates of change. This understanding forms a key part of mastering calculus and its applications. Created by Sal Khan.

## Want to join the conversation?

• Hello, Mr. Sal.
I wondered, how the rate of change of a circle's area, can be constant and equal to 6π? I mean, just image what happens when radius reaches hundreds of meters, the area gonna increase much more thant 6π sqcm/s.
If r is a function of time with rate of change 1 cm/s, then we can define this function as
r = t + 3. A is a function of r and r is function of time, so A can be written as a function of time also. A = π( t + 3)² = π t² + 6π t + 9. As we see from square, A is increasing not constantly. We can find the function which defines it's rate of change. A' = 2π t + 6π .
• You are correct, but you did not realize what Sal was finding. He wanted to know "what is dA/dt when r=3?". You are correct in saying that A does not increase constantly. But at the moment when r=3 cm, dA/dt=6π.
By the way, your result confirms this. You said A'=2πt+6π. When r=3, t=0, and A'=6π.
Don't you love when math works by going in different ways? Sal used chain rule, and you substituted then differentiated. However, both approaches yield the same result. Ya math!
• Hi Khan Academy Community,
I have a question about the rate of change of spheres versus cubes so here goes:
I'm using a calculus book and it asked a few questions about rate of change one of them being what is the rate of change of a sphere's volume, when I found the rate of change it turned out to be the surface area of the sphere, this seemed pretty sensible because when you increase the volume of a sphere by a little bit you are increasing it by the current surface area. Then I ran into a question about the rate of change of a cube's volume I simply assumed that the rate of change would be the the surface area again because that seemed pretty sensible, turns out that it isn't, when you take the derivative of the volume of a cube (s^3), you get 3s^2 (using exponent rule) that isn't equal to the surface area of a cube, which is 6s^2. I asked my teacher why this is, and he said that he didn't know and that it would be an interesting topic to further research. I looked online for an answer to this but couldn't find anything. Is it possible that I overlooked something simple or is there a very complex answer to this?
• Great question! Very thought-provoking. Here's what I think.

When we change the volume of a sphere, we think in terms of expanding the radius, which extends from the center out to the surface of the sphere, so that it expands equally in all directions. But when we change the volume of a cube, we think of expanding an edge of the cube, which is analogous to a diameter of the sphere, not a radius of the sphere.

What happens if we treat a length equal to ½ the cube's edge as its radius? Call it r, so each edge of the cube has a length of 2r, and the volume of the cube is 8r^3. The rate of change is 24r^2, and that's the surface area of the cube because each face has an area of 4r^2.

The key is to get the cube expanding in all directions. When you expand the length of a side, in effect you're expanding it in three directions: west (but not east), north (but not south), height (but not depth). In this case, the rate of change is equal to the surface area of the three faces of the cube that face in the directions you're expanding (so, half the overall surface area). It's as if an increment of change was like adding a layer of paint to three of the cube's sides. When you expand using this pseudo-radius, you expand the cube in all directions, the same as when you expand the radius of a sphere, like putting a layer of paint on all six sides of the cube, so you get a rate of change in volume that's equal to the entire surface area.

I'm not completely sure this analysis holds up in all cases but it's the explanation that makes sense to me.
• I may have missed something very simple here and not be remembering a rule correctly, but why are we able to keep PI when taking the derivative? I thought PI was a constant and anytime we take the derivative of a constant it equates to 0?
• If we're both thinking of the same thing, then the Pi here is a coefficient, so it doesn't equate zero when you take the derivative.
• I am confused.
When they say that dr/dt is 1 cm/sec, does it mean
1) After 1s, the radius has increased by 1 cm.
OR
2) For a small change in time, the radius increases at the rate of 1 cm/s?

If it is the second one, aren't the units weird?Like, shouldn't it be 1cm/("small change in time")? I understand that seconds is a unit of time but saying that the rate is 1 cm/s, does it not imply that the radius increases by one after a second?
• Good question.
The units used for a rate of change does not affect the problem. So you could convert the units to miles per hour and it wouldn't matter because the rate is only true for an instant. So that rate of 1 cm/s does NOT imply the radius increases by one after a second. That would only be true if we know the rate 1 cm/s was constant for at least one second. Hope that helps.
• This might be a dumb question, but I'm still confused about the rate of change mentioned in the video.

If, for example, I have a circle of radius 3 cm growing at 1 cm/second as in the video:

We have no reason to assume the 1cm/second rate will change in 1 second as it's not indicated in the problem. Therefore, it seems reasonable that can treat it as a constant rate of change for the period of time we're evaluating (1 second).

If this is true, then in 1 second the new radius will be 3 cm + 1 cm = 4 cm.

The new area will be A₂ = ((4 cm)²) π = 16 π cm²

The original area is A₁ = ((3 cm)²) π = 9 π cm²

So wouldn't the rate of change of the area simply be the amount by which the area changed in 1 second?

In other words, (16 - 9) π cm²/second = 7 π cm²/second

However we know from the video that the result is 6 cm²/second.

I guess I'm a little bit confused about how the calc math aligns with just looking at the problem from a geometric standpoint. I feel like I'm probably missing something really obvious here.
• In calculus we are looking for instantaneous rates of change. ie what is the rate of change of the area at the very instant that the circle is 3cm in radius. Not the average rate of change for the whole second after.
Try your thought experiment again, this time using 1/10 of a second.
A₂ = 3.1² · π cm² = 9.61 · π cm²
Note this is not per second as you wrote (incorrectly)

Now we have the change in area as (9.61 - 9) π = 0.61π
And the rate of change is 0.61π / (1/10) = 6.1π cm²s⁻¹

Try again this time with 1/100 of a second, and you should get 6.01π cm²s⁻¹
And so on.
What calculus tells us is the limiting value of this process as we shrink our time slice ever smaller. That limit is our instantaneous rate of change: 6π cm²s⁻¹
• Does Sal mean that at the current moment, right then when r = 3cm, that the rate is growing at 1cm/sec, or does he mean that overall the rate of change is 1cm so that, at 1sec, you have r=1, at 2secs, you have r = 3, 3 secs, r = 5 etc?
• Good Question!
He means just at the current moment. Normally, if it is relevant to a part of the solution, the rate of change (in this case, the radius) will be stated. Sometimes the rate of change is constant, and sometimes it is based on a function, which, for other types of problems, you will need to take into consideration.
• Silly question, but when Mr, Khan says, at , that r is a function of t, what does this mean? The way I interpret r is that radius is a constant. It's a value. Also, what would the function look like if r is a function of t, e.g how would you express in terms like y=mx+b.
• To answer your first question, any letter can be used to denote any function. In this case, the r actually does stand for radius; the function r(t) will return the value of the radius when t (time) is equal to a certain value.
To answer your second question, if we assume the radius started at zero, and that the rate at which the radius is increasing is constant (both fairly safe assumptions), we get
r(t) (in centimeters) = 3t
The equation would be different (and more complicated) if the rate that the radius is increasing at is not constant.
• How come Sal is able to use r = 3 as a constant even though it would be a changing variable as r expands. Does this mean that he can only find the rate of change of area at that specific moment?

Or actually, I think that the change in area is accounted for when using the chain rule and the dr/dt. Is that right?
• See what the question gives us, and what it's asking us to solve for. We're given the rate of change of radius, and the radius at a specific point. We need to find the rate of change of area at the same point. Hence, we can substitute r=3, as it is information about the stuff happening at the instant of time we care about.
• Hi Khan Academy Community,
I have a question about the rate of change of surface area of a sphere w.r.t to its radius. Upon differentiating 4* pi r^2, we get 8 pi * r. This got me confused because when we differentiate the area of a circle (w.r.t rad), we get 2*pi*r, which is its circumference. Similarly, does 8*pi*r hold any significance in a sphere's geometry?