- Related rates intro
- Analyzing problems involving related rates
- Analyzing related rates problems: expressions
- Analyzing related rates problems: expressions
- Analyzing related rates problems: equations (Pythagoras)
- Analyzing related rates problems: equations (trig)
- Analyzing related rates problems: equations
- Differentiating related functions intro
- Worked example: Differentiating related functions
- Differentiate related functions
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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- 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 .(158 votes)
- 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!(225 votes)
- 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?(22 votes)
- 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.(31 votes)
- 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?(8 votes)
- 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.(19 votes)
- 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.
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?(5 votes)
- 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.(8 votes)
- 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?(4 votes)
- 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.(5 votes)
- Silly question, but when Mr, Khan says, at6:06, 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.(4 votes)
- 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.(4 votes)
- 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.(3 votes)
- 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(5 votes)
- 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?(3 votes)
- Yes, but only with a sphere that is expanding or contracting.
Where L is the surface area, r is the radius and t is time.
So, 8πr(dr/dt) is the rate at which a sphere's surface area is changing over time.(3 votes)
- At4:04: Why can you rewrite d/Dt [pi*r^2] as pi*d/dt [r(t)?(3 votes)
- Andrew M is exactly right, but to be more explicit: To take a derivative is to measure the rate of change of something with respect to something else. A 'constant' term is just that; constant, therefore it never changes with respect to anything else.(5 votes)
- sal said that pie is a constant. my question is why it is taken out because i thought that the derivative of a constant is 0?(2 votes)
- 𝜋 is a constant, but in this case it's also the coefficient of an expression that is not constant.
𝑔(𝑥) = 𝑘 ∙ 𝑓(𝑥) ⇒ 𝑔'(𝑥) = 𝑘 ∙ 𝑓 '(𝑥)(4 votes)
So let's say that we've got a pool of water and I drop a rock into the middle of that pool of water. And a little while later, a little wave, a ripple has formed that is moving radially outward from where I dropped the rock. So let's see how well I can draw that. So it's moving radially outwards. So that is the ripple that is formed from me dropping the rock into the water. So it's a circle centered at where the rock initially hit the water. And let's say right at this moment the radius of the circle is equal to 3 centimeters. And we also know that the radius is increasing at a rate of 1 centimeter per second. So radius growing at rate of 1 centimeter per second. So given this, right now our circle, our ripple circle has a radius of 3 centimeters. And we know that the radius is growing at 1 centimeter per second. Given that, at what rate is the area growing? At what rate is area of circle growing? Interesting. So let's think about what we know and then what we don't know, what we're trying to figure out. So if we call this radius r, we know that right now r is equal to 3 centimeters. We also know the rate at which r is changing with respect to time. We also know this information right over here. dr dt, the rate at which the radius is changing with respect to time, is 1 centimeter per second. Now what do we need to figure out? Well, they say at what rate is the area of the circle growing? So we need to figure out at what rate is the area of the circle-- where a is the area of the circle-- at what rate is this growing? This is what we need to figure out. So what might be useful here is if we can come up with a relationship between the area of the circle and the radius of the circle and maybe take the derivative with respect to time. And we'll have to use a little bit of the chain rule to do that. So what is the relationship at any given point in time between the area of the circle and the radius of the circle? Well, this is elementary geometry. The area of a circle is going to be equal to pi times the radius of the circle squared. Now what we want to do is figure out the rate at which the area is changing with respect to time. So why don't we take the derivative of both sides of this with respect to time? And let me give myself a little more real estate. Actually, let me just rewrite what I just had. So pi r squared. Area is equal to pi r squared. I'm going to take the derivative of both sides of this with respect to time. So the derivative with respect to time. I'm not taking the derivative with respect to r, I'm taking the derivative with respect to time. So on the left-hand side right over here, I'm going to have the derivative of our area. Actually, let me just write it in that green color. I'm going to have the derivative of our area with respect to time on the left-hand side. And on the right-hand side, what do I have? Well, if I'm taking the derivative of a constant times something, I can take the constant out. So let me just do that. Pi times the derivative with respect to time of r squared. And to make it a little bit clearer what I'm about to do, why I'm using the chain rule, we're assuming that r is a function of time. If r wasn't a function of time then area wouldn't be a function of time. So instead of just writing r, let me make it explicit that it is a function of time. I'll write r of t. So it's r of t, which we're squaring. And we want to find the derivative of this with respect to time. And here we just have to apply the chain rule. We're taking the derivative of something squared with respect to that something. So the derivative of that something squared with respect to the something is going to be 2 times that something to the first power. Let me make it clear. This is the derivative of r of t squared with respect to r of t. The derivative of something squared with respect to that something. If it was a derivative of x squared with respect to x, we'd have 2x. If it was the derivative of r of t squared with respect to r of t, it's 2r of t. But this doesn't get us just the derivative with respect to time. This is just the derivative with respect to r of t. The derivative at which this changes with respect to time, we have to multiply this times the rate at which r of t changes with respect to time. So the rate at which r of t changes with respect to time? Well, we could just write that as dr dt. These are equivalent expressions. And of course, we have our pi out front. And I just want to emphasize this is just the chain rule right over here. The derivative of something squared with respect to time is going to be the derivative of the something squared with respect to the something. So that's 2 times the something, times the derivative of that something with respect to time. I can't emphasize enough. What we did right over here, this is the chain rule. That is the chain rule. So we're left with pi times this is equal to the derivative of our area with respect to time. Now let me rewrite all this again just so it cleans up a little bit. So we have the derivative of our area with respect to time is equal to pi times-- actually let me put that 2 out front. Is equal to 2 times pi times-- I can now switch back to just calling this r. We know that r is a function of t. So I'll just write 2 pi times r times dr dt. Actually, let me make the r in blue. 2 pi r dr dt. Now what do we know? We know what r is. We know that r, at this moment right in time, is 3 centimeters. Right now r is 3 centimeters. We know dr dt right now is 1 centimeter per second. We know this is 1 centimeter per second. So what's da dt going to be equal to? Well, it's going to be equal to-- do that same green-- 2 pi times 3 times 3 times 1 times-- that's purple-- times 1 centimeter per second. And let's make sure we get the units right. So we have a centimeter times a centimeter. So it's going to be centimeters. That's too dark of a color. It's going to be square centimeters, centimeters times centimeters, square centimeter per second, which is the exact units we need for a change in area. So we have da dt is equal to this. da, the rate at which area is changing with respect to time, is equal to 6 pi. So it's going to be a little bit over 18 centimeters squared per second. Right at that moment. Yep, 3 times 2 pi. So 6 pi centimeters squared per second is how fast the area is changing. And we are done.