If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

# Related rates intro

AP.CALC:
CHA‑3 (EU)
,
CHA‑3.D (LO)
,
CHA‑3.D.1 (EK)
,
CHA‑3.D.2 (EK)

## Video transcript

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 I drop a rock in the middle of that pool of water and a little while later a ripple has a little wave a ripple has formed that is moving radially outward from where I drop the rock so let me see how well I can draw that so it's moving radially outwards so that is the ripple that is formed for me dropping in 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 this circle the radius of the circle is equal to 3 centimeters is equal to 3 centimeters and we also know we also know that the radius is increasing at a rate of 1 centimeter per second so radius radius growing growing at rate 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 area of circle 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 are we know that right now R is equal to 3 centimeters we also know the rate at which the radius with the rate at which R is changing with respect to time we also know this information right over here D R DT the rate at which the radius is changing with respect to time is 1 centimeter per second 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 what rate is the area of the circle where a is the area of the circle at what 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 this is elementary geometry the area of a circle is going to be equal to pi times the radius of the circle 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 I'm just rewrite what I just had so pi 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 I'm going to have whoops 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 the derivative with respect to time of R squared and it'd 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 was it a function of time then area wouldn't be a function of time so we let me write 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 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 something is going to be two times two times that something to the first power and we're going to let me make it clear this is the derivative of 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 the derivative of x squared with respect to X we'd have 2x if it's the derivative of R of T squared with respect to R of T it's two R 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 with the derivative with respect to the derivative which this changes with respect to time we have to multiply this times the rate at which the rate at which R of T the rate at which R FG 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 two times two 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 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 what do let me rewrite all of 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 s you only put that two out front is equal to two two 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 two pi times R times D R DT actually let me make the R in blue two pi r dr dt dr d r dt now what do we know we know what R is we know that R at this moment right right in time is three centimeters right now R is three centimeters we know dr/dt right now is one centimeter per second we know this is one centimeter per second so what's da/dt going to be equal to what's going to be equal to do that same green 2 pi times 3 times 3 times 1 times that's purple times 1 times 1 centimetre per second and let's make sure we get the unit's 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 centimeters 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