Current time:0:00Total duration:7:43
0 energy points
Studying for a test? Prepare with these 5 lessons on Applications of derivatives.
See 5 lessons
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. 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.