In the previous video, we got introduced to this idea of related rates. And recall that when doing these types of problems, we took time derivatives on both sides of the equation and then used implicit differentiation to find a missing rate with respect to time. Now this is the process that you're going to need to use for any type of related rates problem. Only in this video, we're going to learn that some of the problems are not as straightforward as simply being given an equation and having to solve it. But don't sweat it, because in this video, I'm going to walk you through how to solve some of these more complicated types of problems.
So let's just get right into things. Now a common problem type that you'll see is a changing geometries problem, and one of those problems would look something like this. Here, we're told in this example that a cube grows at a rate of 2 cubic centimeters per second. And we're asked how fast the side of the cube is growing when it has a length of 0.8 centimeters? Now, you'll notice right off the bat, this is not a situation where we are being given an equation nor are we given any of the rates really that are happening in a straightforward manner.
So in this problem, we're going to have to do a couple of extra steps to figure out what all the rates are that we're dealing with and what equation we need to use. Now a good first step would be to draw and label a picture of the scenario. Now we already have a picture drawn for us right here of a cube. Now we're told specifically this cube is growing at 2 cubic centimeters per second. So that means the volume of this cube is expanding, and the rate at which this is expanding is given to us in this problem as 2.
Now something that I'll mention is this is a very common problem type where you have a shape that will either grow or shrink over time. Now in situations where you are dealing with a shape that is growing, what you want to do is make your rate positive. And since I can see here that this shape is growing, then we have a positive 2 cubic centimeters per second as our rate. If this cube were shrinking, we would need to make this value negative instead. But since it's growing, we'll keep it positive.
Now we're also told that at one instance, it has a side length of 0.8 centimeters. So we're asked for this moment of time where it has a 0.8 centimeter side length, and we want to know at this moment how fast one side of the cube here is growing. So this is what we're trying to figure out. Now this still might seem a bit confusing, but what I can do is extract the variables that we have in this problem. Since I can see here that we have cubic centimeters per second as our rate, that means we have the changing volume with respect to time.
We're told here that it's 2 cubic centimeters per second. Now another variable that we have is the side length at one instant. It's going to be 0.8 centimeters. So these are two variables that we can find from our problem, and what we're looking for is how fast the side of the cube is growing. So this rate would be the side length with respect to time, which we could say is dxdt.
And this is what we're ultimately looking for. Now looking at all my variables here, I've gone ahead and identified and labeled the diagram. So my next step would be to figure out the equations that I need that's going to relate all of the variables together. And the equation that I'm looking for, well, I see here that I have a changing volume with respect to time. I have a side length that I have a changing side length with respect to time.
So I think the equation that's going to do the trick is the volume of a cube. Because I know that the volume of a cube is just x3 where x is one side length of the cube. And what I can do to figure this out is all the steps that we've done already when solving related rates. Because we know from here, we just need to take the time derivative on both sides and then use implicit differentiation. And so let's go ahead and do that step, which we talked about in the previous video.
So I'm going to go ahead and take the time derivative on both sides of the equation, and that's literally just taking the time derivative of every variable that we see. So we're going to have the time derivative of volume on the left side of the equation, and then we'll have the time derivative of x3 on the right side of the equation. Now taking the derivative of volume with respect to time, that's just going to give you 1. Then we need to multiply this using the chain rule by the derivative of v with respect to t, which is going to give you dvdt. Now, dvdt multiplied by 1 is just going to be this rate.
So I'm just going to write dvdt. And then on the right side, the derivative of x3, well, that's going to be 3 x2 using the power rule here. But again, we need to use the chain rule since we're taking the derivative of x with respect to t, and doing that will give you dxdt. So this is going to be the other rate that we get out of this equation. Now notice something here.
Notice when I went ahead and took this time derivative here, notice we got all the variables that we need. I see that we have our dvdt. I see that we have our dxdt, and then I see that we have the x that was the one side length right here. So we know that we used the right equation, so we're in a good position here. Now our next step since we know we did use the right equation would be to isolate the target rate of change.
Now the target rate of change in this example, well, that's going to be this dxdt that we're trying to find right here. So if I want to isolate that on one side, well, what I'm going to go ahead and do is divide this 3 x2 on both sides of the equation. So I'm just going to use some quick algebra here, and that's going to get the 3 x2 to cancel on the right side. So doing this, I'm going to end up with just my rate dxdt on one side of the equal sign. And then on the other side, we're going to have this whole equation right here, which I'm going to write like this, well dvdt on top and that whole thing is going to be divided by 3 x2.
So this is the equation. Now at this point, what I need to do for my last step is plug in all of the known variables and solve. Now I can see we have the dxdt solved for on one side of the equation. So I can see we have dvdt, which is given to us as 2 cubic centimeters per second. We already figured this out in the problem.
And I can divide that by 3 times x2, which in this case, x is 0.8 centimeters. So we have 2 divided by 3 times 0.8 squared. And if you do this on a calculator, all of this should come out to about 1.04. Now on your calculator, it's probably going to read as 1.0416666, and this is just going to keep repeating here. Let's go ahead and round this off.
I'm just going to say it's about 1.04. And since it's a length with respect to time, we're going to have centimeters per second. So this right here is going to be the missing rate and that is the solution to this problem. So as you can see, this process really is the same idea as what we did in the previous video. The main difference when dealing with these types of problems is notice in this situation, we had to first identify all the missing variables, and then we had to find our equation.
In the previous video, we had all of this information up here given to us, but here we need to extract it from the problem. But if you're able to draw out a diagram and figure out what exactly is happening, then doing this should be a pretty straightforward process to find exactly the equation that you need. So hope you found this video helpful, and let's go ahead and try getting some more practice.