 A water trough is 10 meters long and a cross section has the shape of the soclest trapezoid that is 30 centimeters wide at the bottom, 80 centimeters wide at the top and has a height of 50 centimeters. If the trough is being filled with water at a rate of 0.2 meters cubic meters per minute, how fast is the water level rising when the water is 30 centimeters deep? So what we see here, let's think about what are the derivatives we need to find, okay? So we're giving some information about this derivative. How do I know it's a derivative? It's because, notice, I mean, by the units here, I'm measuring volume per minute, volume per time. We could think of it that way. And so this information, if I just pull that out for a moment, I have this information about the change of volume with respect to time. This is a constant 0.2 cubic meters per minute where I'm measuring distance in meters and I'm measuring time in minutes, excuse me. The other thing, then we can see our words like fast and rising. These are words that describe a rate of change, the rate of change of what of the water level. If we call the water level, for example, just H right here, how deep is the water at any given time, then we are trying to figure out what is the change of height with respect to time at the moment that the height is equal to 30 centimeters for which we see right here. This is the unknown that we're trying to find. So we have a known derivative with respect to volume and we have an unknown derivative with respect to the height right there. And so we have to find a connection that puts the, we have to find a relationship that connects the volume and the height together. And so when we think of this water trough right here, what we're thinking of is kind of like something, it would look kind of like a really long rain gutter or something like that. We see this kind of picture right here. And so we're trying to find the volume, the volume of this thing right here for which then the height would be in connection right there. So yeah, it seems like the volume formula might be the right way to approach this thing. So I wanna mention that in general, if you have some type of prism, it doesn't matter how complicated the face is, if you stretch this face in the third dimension, right? The volume of the prism is just gonna equal the area of a single face times the length of that said prism, right? So area times length, that gives you the volume of a prism. Now that doesn't matter if you have a cylinder, which of course is just a circular prism, rectangular prism, triangular prism, or in this case this water trough is a trapezoidal prism. We need to take the area of the face, our face of course is gonna be this trapezoid right here, times it by the length. Now the length we already know, that's gonna be 10. So we need to times that by the area. And so notice how the length of this water trough, that's that length of the prism, and that's gonna be 10 meters long. So let's try to figure out the cross sectional area a little bit better before going on. One other thing I should mention here is that you'll notice that the volume, the derivative there was being measured in meters cube per minute. We also have the length in 10 meters. We have these other measurements in centimeters right here, right? We have this mismatch of lengths. We need to convert everything to the same length so avoid some potential calculational errors that could creep up later on. The easiest way to do that is to actually convert all of the centimeters into meters. So instead of 80 centimeters, we're gonna get 0.8 meters. Instead of 30 centimeters, we're gonna get 0.3 meters. And instead of 50 centimeters, we're gonna get 0.5 meters like so. You can also see this diagram. I've shown you what a cross section is gonna look like where we have a typical water level at some moment of time right going on right here. If we wanna find the area of this trapezoid, so we need to find the area of just not the whole thing, but we need the area of just the water where the water is gonna be. That would give us then the volume because we're inserting volume over time. Okay, so how do you find the volume of trapezoid? Well, it's significant that it's a sosceles trapezoid. What does that mean? Well, kind of like an sosceles triangle, it means that two of the sides are congruent to each other. And given that as trapezoid, the two not necessarily parallel sides, because trapezoids always have at least one pair of parallel lines there. The two not necessarily parallel sides, those are gonna be congruent. That's what an sosceles trapezoid means in this situation. And that symmetry is gonna be useful for us. So, because notice if you were to snip off this triangle right here, right? You snip it off and you move it over here, then you actually form what turns out to be a rectangle for which the length of the rectangle is equal to the height of this original trapezoid. And then the width is equal to just the shorter side plus this extra dimension A. So let's investigate what that means a little bit here. And I'm gonna write this over here. That's we get 10 times, well again, one dimension of that rectangle is gonna be H, the height, the original height. But then we have to multiply it by this 0.3 plus the A value. That was the other dimension. So how do we figure out the A value? Well, looking at their diagram right here, I'm gonna try to make an argument using similar triangles, where you have this little triangle right here that involves the A on top and the H to the side. So let me copy this over here. So you have H over here and A over here. And then I'm gonna compare that to a triangle which has the same angle measurements, which you're gonna see that triangle over here is exactly gonna do it. You'll notice that both of these triangles are right triangles, so they both have a 90-degree angle. And then this angle right here is identical to itself, so they're gonna be equal. And then the other angles would have to be congruent as the angle sum will add up to 180 degrees. So therefore we have similar triangles and we have this bigger triangle for which this side right here will just be the 0.5 that we see. So we have this side on the top. We'll notice that if you look at the entire top of the trapezoid, you have 0.8 going on right here. And then of that 0.8, 0.3 of them are being taken care of right here. If this is called X, the fact that it's an isosceles trapezoid is significant because by symmetry we get X over here. So we're gonna get that two, basically we get that 2X plus 0.3 is equal to 0.8. If you solve for X, you're gonna get X is equal to 0.25. So we put that in right here. And therefore if we compare these two triangles together, these two triangles are proportional to each other, we're gonna get that A over H coincides with 0.25 over 0.5 which you can actually simplify that fraction that just becomes 1.5. Therefore A is equal to 1.5 H or I'm actually gonna prefer this as 0.5 H and we're gonna plug this in for A right here. Therefore the volume of this trough turns out to be 10H times 0.3 plus 0.5 H. And then just to kind of simplify the arithmetic, I'm gonna distribute this 10 through because multiply by 10 just moves the decimal place by one, right? So we get that the volume equals H times three plus five H. I'm also gonna distribute the H through and so we get three H plus five H squared. We're now ready to take the derivative of both sides and we're gonna take the derivative with respect to time, DDT, like so. So the derivative of volume, that's just gonna be V prime, that's all we need to know about that one. And then on the right hand side by the usual rules of derivatives, the derivative is gonna look like taking the derivative with respect to time here, we're gonna get three H prime plus 10 H H prime. So when you take the derivative of the five H squared, the outer derivative is gonna be five times two H given us the 10 H and then the inner derivative is gonna be the H prime. I'm gonna factor out the H prime because H prime is what we're trying to solve in the situation. If you divide both sides by the three plus 10 H, you then have solved for H prime in that situation. So H prime is equal to V prime over three plus 10 H. All right, so we aren't after any old H prime, we're after a specific H prime. We need to find the derivative of H with respect to time when the height remember was 30 centimeters, but we need the units to be consistent. So we'll actually do 0.3 meters in this situation. So then remember the change of volume was 0.2 cubic meters per minute. So we're gonna get 0.2 on top. You get three plus 10 times the 0.3. And now just try to simplify this thing. You get 0.2 over three plus three. Three plus three of course is a six. So we end up with 0.2 over six. If you wanna move the decimal place there, you're gonna get two over 60, or in other words, one over 30. One over 30 what? Well, this is going to be measuring, of course, meters per minute. Those are our units right here. For which then if we approximate that, that would be more useful in this situation. 1.30 becomes 0.033 meters per minute. But as we were originally measuring the height in centimeters, it might make sense to switch back to centimeters per minute. So times this by 100, you get 3.3 centimeters per minute. And so that tells us how quickly the water is rising inside of our water trough at the moment that the height or the depth, whatever you wanna call it, is currently 30 centimeters.