A 13-foot tall silo is used to store grain. The area of the cross section of the silo at height ℎ feet is given by the function 𝐴, where 𝐴 of ℎ is measured in square feet. The function 𝐴 is continuous and increases as the value of ℎ increases. Selected values of 𝐴 of ℎ are given in the table. Part one, use a right Riemann sum with the five subintervals indicated by the data in the table to approximate the volume of the silo. Indicate the units of measure.
There are three other parts to this question, which we’ll come to later. Before we jump in and try to solve part one, let’s understand the scenario that we were given. The question involves a 13-foot tall silo. And if you didn’t already know, a silo is a building size container for storing grain. We’re told about this function 𝐴 of ℎ, which is the area of the cross section of the silo at height ℎ feet. So ℎ feet above the ground. The cross section will have area 𝐴 of ℎ square feet. We’re also told that the function 𝐴 is continuous and increases as the value of ℎ increases. As height ℎ increases then and we go up the silo, the cross sectional area 𝐴 of ℎ increases. And so the silo gets wider.
It might therefore be a good idea to update our diagram. We’re given a table of some values of 𝐴 of ℎ. And so we know the cross sectional area at various different heights. For example, at height zero, the cross sectional area is 320 square feet. So the area of the base on the ground is 320 square feet. Similarly, at height 13 feet and as the silo is 13 feet tall, we know that this is the top of the silo. The cross sectional area is 450.4 square feet. We also get four intermediate values at four other heights. Now, we have some better understanding of our scenario. We move on to part one, where we have to use a right Riemann sum to approximate the volume of the silo.
Now we’re given a table of values of a function. And we’re told to compute a right Riemann sum. Now, given such a table of values, we can compute the right Riemann sum with no issues. We don’t even have to understand this scenario. But it’s important to understand why this will help us approximate the volume of the silo. This is essential not only for this part of the question, but for later parts. The reason is that the volume is the integral of cross sectional area with respect to height.
We use a notation in the question to write this as the integral of 𝐴 of ℎ with respect to ℎ. This hopefully makes sense. We can cut our silo into thin cylindrical slices, each one having heights Δℎ and base area 𝐴 of ℎ. And hence volume 𝐴 of ℎ times Δℎ. We add up the volumes of the slices to get an approximation for the volume of the silo. And as Δℎ tends to zero, this sum becomes an integral. And the approximation becomes exact.
Now we see why the right Riemann sum we’re about to compute is an approximation for the volume of the silo. Now that we’re thinking about the integral of the function 𝐴 of ℎ, it might be worth drawing its graph. There are six values of ℎ, for which we know the value of 𝐴 of ℎ. We can sketch the data points that we were given and use the fact that the function 𝐴 is continuous and increasing to sensibly fill in the gaps. Now the volume that we’re looking for is the area under this graph. And we approximate this using a Riemann sum, which involves using rectangles. We get five rectangles from the five subintervals between the six data points. And as we use a right Riemann sum, the height of each rectangle should be the value of the function at the right endpoint of each interval.
So for the first rectangle, the height is the function 𝐴 evaluated at the right-hand endpoint. That’s two. Its height is 𝐴 of two. And its width is two minus zero. So its area is their product. The right Riemann sum is the sum of all such areas. For the second subinterval, its height is 𝐴 of four. And its width is four minus two. For the third, the height is 𝐴 of seven. And the width is seven minus four. And the last two rectangles go similarly. This suggests a formula which you may be aware of.
In any case, we can now evaluate this sum. We read off the value to the function from the table. For example, 𝐴 of two is 349.1. And we can also easily evaluate the width. For example, two minus zero is just two. And we proceed similarly with the other values. We can then evaluate this using our calculators to get a value of 5297.3. And remember, we need to indicate the units as well. This is a volume we got by multiplying heights in feet by areas in square feet. So our answer is 5297.3 cubic feet. Now we move on to part two.
Does the approximation in part one overestimate or underestimate the volume of the silo? Give a reason for your answer.
Well, if we turn back to our graph, the volume we were looking for was the area under this graph. And we estimated it by using the areas of the rectangles. As the rectangles cover not only the area under the graph, but also some small regions above the graph. We see that their total area must be greater than the total area under the graph. And so our Riemann sum approximation is greater than the actual volume. It’s an overestimate. We also need to give a reason. We can put our explanation involving rectangles into words or use something more formal sounding. Such as 𝐴 is an increasing function. And the right Riemann sum of an increasing function is always an overestimate for its integral. Now we move on to part three.
The cross sectional area of the silo, in square feet, at height ℎ feet can be modeled by the function 𝑓 of ℎ equals 160 times the square root of ℎ plus eight over 0.05ℎ plus two. Use this model to find the volume of the silo.
Well, we know that the volume of the silo is the integral of cross sectional area with respect to height. Now we have a new model for this cross sectional area function. We just substitute this function then. Now what should our limits be for this integral? Well, we’re looking for the total volume of the silo between the ground, which is at height zero, and the highest point, which is at height 13 feet as it is 13 feet tall. This is now a definite integral that we can evaluate very accurately using our graphing calculator. How exactly you do this will depend on the model of calculator you have. But on my TI84, I press math and then nine and then enter the integral pretty much as you see it. Doing so, we get a value of about 5134. And remember, this is measured in cubic feet. This is very close to the estimate we got in part one of 5297.3 cubic feet. And indeed as part two says, the estimate we got in part one was an overestimate. It’s greater than the more accurate value we get in part three.
So if you didn’t know which way to argue in part two, whether to argue for an overestimate or an underestimate. Well, the values in parts three and one give you some hint. Let’s move on to the fourth and final part.
A farmer is storing grain in the silo. When the height of the grain in the silo is three feet, the height increases at a rate of 0.1 feet per minute. Using the model from part three, find the rate of change of the volume of the grain in the silo with respect to time when the height of the grain is three feet. Indicate the units of measure.
When the height of grain in the silo is three feet, the height increases at a rate of 0.1 feet per minute. So the height increasing suggests a derivative, the rate of change of height. We’re told that this is 0.1 feet per minute when the height is three feet. That’s something we’re given. We’re also told to use the model from part three. I’ve erased this from the screen. But the model was that the cross sectional area at height ℎ was 160 times the square root of ℎ plus eight all over 0.05ℎ plus two. Now what we require is the rate of change of the volume of the grain in the silo with respect to time when the height of the grain is three feet.
Okay, so how do we find this? The key inside is that the cross sectional area at height ℎ is the rate of change of volume with respect to height. Using the fundamental theorem of calculus, this is essentially the same as stating that the volume is the integral of this area with respect to height. Now we can apply the chain rule. d𝑣 by d𝑡 is d𝑣 by dℎ times dℎ by d𝑡. We are interested in this when the height ℎ is three. And so we need to evaluate d𝑣 by dℎ when ℎ is three, which we do by substituting three for ℎ in the formula we have. d𝑣 by dℎ at ℎ equals three is therefore 160 times the square root of three plus eight over 0.05 times three plus two. And we know what dℎ by d𝑡 at ℎ equals three is. It’s 0.1.
Putting this into a calculator, we get an answer of 36.1907 and so on, which is about 36. And remember, we need the units of measure as well. This is a rate of change of volume with respect to time. The volume is measured in cubic feet and the time in minutes. And so we have 36 cubic feet per minute.