a-cross-section-of-an-airplane-wing-is-shown-measurements-of-the-thickness-of-the-wing-in-centimet-2

Name: a cross section of an airplane wing is shown measurements of the thickness of the wing in centimet 2
Uploaded: 2020-02-02
Duration: 6 min 37 s
Channel: Numerade
Description: A cross-section of an airplane wing is shown. Measurements of the thickness of the wing, in centimeters, at 20 -centimeter intervals are $5.8,20.3,26.7,29.0,27.6,27.3,$ $23.8,20.5,15.1,8.7,$ and $2.8 .$ Use the Midpoint Rule to estimate the area of the wing's cross-section.

Question

A cross-section of an airplane wing is shown. Measurements of the thickness of the wing, in centimeters, at 20 -centimeter intervals are $5.8,20.3,26.7,29.0,27.6,27.3,$ $23.8,20.5,15.1,8.7,$ and $2.8 .$ Use the Midpoint Rule to estimate the area of the wing's cross-section.

Lucas Everson · Accepted Answer

Hello. My name is Lucas, and today we&#x27;re gonna be taking a look at this calculus. Question is shown right here, and hopefully we all will be arriving at this answer shown right here. If you&#x27;re following along the textbook, this is chapter six, Section one, question 19. And if you want to pause the video and write this down, feel free, but to summarize, we have this shape right here. It is 200 centimeters wide and were given the height of the shape or the context of the problem. The thickness of the wing at each 20 centimeter interval. And those numbers are given right here. With this information, we have to use the midpoint rule to estimate the area of this ship. We&#x27;re going to begin with some conceptual things here. Um, a big part of calculus is finding the area under the curve because, as we will learn, uh, area under the curve has a lot of applications and calculus. Um, in many cases, such as this one, we&#x27;re gonna be given very irregular looking shape, and we&#x27;re not exactly gonna know how to find the exact area off this shape to get around this. We&#x27;re gonna use approximation and so they were gonna be learning. One of these methods is called Raymond Some. And this is essentially using a bunch of rectangles whose areas we do know how to find. It&#x27;s just a simple base, times height. And to visualize this, we&#x27;re gonna be essentially putting a bunch of rectangles next to each other spanning the width of this shape, adding all of their areas together toe hopefully find a pretty good approximation of what the true area of this shape is. So translate what we&#x27;re given in the textbook onto paper here, Um, this is a very poorly drawn. I&#x27;m sorry about that, but in black we have the approximate shape of what we&#x27;re given. It&#x27;s 200 centimeters wide and at each 20 centimeter partition, we have, ah than the height, the heights that were given in the problem. And in green. Here are the rectangles we&#x27;re going to be using to add together to ultimately find approximation off our shape. And to do this, we have to use the midpoint rule formula, which is shown here. It does look like a lot to look at it first, but it&#x27;s essentially it&#x27;s essentially a fancy way of saying we&#x27;re gonna be calculating the areas of a bunch of rectangles and adding them together. Ah, in this case, um, another way to look at this is we can view Delta X as the width of the rectangles wreck triangles. As I wrote there, that&#x27;s a mistake. Sorry. Um, and in this case, Delta X er, as we can look at the problem, we can have five, uh, rectangles of equal length and each vehicle with I&#x27;m sorry, and each winth will be 40 centimeters 20 plus 20. So Delta X, in this case in our problem will be 40 centimeters and n in this case will be five. The number of rectangles that we will be using and f off What&#x27;s going on here is essentially finding the height of the function between two given points. And what I mean by that is we have fo zero. The function at zero is 5.8 and the function at X equals 40. Right here is 26.7. We want to find the mid point of that. So the average of zero and 40 is 20 and the function at 20 in this case is 20.3. So in our first rectangle that we will be calculating. Here, Um, we have 20.3 here and in on this page here. Green represents the heights of the triangles. Rectangles. Sorry that we will be calculating, and Rand represents Delta X or the width of the rectangles. So for our first rectangle, our height will be f of 20. Wish, as we know, is 20.3. We&#x27;ll be multiplying that by Delta X, which is 40 and once again shown right here are next rectangle shown here is 29 times 40 29 is shown right here, which is F of 60 the midpoint between 40 and 80. And we keep on going, you know, f of 100 half of 140 half of 180 each multiplied by Delta X, which is always going to be 40 centimeters in this case in this problem. So if we, uh, grab our calculators, we can find that each of these rectangles have areas shown here, and if we add all of these rectangles together, we should hopefully get 4232. And in this case, it&#x27;s always. Well, it&#x27;s always important to include units in this case. That will be centimeters squared if we take a look at what our rectangles look like in relation to the curve, Um, we would hope, and we should see that this is actually a pretty good approximation of what it actually is. Because, as we see here, we have some area of the rectangle that isn&#x27;t included in the curve, but at the same time, in the same rectangle. We have this area here which is being excluded, and these should hopefully cancel each other out pretty well. And same thing goes for each of these rectangles shown here. And as we calculated, um, it all cancels out to a pretty good approximation. Well, if once again, little over 4000 centimeter square. Thank you.

Kenneth Kobos · Answer

a cross section of an airplane wing is shown. Measurements of the thickness of the wing in centimeters at 20 centimeter intervals are 5.82 point 20.3 26.7 29 27.6 27.3 23.8 20.5 15.1 point seven and 2.8 used the midpoint rule to estimate the area of the wings cross section. Okay, since we know measurements at every 22nd centimeter intervals and we want to use the midpoint rule, what we want to do is we want to break up this wing into various, um, smaller areas and then approximate the area of those, uh, areas. So we want to make it these the size of these intervals so that we can use the mid points. Since we have measurements at every 20 centimeters, we want to break it up into areas of with of 40 centimeters. So that&#x27;s going to give us 12 three for in five. So this happens at 40 centimeters 81 20 1 60 and this last one at 200. Of course, this 1st 1 is at zero. Okay, So now we want us Look at what are the mid points. So that&#x27;s going to be a 20. This is going to be It&#x27;s 60. This one&#x27;s at 100 1 30 and one 80. Okay, so we&#x27;re actually not interested in all of our measurements since we&#x27;re using the midpoint rule S O, we have measurements at 20 centimeter intervals. So we&#x27;re interested in the first value, which happens at 20. We&#x27;re not interested in the value that happens at 40. So we skipped this number. We&#x27;re interested in the number that happens at 60 centimetres. 100 centimetres come. So I&#x27;m in a mistake. This is 1 40 over here. So this is 1 40 centimeters and this is 1 80 centimeters. So we&#x27;re actually not going to be using all of our measurements to approximate the area area is approximately. So What we do is we take the midpoint. So looking at this first region over here between zero and 40 are we take the measurements at the midpoint at 20. So we have 5.8 centimeters and then we multiply it by the width of this region, which is 40 centimeters. Since we&#x27;re going from 0 to 40. So that&#x27;s how approximation of the first region. And we&#x27;re going to do the same thing for all the other four regions. So the midpoint of the second region is has a width of 26.7 centimeters and the width of this region is 40 centimeters. So we just carry on 27.6. I&#x27;m going to drop all the centimeters since everything here is our in centimeters for 40 centimeters plus 23.8 times 40 plus 15.1. I&#x27;m multiplied by 40. So after we add up all of these, well, we multiply first. Then we add up. Ah, we get the final answer of 4232 and our units are centimeters squared since we are multiplying centimeters, time, centimeters in each of these terms. So our approximation for the area of this wing is 4232 centimeters squared

Bahar Tehranipoor · Answer

this question were given to the design of a new airplane choirs, gasoline, tank of constant cross section area, and each win a scale drawing of a cross section is shown here. The tank must hold 5000 pounds of gasoline, which has a density of 42 pounds per cubic feet as straight the length of the tanks were given a lot of information here, so let&#x27;s just write some of it down so that we have it close by, so we know that it has to hold 5000 pounds. One of the density is 42 pounds her feet cute. Who wants his time? City equals. And then we&#x27;re given this cross section area. So let&#x27;s use Simpsons Rule to estimate this cross section area and all I have to do there is get information that we have here. We can say s equals me knows h over three. I&#x27;m giving it. The horizontal spacing is one foot, so H equals one have won over three and we have y 1.5 plus four times. Why one so four times 1.6 plus two times y to 1.8 me. Just move this real quick. Oh, not that this. See that spot right there? You see? I mean, plus four times. Why? Three? 1.9 plus two times two point. Oh, plus four times 2.1. And until last, we have y six. We just added close to 60.1. So now let&#x27;s go into our calculators and plug that in. So we get 1.5 plus four times 1.6 plus two times 1.8 plus four times 1.9 plus two times two plus four times 2.1 plus 2.1. So inside this big parentheses, we have 33.6 and then we had this times 1/3. So we divide that by three. If we divide this by three, we get s equals 11.2. So now we have his 11.2 square feet because we just found the area. So we have how many pounds we need. We have the density. We have the cross sectional area. What we&#x27;re looking for is the length of the tank. So let&#x27;s start crossing things out that we can cross out. We know that we say lengthy equals. We want to be left with just feet. So if we want to be left with just feet, we take our 5000 pounds, 5000 pounds and we multiply by one over the density because our density will not be one cubic foot 4th 42 pounds. So now you can get rid of our pants. We still have cubic feet. And what we want is just good. So he multiplied by one over our cross section area, 11.2 square feet. So this cubic feet over square feet cancels out and gives us just feet. And so we have the length of our tank is 10 point sixthree, approximately 10.63 feet long.

Holly Miner · Answer

So here we&#x27;re told that herons and storks have comparable wingspans, so we&#x27;re going to assume that their wingspans are equal. S O R word ratio is wing span, uh, to the wing whip. Okay, that&#x27;s our word ratio. So we&#x27;re gonna set up what we know. We know that the gray hair on has a span of 180 centimeters, and it we know that, uh, the hair on also has a width of 24 centimeters, and so this is going to be equal to the storks ratio. So we know that the stork span is 200 uh, centimeters, But we don&#x27;t know. The whip says what we&#x27;re trying to solve for So in order to solve for this, we need to find a scale factor. We&#x27;re stretching where these air equal ratios. I want to know. Ah, 180 times. What is 200? We&#x27;re gonna apply that same scale factor here. Hey, so it&#x27;s gonna be multiplied by 10 nights, and so we&#x27;re gonna multiply 24 by 10 nights, keep him equal. So 24 times 10 is 240. So we&#x27;ve got 249th which can be reduced. Three king, one of the top and three can go into the bottom. So we get 83rd and 83rd is the same as 20 six. Uh, and 2/3 centimeters. Okay, which makes sense. If 24 centimeters is to 180 we would assume something slightly larger is to 200 centimeters.

Elizabeth Xu · Answer

What we want to do is find the area of one of the triangle the wings. So we know that this link here is 19.2 feet. You have 68 degrees here and 37 26 feet here, so we can use our formula that A is equal to a times B for what, Half a times b times these signs of C so we get is 1/2 times 19.2 times 37.6 tons of sign of 68. And our answer here, this is equal to 334.7 square feet.

Gerry Chen · Answer

all right. And this problem, we&#x27;re basically asked to find the, um the size of the length of a way that&#x27;s needed in order to allow the wing toe hold a certain amount of volume. So the amount of bullying that we have to hold is given by this caught cross constant cross sectional area times the length of the wings. So we&#x27;re kind of assuming that the wing is prismatic here. Um, so we&#x27;re given. So I guess the volume of wing it is equal to, um a this cross sectional area exit will call this a and then times the length of the wing on that were given that this has to hold a certain amount of gas in the amount of gasoline that has to hold is £5000 worth, and the density is £42 per cubic foot. So, um, so let&#x27;s see here. Oh, how do you do some unit conversion here. So we have one cubic foot. Her £42 of gasoline. Good. Um, and we need to hold £5000 worth. So this comes out to roughly 1 19 It&#x27;s approximately 119 cubic feet. Okay, so the volume has to be roughly 1 19 cubic feet. Now, in order to find the length we&#x27;re solving for the length, length is going to be equal to 1 19 over the area. So the volume over the area, um, that means that we have to find the area because we already found the volume. So now I just find the area they&#x27;re giving us that the area, Um that they&#x27;re trying todo let me make a new sheet here. They&#x27;re saying that the wing is kind of this shape and they&#x27;re saying, Oh, my gosh, uh, they&#x27;re they&#x27;re giving us a couple of these lengths in a table for him. So they say the each of these is why. So this is why zero. This is why one this is why to dot, dot, dot And then this is why six and they say why zero is 1.5 feet. Why one is 1.6 feet. Why two is 1.8 feet and so on and so forth. And they&#x27;re telling us to approximate is total area using trappers little rule. And they say that the horizontal spacing. So that&#x27;s kind of like Delta X. So That&#x27;s, uh, this This might hear This is Delta X, and they&#x27;re saying that that is a constant one foot. So using traps, little rule, the area of this guy is going to be Delta X over two. So that&#x27;s one over two, sometimes y zero. So that&#x27;s 1.5 plus twice. Why one? So that&#x27;s 3.2 plus twice. Why two? That&#x27;s 3.6 plus twice. Why three? That&#x27;s 3.8 was twice. Why for sorry. That&#x27;s four plus twice. Why five? That&#x27;s 4.2 and then Plus why six is 2.1. That&#x27;s 1/2 times, Let&#x27;s see here. So this is, uh, 37 Support the calculator on 25 scoop on TV. Sleep on six plus three on a bus four plus 4.2 plus 2.1 That&#x27;s 20.4. So this comes out to 22.4 and 1/2 of that. There&#x27;s 11.2, So this is what is so now. Finally, we can find El Ellis de over a So l is the over a The was roughly 1 19 Maybe we should find the exact value for that, and A is 11 point you, Uh, some approximate signs here. Mmm. Um, let me just evaluate that real quick. So the exactly like this 1 19 was actually more like £5 over 42 and then divided by 11.2. That comes out to 10.63 And let&#x27;s just make sure that that agrees with the answer we have. And yes, indeed. That is the enter that the book is. So the answer here is 10.63 feet. Uh, jeez. All right. You get the idea. 10.63 feet.

Problem

The widths (in meters) of a kidney-shaped swimmin…

Problem 19 Hard Difficulty

A cross-section of an airplane wing is shown. Measurements of the thickness of the wing, in centimeters, at 20 -centimeter intervals are $5.8,20.3,26.7,29.0,27.6,27.3,$ $23.8,20.5,15.1,8.7,$ and $2.8 .$ Use the Midpoint Rule to estimate the area of the wing's cross-section.

Answer

4232 $\mathrm{cm}^{2}$

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4232 $\mathrm{cm}^{2}$

Biocalculus Calculus for the Life Sciences

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Integration Review - Intro

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