course Mth 151 â”äsŠcÓ™ïǬ²sψ³Éê™{ª¬íƸw¥¬assignment #002 002. Volumes qa areas volumes misc 08-26-2008
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14:53:15 `q001. There are 9 questions and 4 summary questions in this assignment. What is the volume of a rectangular solid whose dimensions are exactly 3 cm by 5 cm by 7 cm?
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RESPONSE --> The volume is 105m^3, you multiply the answers together. In future problems like this I can simply multiply those numbers together to get the full volume. I am positive that this answer is the correct one, at first I was going to put 105 m, but remember that the cm would be multiplied together as well. confidence assessment: 2
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14:55:30 If we orient this object so that its 3 cm dimension is its 'height', then it will be 'resting' on a rectangular base whose dimension are 5 cm by 7 cm. This base can be divided into 5 rows each consisting of 7 squares, each 1 meter by 1 meter. There will therefore be 5 * 7 = 35 such squares, showing us that the area of the base is 35 m^2. Above each of these base squares the object rises to a distance of 3 meters, forming a small rectangular tower. Each such tower can be divided into 3 cubical blocks, each having dimension 1 meter by 1 meter by 1 meter. The volume of each 1-meter cube is 1 m * 1 m * 1 m = 1 m^3, also expressed as 1 cubic meter. So each small 'tower' has volume 3 m^3. The object can be divided into 35 such 'towers'. So the total volume is 35 * 3 m^3 = 105 m^3. This construction shows us why the volume of a rectangular solid is equal to the area of the base (in this example the 35 m^2 of the base) and the altitude (in this case 3 meters). The volume of any rectangular solid is therefore V = A * h, where A is the area of the base and h the altitude. This is sometimes expressed as V = L * W * h, where L and W are the length and width of the base. However the relationship V = A * h applies to a much broader class of objects than just rectangular solids, and V = A * h is a more powerful idea than V = L * W * h. Remember both, but remember also that V = A * h is the more important.
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RESPONSE --> In my mind, I had the object raised so I could figure out the altitude, otherwise I wouldn't have been able to solve the problem. I did not bother to think about dividing the rows, just basically assumed that the area would be 35 m^2 after finding the altitude. I will remember that V=A * h applies to more subjects then just rectangular solids. self critique assessment: 2
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14:57:20 `q002. What is the volume of a rectangular solid whose base area is 48 square meters and whose altitude is 2 meters?
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RESPONSE --> If the base area is 48m^2 and the altitude is 2 m, then you proceed like this V=A * h V=48m^2 * 2m Multiplying those together, one can't forget to multiply the meters as well, so after applying this, you would get: V=96m^3 confidence assessment: 3
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14:57:32 Using the idea that V = A * h we find that the volume of this solid is V = A * h = 48 m^2 * 2 m = 96 m^3. Note that m * m^2 means m * (m * m) = m * m * m = m^2.
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RESPONSE --> OK self critique assessment: 3
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14:58:47 `q003. What is the volume of a uniform cylinder whose base area is 20 square meters and whose altitude is 40 meters?
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RESPONSE --> Using the previous methods I have used in the other problems, I can deduce that V= 20m^2 * 40m Multiplying these gives us V=800m^3 confidence assessment: 3
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14:59:29 V = A * h applies to uniform cylinders as well as to rectangular solids. We are given the altitude h and the base area A so we conclude that V = A * h = 20 m^2 * 40 m = 800 m^3. The relationship V = A * h applies to any solid object whose cross-sectional area A is constant. This is the case for uniform cylinders and uniform prisms.
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RESPONSE --> OK, I will remember that V= A * h applies to objects with a constant cross-sectional A self critique assessment: 2
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15:01:10 `q004. What is the volume of a uniform cylinder whose base has radius 5 cm and whose altitude is 30 cm?
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RESPONSE --> Well since the cylinder is uniform, it's area is constant, so we can use the formula V= A*H In that case V= 5cm*30cm V=150cm^2 I am still sketchy on the subjects of radius', but this is how I arrived at my answer, but I'm not fully confident about it confidence assessment: 1
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15:03:44 The cylinder is uniform, which means that its cross-sectional area is constant. So the relationship V = A * h applies. The cross-sectional area A is the area of a circle of radius 5 cm, so we see that A = pi r^2 = pi ( 5 cm)^2 = 25 pi cm^2. Since the altitude is 30 cm the volume is therefore V = A * h = 25 pi cm^2 * 30 cm = 750 pi cm^3. Note that the common formula for the volume of a uniform cylinder is V = pi r^2 h. However this is just an instance of the formula V = A * h, since the cross-sectional area A of the uniform cylinder is pi r^2. Rather than having to carry around the formula V = pi r^2 h, it's more efficient to remember V = A * h and to apply the well-known formula A = pi r^2 for the area of a circle.
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RESPONSE --> I didn't take into account the formula involved in radius', and was right to be unsure about my answer. I will remember that the formula for that is pi r^2 self critique assessment: 1
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15:06:44 `q005. Estimate the dimensions of a metal can containing food. What is its volume, as indicated by your estimates?
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RESPONSE --> I am actually not sure, one would measure food by inches. So say we have a can that has a base of 7 inches, and an altitude of 8 inches, and it's a uniform cylinder, so that means its area is constant. V= A * h V= 7in * 8in V= 56 inches I realize there are many answers to this problem, and that it exercises my critical thinking. confidence assessment: 2
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15:08:51 People will commonly estimate the dimensions of a can of food in centimeters or in inches, though other units of measure are possible (e.g., millimeters, feet, meters, miles, km). Different cans have different dimensions, and your estimate will depend a lot on what can you are using. A typical can might have a circular cross-section with diameter 3 inches and altitude 5 inches. This can would have volume V = A * h, where A is the area of the cross-section. The diameter of the cross-section is 3 inches so its radius will be 3/2 in.. The cross-sectional area is therefore A = pi r^2 = pi * (3/2 in)^2 = 9 pi / 4 in^2 and its volume is V = A * h = (9 pi / 4) in^2 * 5 in = 45 pi / 4 in^3. Approximating, this comes out to around 35 in^3. Another can around the same size might have diameter 8 cm and height 14 cm, giving it cross-sectional area A = pi ( 4 cm)^2 = 16 pi cm^2 and volume V = A * h = 16 pi cm^2 * 14 cm = 224 pi cm^2.
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RESPONSE --> Apart from my lack of making the inches square after their multiplication, which is a careless error I will have to rectify, I understand the way to achieve the various answers. self critique assessment: 2
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15:11:07 `q006. What is the volume of a pyramid whose base area is 50 square cm and whose altitude is 60 cm?
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RESPONSE --> I am positive this is not a correct answer, but from knowing what I have previously refreshed in my mind, since it's not a cylinder, you would use a different formula, obviously, but I can't currently remember that formula at the moment. confidence assessment: 1
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15:12:09 We can't use the V = A * h idea for a pyramid because the thing doesn't have a constant cross-sectional area--from base to apex the cross-sections get smaller and smaller. It turns out that there is a way to cut up and reassemble a pyramid to show that its volume is exactly 1/3 that of a rectangular solid with base area A and altitude h. Think of putting the pyramid in a box having the same altitude as the pyramid, with the base of the pyramid just covering the bottom of the box. The apex (the point) of the pyramid will just touch the top of the box. The pyramid occupies exactly 1/3 the volume of that box. So the volume of the pyramid is V = 1/3 * A * h. The base area A is 30 cm^2 and the altitude is 60 cm so we have V = 1/3 * 50 cm^2 * 60 cm = 1000 cm^3.
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RESPONSE --> I will remember that the formula involved in dealing with pyramids is V= 1/3 *A * h, and utilize that in future problems. self critique assessment: 2
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15:14:57 `q007. What is the volume of a cone whose base area is 20 square meters and whose altitude is 9 meters?
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RESPONSE --> A cone is basically in a slimmer pyramid form, so I would imagine you would use the pyramid formula V= 1/3 * A * h Which, given the dimensions of the cone, and the order of operations, gives us V= 1/3 *120m^3 V= 60m^3 confidence assessment: 2
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15:15:27 Just as the volume of a pyramid is 1/3 the volume of the 'box' that contains it, the volume of a cone is 1/3 the volume of the cylinder that contains it. Specifically, the cylinder that contains the cone has the base of the cone as its base and matches the altitude of the cone. So the volume of the cone is 1/3 A * h, where A is the area of the base and h is the altitude of the cone. In this case the base area and altitude are given, so the volume of the cone is V = 1/3 A * h = 1/3 * 20 m^2 * 9 m = 60 m^3.
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RESPONSE --> I am pleased that my refreshment of the pyramid formula aided in my solution of the given problem, but I can always delve into the problem deeper then I do currently. self critique assessment: 3
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15:16:49 `q008. What is a volume of a sphere whose radius is 4 meters?
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RESPONSE --> I can't currently remember the formula for a sphere. I would imagine the radius would be pi 16m^2 though. confidence assessment: 0
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15:18:44 The volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere. In this case r = 4 m so V = 4/3 pi * (4 m)^3 = 4/3 pi * 4^3 m^3 = 256/3 pi m^3.
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RESPONSE --> I have taken note of the volume of a sphere, and I have gained more confidence in dealing with future problems with that formula. These questions, even though some I can't answer, are excellent exercise in refreshing my memory. self critique assessment: 2
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15:22:40 `q009. What is the volume of a planet whose diameter is 14,000 km?
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RESPONSE --> The planets are in sphere shape, breaking it down in that simplified way, you can say that V = 4/3 pi r^3 I remember that to find the radius, you divide the diameter in half, so that gives us V= 4/3 pi *(7000)^3 V= 4/3 pi * 343,000,000,000 km^3 V= 1,372,000,000,000 / 3 * pi km^3. confidence assessment: 2
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15:22:54 The planet is presumably a sphere, so to the extent that this is so the volume of this planet is V = 4/3 pi r^3, where r is the radius of the planet. The diameter of the planet is 14,000 km so the radius is half this, or 7,000 km. It follows that the volume of the planet is V = 4/3 pi r^3 = 4/3 pi * (7,000 km)^3 = 4/3 pi * 343,000,000,000 km^3 = 1,372,000,000,000 / 3 * pi km^3. This result can be approximated to an appropriate number of significant figures.
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RESPONSE --> OK self critique assessment: 3
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15:23:57 `q010. Summary Question 1: What basic principle do we apply to find the volume of a uniform cylinder of known dimensions?
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RESPONSE --> That when the cylinder is constant and or uniform, that the volume used is V= A * h A is the area, and h is the altitude, basically length times width. confidence assessment: 2
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15:24:08 The principle is that when the cross-section of an object is constant, its volume is V = A * h, where A is the cross-sectional area and h the altitude. Altitude is measure perpendicular to the cross-section.
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RESPONSE --> OK self critique assessment: 3
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15:25:10 `q011. Summary Question 2: What basic principle do we apply to find the volume of a pyramid or a cone?
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RESPONSE --> The volume is V = 1/3 A * h, where A is the area and h is the altitude confidence assessment: 3
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15:25:45 The volumes of these solids are each 1/3 the volume of the enclosing figure. Each volume can be expressed as V = 1/3 A * h, where A is the area of the base and h the altitude as measured perpendicular to the base.
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RESPONSE --> I understand the basics of the formula, but I still need to look deeper as to why those formulas are what they are. So I only feel I got the answer half right. self critique assessment: 2
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15:26:38 `q012. Summary Question 3: What is the formula for the volume of a sphere?
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RESPONSE --> The volume of a sphere is V = 4/3 pi r^3, r is the radius, and I'm not entirely sure why the measurement 4/3 is used, perhaps because the sphere can be divided that way? confidence assessment: 2
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15:26:45 The volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere.
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RESPONSE --> OK self critique assessment: 3
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15:28:32 `q013. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> The principles were stored in the very back of my mind, and at first, these questions were bewildering, but after concentrating hard enough, the formula V= a * h came to mind after thinking about the properties of cylinders. Until these exercises, these principles were not organized AT ALL. But now they feel more prominent in my mind, and will be utilized in future problems involving the principles. confidence assessment: 3
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15:28:39 This ends the second assignment.
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RESPONSE --> ok confidence assessment: 3
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