course Mth 158 ?x?????x??????assignment #002
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10:00:58 `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 --> Volume is equal to length * width * height. Volume= (3 cm) * (5 cm) * (7 cm). V= 105 cm ^3. confidence assessment: 2
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10:05:08 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 --> I got 105, but why do he units change from cm to m? self critique assessment: 2
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10:06:48 `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 --> In the previous question, I learned that the volume of any rectangular solid is equal to the area of the base * altitude. V= (48 m^2) * ( 2 m). Volume= 96 m^3 confidence assessment: 2
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10:07:24 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 --> That is exactly how I worked it. self critique assessment: 3
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10:09:53 `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 --> V= base area * altitude. V= (20 m^2) * 40 m. Volume= 800 m^3 confidence assessment: 3
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10:12:01 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. self critique assessment: 3
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10:16:11 `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 --> We first must find the base area. Since the base of a cylinder is a circle, we use the formula A=pi*r^2. A= pi* (5 cm)^2. A= 3.14 * 25 cm^2. Area=78.5 cm^2. Then, we take the base area and multiply it by the altitude. V= (78.5 cm^2) *(30 cm). Volume= 2,355 cm^3. confidence assessment: 2
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10:17:51 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 used the formula V= A*h. Would it be correct to go ahead and work out the problem, or should I have just left it 750 pi cm^3? self critique assessment: 2
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10:23:03 `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 would estimate the can as having dimensions of 8 cm wide and 12 cm high. Using my estimates I would first find the base area. Since the can is 8 cm wide, I would divide 8 by 2 to find the radius. Radius= 4 cm. Then I would use the formula A=pi*r^2. A=pi *4cm^2. A=16 cm^2 pi. Then I use the formula V= A*h to find the volume. V= 16 cm^2 pi * 12 cm. Volume= 192 cm^3 pi confidence assessment: 2
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10:24:12 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 --> I think I did it correctly. I'm still not sure when it is appropriate to leave the answer in terms of pi or to multiply it out. self critique assessment: 2
Also it 's generally easier to connect a number like 25 pi or 9 pi to the information given in the problem. You can recognize 25 as the square of 5, whereas you probably won't recognize any connection between 78.5 (which is 25 pi) and 5.
However when the information given is approximate in nature, or when an approximate solution fits the situation, it's fine to multiply the quantities out. The number of significant figures you use when you do this depends on the situation; usually 3 or 4 figures is sufficient. &#
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10:26:04 `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 --> V= base area * altitude. V= 50 cm^2 ( 60 cm). Voulme= 3,000 cm^3. confidence assessment: 2
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10:30:44 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 --> OK, so you can't use V= A*h to find the volume of a pyramid because the cross- section isn't constant. The visual of a pyramid in a box is helpful, I can see how it would only occupy 1/3 of the volume of the box. self critique assessment: 2
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10:38:34 `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 a triangle with a semicircle on top of it. You can't use V= A*h because the cross-section isn't constant. The cone could be like the pyramid and be reassembled to show that the volume is 1/3 that of a rectangle. V= 1/3* 20 m^2* 9 m. Volume=60 m^3. confidence assessment: 1
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10:39:42 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 was right about the cone and the pyramid having 1/3 the volume of a rectangle. self critique assessment: 3
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10:43:22 `q008. What is a volume of a sphere whose radius is 4 meters?
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RESPONSE --> I know that a sphere is a 3 dimensional circle, but I don't have any idea how to find the volume of one. confidence assessment: 0
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10:44:30 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 --> Well if I knew the formula I could've worked the problem. So the volume of a sphere is 4/3 pi r^3. Ok I can remember that. self critique assessment: 2
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10:50:18 `q009. What is the volume of a planet whose diameter is 14,000 km?
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RESPONSE --> A planet is in the shape of a sphere. First I need to find the radius. 14,000 km / 2 is 7,000 km. Then I use the formula V= 4/3*pi*r^3. V= 4/3*pi*(7,000km)^3. Volume= 4.573 *10^11 km^3 pi confidence assessment: 1
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10:51:36 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 --> I worked the problem correctly. Would it be correct to write the answer in scientific notation? self critique assessment: 2
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10:52:32 `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 --> We use the principle of V= base area * altitude to find the volume of a cylinder. confidence assessment: 3
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10:53:22 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 --> I got it. I know that this formula can only be used when the cross section is constant. self critique assessment: 2
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10:54:34 `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 --> We think about a pyramid or cone having 1/3 the volume of a rectangle, so we use the formula V=1/3 * base area * altitude. confidence assessment: 3
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10:54:44 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 got it. self critique assessment: 3
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10:55:12 `q012. Summary Question 3: What is the formula for the volume of a sphere?
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RESPONSE --> The formula for volume of a sphere is 4/3 pi * r^3. confidence assessment: 3
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10:55:21 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 --> I got it. self critique assessment: 3
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10:55:55 `q013. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> I organized my knowledge based on the formulas for the various shapes. confidence assessment: 3
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