course Mth 164 1/24 9pmThis work was also submitted on 1/22 but it still has not been posted to my access page. If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `aIf 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. STUDENT QUESTION I guess I am confused at what the length and the width are???? I drew a rectangle I made the top length 5 and the bottom lenghth 7 then the side 3. So the 7 and the 5 are both width and the 3 is the height?????? INSTRUCTOR RESPONSE You can orient this object in any way you choose. The given solution orients it so that the base is 5 cm by 7 cm. The area of the base is then 35 cm^2. In this case the third dimension, 3 cm, is the height and we multiply the area of the base by the height to get 105 cm^3. Had we oriented the object so that it rests on the 3 cm by 5 cm rectangle, the area of the base would be 15 cm^2. The height would be the remaining dimension, 7 cm. Multiplying the base by the height we would be 15 cm^2 * 7 cm = 105 cm^3. We could also orient the object so its base is 3 cm by 7 cm, with area 21 cm^2. Multiplying by the 5 cm height we would again conclude that the volume is 105 cm^3. All these results can be visualized in terms of 1-cm squares and 1-cm cubes, as explained in the given solution. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did not mention the formula V=L*W*h in my solution because I was taught that you use the area and then multiply that by the height but I do understand that it would be the same in this particular problem. ------------------------------------------------ Self-critique rating #$&*: 3 ********************************************* Question: `q002. What is the volume of a rectangular solid whose base area is 48 square meters and whose altitude is 2 meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the volume of a solid rectangle that has a base of 48 m^2 and a height of 2m I would use the formula V=A*h. Therefore the equation would be V=48m^2*2m which would equal 96M^3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aUsing 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&*: ********************************************* Question: `q003. What is the volume of a uniform cylinder whose base area is 20 square meters and whose altitude is 40 meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the volume of a uniform cylinder whose base area is 20m^2 and height is 40m I would use the formula V=A*h. Therefore the equation would be V=20m^2*40m which equals 800m^3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aV = 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating Question: `q004. What is the volume of a uniform cylinder whose base has radius 5 cm and whose altitude is 30 cm? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The volume of a uniform cylinder whose base has a radius of 5cm and a height of 30 cm could be calculated by using the formula V=A*h. In order to use this formula I need to also use a formula to get the area of the cylinders base. This formula would be A=pir^2. When I use A=pir^2 the equation is A=pi(5cm)^2 which equals A=pi25cm^2 or A=25picm^2. Now I can use this information to complete the formula V=A*h where V=25picm^2*30cm which would equal V=750picm^3.
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Given Solution: `aThe 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. STUDENT QUESTION why do we not calculate the pi times the radius and then the height or calculate the pi after the height why do we just leave the pi in the answer? INSTRUCTOR RESPONSE pi cannot be written exactly in decimal form; it's an irrational number and any decimal representation is going to have round-off error. 750 pi cm^3 is the exact volume of a cylinder with radius 5 cm and altitude 30 cm. 750 pi is approximately 2356. However 2356 has two drawbacks: • 2356 is a 4-significant-figure approximation of 750 pi. It's not exact. This might or might not be a disadvantage, but we're better off expressing the result as a multiple of pi, which we can then calculate to any desired degree of precision, than in using 2356, which already contains a roundoff error. • It's hard to look at 2356 and see how it's related to 5 and 30. You probably can't calculate that in your head. However it's not difficult to see that 30 * 5^2 is 30 * 25 or 750. When in doubt, we use the exact expression rather than the approximation. It's fine to give an answer like the following: The volume is 750 pi cm^3, which is approximately 2356 cm^3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&*: ********************************************* Question: `q005. Estimate the dimensions of a metal can containing food. What is its volume, as indicated by your estimates? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To estimate the dimensions of a metal can containing food and then provide the volume I would estimate that most typical food cans are approximately 8cm in diameter and 14cm in height. Based upon these dimension I would use the formula V=A*h to find the volume of the can. To do this I would need to use the cans 8cm diameter to find its base area. A=pir^2 so I need to divide the diameter of the can by two in order to get the radius, this would make the radius 4cm. Now I can find the area [A=pi(4cm)^2] or A=16picm^2. With the area I can now insert this information into the formula for volume to complete the problem. V=A*h would be V=16picm^2*14cm which would equal V= 224picm^3. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aPeople 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&*: ********************************************* Question: `q006. What is the volume of a pyramid whose base area is 50 square cm and whose altitude is 60 cm? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the volume of a pyramid whose base area is 50 cm^2 and height is 60 cm I would use the formula V=1/3*A*h. By using this formula I would simply set the equation up to read V=1/3*50cm^2*60cm. Therefore, the volume of this pyramid would be V=1/3*3000cm^3 which would equal V=1000cm^3. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aWe 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I had never thought about the fact that a pyramid could be considered to be 1/3 of a rectangle. I had really never wondered why the 1/3 was 1/3. It makes sense and helps connect some loose dots for me. ------------------------------------------------ Self-critique rating #$&*: 3 ********************************************* Question: `q007. What is the volume of a cone whose base area is 20 square meters and whose altitude is 9 meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the volume of a cone whose base area is 20m^2 and height is 9m I would use the formula V=1/3A*h. Since we already have all of the components to the formula we would simply set up the equation to read V=1/3*20m^2*9m which would equal V=1/3*180m^3 or V=60m^3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aJust 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&*: ********************************************* Question: `q008. What is a volume of a sphere whose radius is 4 meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I do not know how to calculate the volue of a sphere. I understand what is being asked but I do not know the formula to use to get this measurement. confidence rating #$&*:0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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. STUDENT QUESTION: How does a formula come up with multiplying by pi? I understand how to work a formula, but don’t know how to calculate the formula. Does that make sense? INSTRUCTOR RESPONSE: It makes perfect sense to ask that question. However the answer is beyond the scope of your course. (one answer, which will not make sense to anyone until at least the midway point of their third semester of a challenging calculus sequence, is that the volume of a sphere of radius R is the integral of rho^2 sin (phi) cos(theta) from rho = 0 to R, phi from 0 to pi and theta from 0 to 2 pi; also the surface area of a sphere of radius R is double the double integral of r / secant(theta), integrated in polar coordinates from r = 0 to R and theta from 0 to 2 pi) . (there is another way of figuring this out using solid geometry, a topic with which few students are familiar). In other words, at this point your best recourse is to just learn the formulas. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): If I use the formula V=4/3pir^3 to get the volue of a sphere that has a radius of 4m the formula would be set up to read V=4/3pi(4m)^3 which would be V=4/3pi*4^3m^3. ####In the given solution this is worked out further read 256/3pim^3, I do not understand how to get the 256/3? Could you please explain this to me?#### 4/3 pi * 4^3 = 4/3 pi * 64 = 4 * pi * 64 / 3 = 256 * pi / 3. ------------------------------------------------ Self-critique rating #$&*:2 ********************************************* Question: `q009. What is the volume of a planet whose diameter is 14,000 km? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To calculate the volume of a planet whose diameter is 14,000km I would use the formula V=4/3pir^3. However, in order to use this formula I will need to divide the given diameter by 2 in order to have the radius measurement to use in the volume formula. r=1/2d so r=7,000km. Therefore, V=4/3pi(7,000km)^3 which is V=4/3pi*343,000,000,000km^3. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):####I do not understand how to take V=4/3pi*343,000,000,000km^3 and get to your given solution of 1,372,000,000,000/3*pikm^3. Could you please explain?####
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Given Solution: `aThe 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. STUDENT QUESTION What does it mean “when the cross-section of an object is constant”? When would it not be constant? INSTRUCTOR RESPONSE For example the cross-sectional area of a cone, which tapers, is not constant; nor is the cross-sectional area of a sphere. STUDENT QUESTION And why is altitude measured perpendicular to the cross-section? INSTRUCTOR RESPONSE This is for essentially the same reason the altitude of a parallelogram is measured perpendicular to its base. If you imagine nailing four sticks together to make a rectangle, then imagine partially 'collapsing' the rectangle into a parallelogram, you will see that the altitude of the resulting parallelogram is less than that of the original rectangle, and its area is correspondingly less. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): In my solution I did not mention that we would use a cross-sectional area to obtain the area and the hight or altitude. I also did not mention that altitude is a measure that is perpendicular to the cross-section. However, I do understand these concepts. ------------------------------------------------ Self-critique rating #$&*: 3 ********************************************* Question: `q011. Summary Question 2: What basic principle do we apply to find the volume of a pyramid or a cone? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the volume of a pyramid or a cone I would use the formula V=1/3A*h. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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. STUDENT QUESTION I thought I had the right idea but I got lost. I’m not sure how to handle the square roots, even after reading the solution, I am confused about this one. INSTRUCTOR RESPONSE Think of a simple example, the equation x^2 = 25. It should be clear that x = 5 is a solution to this equation, as is x = -5. Now 5 is the square root of 25, since 25 is the square of 5. In notation, the same sentence would read 5 = sqrt(25) since 25 = 5^2. So the solutions to this equation are x = sqrt(25) and x = -sqrt(25). We often write that as x = +- sqrt(25), where the '+-' means 'plus or minus'. More generally, if c is any positive number, the equation x^2 = c has solutions x = +- sqrt(c). Now sometimes only one of the two solutions makes sense. In the present problem A radius is a distance, and a distance can't be negative. So after finding the two solutions, we discard the negative solution. However we always find both solutions before discarding everything, in order to make sure we don't throw out something important &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&*: ********************************************* Question: `q012. Summary Question 3: What is the formula for the volume of a sphere? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The formula for volume of a sphere is V=4/3pir^3. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did not mention that r represents the radius of the sphere but I do understand that. ------------------------------------------------ Self-critique rating #$&*: 3 ********************************************* Question: `q013. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: I have organized my knowledge of the principles illustrated by the exercises in this assignment by refreshing my memory about volumes and I have taken notes so that I can go back and refer to them if necessary. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok ------------------------------------------------ Self-critique rating #$&*:
course Mth 164 1/24 9pmThis work was also submitted on 1/22 but it still has not been posted to my access page. If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `aIf 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. STUDENT QUESTION I guess I am confused at what the length and the width are???? I drew a rectangle I made the top length 5 and the bottom lenghth 7 then the side 3. So the 7 and the 5 are both width and the 3 is the height?????? INSTRUCTOR RESPONSE You can orient this object in any way you choose. The given solution orients it so that the base is 5 cm by 7 cm. The area of the base is then 35 cm^2. In this case the third dimension, 3 cm, is the height and we multiply the area of the base by the height to get 105 cm^3. Had we oriented the object so that it rests on the 3 cm by 5 cm rectangle, the area of the base would be 15 cm^2. The height would be the remaining dimension, 7 cm. Multiplying the base by the height we would be 15 cm^2 * 7 cm = 105 cm^3. We could also orient the object so its base is 3 cm by 7 cm, with area 21 cm^2. Multiplying by the 5 cm height we would again conclude that the volume is 105 cm^3. All these results can be visualized in terms of 1-cm squares and 1-cm cubes, as explained in the given solution. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did not mention the formula V=L*W*h in my solution because I was taught that you use the area and then multiply that by the height but I do understand that it would be the same in this particular problem. ------------------------------------------------ Self-critique rating #$&*: 3 ********************************************* Question: `q002. What is the volume of a rectangular solid whose base area is 48 square meters and whose altitude is 2 meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the volume of a solid rectangle that has a base of 48 m^2 and a height of 2m I would use the formula V=A*h. Therefore the equation would be V=48m^2*2m which would equal 96M^3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aUsing 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&*: ********************************************* Question: `q003. What is the volume of a uniform cylinder whose base area is 20 square meters and whose altitude is 40 meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the volume of a uniform cylinder whose base area is 20m^2 and height is 40m I would use the formula V=A*h. Therefore the equation would be V=20m^2*40m which equals 800m^3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aV = 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating Question: `q004. What is the volume of a uniform cylinder whose base has radius 5 cm and whose altitude is 30 cm? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The volume of a uniform cylinder whose base has a radius of 5cm and a height of 30 cm could be calculated by using the formula V=A*h. In order to use this formula I need to also use a formula to get the area of the cylinders base. This formula would be A=pir^2. When I use A=pir^2 the equation is A=pi(5cm)^2 which equals A=pi25cm^2 or A=25picm^2. Now I can use this information to complete the formula V=A*h where V=25picm^2*30cm which would equal V=750picm^3.
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Given Solution: `aThe 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. STUDENT QUESTION why do we not calculate the pi times the radius and then the height or calculate the pi after the height why do we just leave the pi in the answer? INSTRUCTOR RESPONSE pi cannot be written exactly in decimal form; it's an irrational number and any decimal representation is going to have round-off error. 750 pi cm^3 is the exact volume of a cylinder with radius 5 cm and altitude 30 cm. 750 pi is approximately 2356. However 2356 has two drawbacks: • 2356 is a 4-significant-figure approximation of 750 pi. It's not exact. This might or might not be a disadvantage, but we're better off expressing the result as a multiple of pi, which we can then calculate to any desired degree of precision, than in using 2356, which already contains a roundoff error. • It's hard to look at 2356 and see how it's related to 5 and 30. You probably can't calculate that in your head. However it's not difficult to see that 30 * 5^2 is 30 * 25 or 750. When in doubt, we use the exact expression rather than the approximation. It's fine to give an answer like the following: The volume is 750 pi cm^3, which is approximately 2356 cm^3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&*: ********************************************* Question: `q005. Estimate the dimensions of a metal can containing food. What is its volume, as indicated by your estimates? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To estimate the dimensions of a metal can containing food and then provide the volume I would estimate that most typical food cans are approximately 8cm in diameter and 14cm in height. Based upon these dimension I would use the formula V=A*h to find the volume of the can. To do this I would need to use the cans 8cm diameter to find its base area. A=pir^2 so I need to divide the diameter of the can by two in order to get the radius, this would make the radius 4cm. Now I can find the area [A=pi(4cm)^2] or A=16picm^2. With the area I can now insert this information into the formula for volume to complete the problem. V=A*h would be V=16picm^2*14cm which would equal V= 224picm^3. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aPeople 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&*: ********************************************* Question: `q006. What is the volume of a pyramid whose base area is 50 square cm and whose altitude is 60 cm? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the volume of a pyramid whose base area is 50 cm^2 and height is 60 cm I would use the formula V=1/3*A*h. By using this formula I would simply set the equation up to read V=1/3*50cm^2*60cm. Therefore, the volume of this pyramid would be V=1/3*3000cm^3 which would equal V=1000cm^3. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aWe 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I had never thought about the fact that a pyramid could be considered to be 1/3 of a rectangle. I had really never wondered why the 1/3 was 1/3. It makes sense and helps connect some loose dots for me. ------------------------------------------------ Self-critique rating #$&*: 3 ********************************************* Question: `q007. What is the volume of a cone whose base area is 20 square meters and whose altitude is 9 meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the volume of a cone whose base area is 20m^2 and height is 9m I would use the formula V=1/3A*h. Since we already have all of the components to the formula we would simply set up the equation to read V=1/3*20m^2*9m which would equal V=1/3*180m^3 or V=60m^3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aJust 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&*: ********************************************* Question: `q008. What is a volume of a sphere whose radius is 4 meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I do not know how to calculate the volue of a sphere. I understand what is being asked but I do not know the formula to use to get this measurement. confidence rating #$&*:0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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. STUDENT QUESTION: How does a formula come up with multiplying by pi? I understand how to work a formula, but don’t know how to calculate the formula. Does that make sense? INSTRUCTOR RESPONSE: It makes perfect sense to ask that question. However the answer is beyond the scope of your course. (one answer, which will not make sense to anyone until at least the midway point of their third semester of a challenging calculus sequence, is that the volume of a sphere of radius R is the integral of rho^2 sin (phi) cos(theta) from rho = 0 to R, phi from 0 to pi and theta from 0 to 2 pi; also the surface area of a sphere of radius R is double the double integral of r / secant(theta), integrated in polar coordinates from r = 0 to R and theta from 0 to 2 pi) . (there is another way of figuring this out using solid geometry, a topic with which few students are familiar). In other words, at this point your best recourse is to just learn the formulas. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): If I use the formula V=4/3pir^3 to get the volue of a sphere that has a radius of 4m the formula would be set up to read V=4/3pi(4m)^3 which would be V=4/3pi*4^3m^3. ####In the given solution this is worked out further read 256/3pim^3, I do not understand how to get the 256/3? Could you please explain this to me?#### 4/3 pi * 4^3 = 4/3 pi * 64 = 4 * pi * 64 / 3 = 256 * pi / 3. ------------------------------------------------ Self-critique rating #$&*:2 ********************************************* Question: `q009. What is the volume of a planet whose diameter is 14,000 km? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To calculate the volume of a planet whose diameter is 14,000km I would use the formula V=4/3pir^3. However, in order to use this formula I will need to divide the given diameter by 2 in order to have the radius measurement to use in the volume formula. r=1/2d so r=7,000km. Therefore, V=4/3pi(7,000km)^3 which is V=4/3pi*343,000,000,000km^3. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):####I do not understand how to take V=4/3pi*343,000,000,000km^3 and get to your given solution of 1,372,000,000,000/3*pikm^3. Could you please explain?####
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Given Solution: `aThe 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. STUDENT QUESTION What does it mean “when the cross-section of an object is constant”? When would it not be constant? INSTRUCTOR RESPONSE For example the cross-sectional area of a cone, which tapers, is not constant; nor is the cross-sectional area of a sphere. STUDENT QUESTION And why is altitude measured perpendicular to the cross-section? INSTRUCTOR RESPONSE This is for essentially the same reason the altitude of a parallelogram is measured perpendicular to its base. If you imagine nailing four sticks together to make a rectangle, then imagine partially 'collapsing' the rectangle into a parallelogram, you will see that the altitude of the resulting parallelogram is less than that of the original rectangle, and its area is correspondingly less. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): In my solution I did not mention that we would use a cross-sectional area to obtain the area and the hight or altitude. I also did not mention that altitude is a measure that is perpendicular to the cross-section. However, I do understand these concepts. ------------------------------------------------ Self-critique rating #$&*: 3 ********************************************* Question: `q011. Summary Question 2: What basic principle do we apply to find the volume of a pyramid or a cone? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the volume of a pyramid or a cone I would use the formula V=1/3A*h. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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. STUDENT QUESTION I thought I had the right idea but I got lost. I’m not sure how to handle the square roots, even after reading the solution, I am confused about this one. INSTRUCTOR RESPONSE Think of a simple example, the equation x^2 = 25. It should be clear that x = 5 is a solution to this equation, as is x = -5. Now 5 is the square root of 25, since 25 is the square of 5. In notation, the same sentence would read 5 = sqrt(25) since 25 = 5^2. So the solutions to this equation are x = sqrt(25) and x = -sqrt(25). We often write that as x = +- sqrt(25), where the '+-' means 'plus or minus'. More generally, if c is any positive number, the equation x^2 = c has solutions x = +- sqrt(c). Now sometimes only one of the two solutions makes sense. In the present problem A radius is a distance, and a distance can't be negative. So after finding the two solutions, we discard the negative solution. However we always find both solutions before discarding everything, in order to make sure we don't throw out something important &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&*: ********************************************* Question: `q012. Summary Question 3: What is the formula for the volume of a sphere? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The formula for volume of a sphere is V=4/3pir^3. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did not mention that r represents the radius of the sphere but I do understand that. ------------------------------------------------ Self-critique rating #$&*: 3 ********************************************* Question: `q013. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: I have organized my knowledge of the principles illustrated by the exercises in this assignment by refreshing my memory about volumes and I have taken notes so that I can go back and refer to them if necessary. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok ------------------------------------------------ Self-critique rating #$&*: