#$&* course Mth 174 6/3 3 003. Misc: Surface Area, Pythagorean Theorem, Density
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Given Solution: `aA rectangular solid has six faces (top, bottom, front, back, left side, right side if you're facing it). The pairs top and bottom, right and left sides, and front-back have identical areas. This solid therefore has two faces with each of the following dimensions: 3 m by 4 m, 3 m by 6 m and 4 m by 6 m, areas 12 m^2, 18 m^2 and 24 m^2. Total area is 2 * 12 m^2 + 2 * 18 m^2 + 2 * 24 m^2 = 108 m^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q002. What is the surface area of the curved sides of a cylinder whose radius is five meters and whose altitude is 12 meters? If the cylinder is closed what is its total surface area? Your solution: The side surface area of a cylinder can be written as a rectangle with a length of the circumference of the circle of its base and top and with a width of the height. So we have S = 2pi*r * h If r = 5 m and h = 12 m S = 10 * pi m * 12m = 120*pi m^2 (surface area with sides) To get surface area with the tops we just add the area of the top and bottom circle: Circle = pi * (5m)^2 = 25pi m^2 (we need to of these) so we have: Total S.A. = 120 * pi m^2 + 50 * pi m^2= 170 pi m^2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The circumference of this cylinder is 2 pi r = 2 pi * 5 m = 10 pi m. If the cylinder was cut by a straight line running up its curved face then unrolled it would form a rectangle whose length and width would be the altitude and the circumference. The area of the curved side is therefore A = circumference * altitude = 10 pi m * 12 m = 120 pi m^2. If the cylinder is closed then it has a top and a bottom, each a circle of radius 5 m with resulting area A = pi r^2 = pi * (5 m)^2 = 25 pi m^2. The total area would then be total area = area of sides + 2 * area of base = 120 pi m^2 + 2 * 25 pi m^2 = 170 pi m^2. Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q003. What is surface area of a sphere of diameter three cm? Your solution: The only way that I have been able to think about the surface area of a sphere in the past is to memorize the formula. The formula is: S = 4 * pi * r^2 S = 4 * pi * (3/2 cm)^2 S = 4 * pi * 9/4 cm^2 S = pi * 9 * pi cm^2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe surface area of a sphere of radius r is A = 4 pi r^2. This sphere has radius 3 cm / 2, and therefore has surface area A = 4 pi r^2 = 4 pi * (3/2 cm)^2 = 9 pi cm^2. NOTE TO STUDENT: While your work on most problems has been good, you left this problem blank and didn't self-critique. You should self-critique here. • For example you should acknowledge having made note of the formula for the surface area of the sphere, which I expect you didn't know before. I expect from your previous answers that you are very capable of applying the formula once you have it, and based on this history you probably wouldn't need to self-critique that aspect of the process. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ???The only way that I know to solve this problem is by brute memory of the formula. Is this pretty much the only way to do this problem, or is there a deeper way to think about the surface area of a sphere??? ------------------------------------------------ Self-critique Rating: 3
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Given Solution: `aThe Pythagorean Theorem says that the hypotenuse c of a right triangle with legs a and b satisfies the equation c^2 = a^2 + b^2. So, since all lengths are positive, we know that c = sqrt(a^2 + b^2) = sqrt( (5 m)^2 + (9 m)^2 ) = sqrt( 25 m^2 + 81 m^2) = sqrt( 106 m^2 ) = 10.3 m, approx.. Note that this is not what we would get if we made the common error of assuming that sqrt(a^2 + b^2) = a + b; this would tell us that the hypotenuse is 14 m, which is emphatically not so. There is no justification whatsoever for applying a distributive law (like x * ( y + z) = x * y + x * z ) to the square root operator. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q005. If the hypotenuse of a right triangle has length 6 meters and one of its legs has length 4 meters what is the length of the other leg? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using the same as above: a^2 + b^2 = c^2 16 m^2 + b^2 = 36 m^2 b^2 = 20 m^2 b = sqrt(20) meters = approximately 4.5 meters confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIf c is the hypotenuse and a and b the legs, we know by the Pythagorean Theorem that c^2 = a^2 + b^2, so that a^2 = c^2 - b^2. Knowing the hypotenuse c = 6 m and the side b = 4 m we therefore find the unknown leg: a = sqrt( c^2 - b^2) = sqrt( (6 m)^2 - (4 m)^2 ) = sqrt(36 m^2 - 16 m^2) = sqrt(20 m^2) = sqrt(20) * sqrt(m^2) = 2 sqrt(5) m, or approximately 4.4 m. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q006. If a rectangular solid made of a uniform, homogeneous material has dimensions 4 cm by 7 cm by 12 cm and if its mass is 700 grams then what is its density in grams per cubic cm? Your solution: Using knowledge in past courses, density = mass / volume Volume = L * W * H = 7cm * 4cm * 12cm = 336 cm^3 or 336 mL Density = 700 g / 336 mL = approximately 2.08 g / mL confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe volume of this solid is 4 cm * 7 cm * 12 cm = 336 cm^3. Its density in grams per cm^3 is the number of grams in each cm^3. We find this quantity by dividing the number of grams by the number of cm^3. We find that • density = 700 grams / (336 cm^3) = 2.06 grams / cm^3. Note that the solid was said to be uniform and homogeneous, meaning that it's all made of the same material, which is uniformly distributed. So each cm^3 does indeed have a mass of 2.06 grams. • Had we not known that the material was uniform and homogeneous we could have said that the average density is 2.06 grams / cm^3, but not thatthe density is 2.06 grams / cm^3 (for example the object could be made of two separate substances, one with density less than 2.06 grams / cm^3 and the other with density greater than 2.06 g / cm^3, in appropriate proportions; neither substance would have density 2.06 g / cm^3, but the average density could be 2.06 g / cm^3). NOTE TO STUDENT: (in this note the instructor attempts to clarify the idea of 'demonstrating what you do and do not understand about the statement of the problem' and 'giving a phrase-by-phrase analysis of the given solution') You did not respond to the question and did not self-critique. You would be expected to address the question, stating what you do and do not understand. • For example you should understand what a rectangular solid with dimensions 4 cm by 7 cm by 12 cm is, and how to find its volume and surface area. You might not know what to do with this information (for example you might well not understand that it's the volume and not the surface area that's related to density), but from previous work you should understand this much, and should at least mention something along the lines of 'well, I do know that I can find the volume and/or surface area of that solid' in a partial solution. • The word 'density' is clearly very important. Even if you don't know what density is, you could note from the statement of the problem that its units here are said to be 'grams per cubic centimeter'. Having noted these things, you will be much better prepared to understand the information in the given solution. Then you need to address the information in the given solution. A 'phrase-by-phrase' analysis is generally very beneficial: • I expect you understand the first statement from previous knowledge (you should have this understanding from prerequisite courses, and if not you encountered it in the preceding 'volumes' exercise): 'The volume of this solid is 4 cm * 7 cm * 12 cm = 336 cm^3.' It would of course be appropriate to ask a question here if necessary. • It is likely that, as is the case with many students, the concept of density is not that familiar to you. However if this wasn't addressed specifically in prerequisite courses, those courses would be expected to prepare you to understand this concept. The statement 'Its density in grams per cm^3 is the number of grams in each cm^3.' serves as a definition of density. In your self-critique you should have addressed what what this phrase means to you, and what you do or do not understand about it • The next phrase is 'We find this quantity by dividing the number of grams by the number of cm^3.' You would be expected to understand that this phrase is related to the preceding, and as best you can to address the connection. At this point many students would need to ask a question, and it would be perfectly appropriate to do so (or to have done so regarding previous statements). • The subsequent phrase 'density = 700 grams / (336 cm^3) = 2.06 grams / cm^3' is an illustration of the ideas and definitions in the preceding statements. A reasonable self-critique would demonstrate your attempt to understand this statement and its connection to the preceding. Once again questions would also be appropriate and welcome. • The above addresses sufficient information to solve the problem. If you get to this point, you're probably doing OK and you wouldn't necessarily be expected to address the rest of the given solution, which expands on the finer details of the problem and provides additional information. The basic prerequisite courses should have prepared you to understand the information, but students entering Liberal Arts Mathematics, College Algebra and even Precalculus or Applied Calculus (or Physics 121-122) courses probably don't need to address anything beyond the basic solution at this point. Though Precalculus and Applied Calculus students could benefit from doing so, and if time permits would certainly be encouraged to do so, time is also a factor and it would be understandable if these students chose to move on. • Students entering the Mth 173-4 sequence or the Phy 201-202 or 231-232 sequence would be expected to either completely understand all the details of the given solution, or address them in your self-critique. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I do understand the solution to this problem, mostly because of my prerequisite knowledge in chemistry and combining concepts in calculus. I have noted that the substance must be uniformly distributed to be called the exact density. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q007. What is the mass of a sphere of radius 4 meters if its average density is 3,000 kg/cubic meter? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using the density equation again, we will just manipulate it here. Mass = D * V V = (4/3) * pi * r^3 = (4/3) * pi * 64 m^3 = (256/3) * pi m^3 Mass = (3000 kg/m^3) * ((256/3) * pi m^3) = 256,000 * pi kg confidence rating #$&*:: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aA average density of 3000 kg / cubic meter implies that, at least on the average, every cubic meter has a mass of 3000 kg. So to find the mass of the sphere we multiply the number of cubic meters by 3000 kg. The volume of a sphere of radius 4 meters is 4/3 pi r^3 = 4/3 * pi (4m)^3 = 256/3 * pi m^3. So the mass of this sphere is mass = density * volume = 256 / 3 * pi m^3 * 3000 kg / m^3 = 256,000 * pi kg. This result can be approximated to an appropriate number of significant figures. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q008. If we build an object out of two pieces of material, one having a volume of 6 cm^3 at a density of 4 grams per cm^3 and another with a volume of 10 cm^3 at a density of 2 grams per cm^3 then what is the average density of this object? Your solution: We are given volumes and densities of two materials. To find the average density, I will find the grams of each, add them up and then divide by the sum of the volumes. I have written it as mL instead of cm^3 because it is easier to type mL. 6mL * (4g/mL) = 24 g material one 10mL * (2g/mL) = 20 g material two 24g * 20g = 44g total 6mL + 10mL = 16mL total 44g / 16mL = approximately 2.8 g / mL or 2.8 g / cm^3 (to two sig figs.) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe first piece has a mass of 4 grams / cm^3 * 6 cm^3 = 24 grams. The second has a mass of 2 grams / cm^3 * 10 cm^3 = 20 grams. So the total mass is 24 grams + 20 grams = 44 grams. The average density of this object is average density = total mass / total volume = (24 grams + 20 grams) / (6 cm^3 + 10 cm^3) = 44 grams / (16 cm^3) = 2.75 grams / cm^3. Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `q009. In a large box of dimension 2 meters by 3 meters by 5 meters we place 27 cubic meters of sand whose density is 2100 kg/cubic meter, surrounding a total of three cubic meters of cannon balls whose density is 8,000 kg per cubic meter. What is the average density of the material in the box? Your solution: Using the same concept as above we have: 2100 kg/m^3 * 27 m^3 = 56,700 kg (sand) 8000 kg/m^3 * 3 m^3 = 24,000 kg (cannon balls) 56,700kg + 24,000kg = 80,700kg (total mass) 27 m^3 + 3 m^3 = 30 m^3 Avg. Density = 80,700 kg / 30 m^3 = approximately 2690 kg/m^3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aWe find the average density from the total mass and the total volume. The mass of the sand is 27 m^3 * 2100 kg / m^3 = 56,700 kg. The mass of the cannonballs is 3 m^3 * 8,000 kg / m^3 = 24,000 kg. The average density is therefore average density = total mass / total volume = (56,700 kg + 24,000 kg) / (27 m^3 + 3 m^3) = 80,700 kg / (30 m^3) = 2,700 kg / m^3, approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ??? I have followed similar steps in my solution; however I got 2690 when I divided 80,700 kg by 30 m^3. Is this a sig fig error on my part???
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Given Solution: `aThe volume of the slick is V = A * h, where A is the area of the slick and h the thickness. This is the same principle used to find the volume of a cylinder or a rectangular solid. We see that the volume is V = A * h = 1,700,000 m^2 * .015 m = 25,500 m^3. The mass of the slick is therefore mass = density * volume = 860 kg / m^3 * 25,500 m^3 = 21 930 000 kg. This result should be rounded according to the number of significant figures in the given information. STUDENT QUESTION I didn’t round to the most significant figure. ???? How important is this? INSTRUCTOR RESPONSE It will be important. This document is preliminary; the issue of significant figures will be addressed more specifically as we move into the course. Right now I just want you to be aware of the general idea. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q011. Part 1 Summary Question 1: How do we find the surface area of a cylinder? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We combine the ideas of the equation of a rectangle which is: A = L * W We then substitute the Length with the circumference of the base circle, and then the width with the height of the cylinder to have arrive at: S.A. = 2*pi*r * ( h) (only the side surface area). If the question asks for the total surface area, including the top and bottom, we add 2 times the area base circle. This equation would read: S.A. = 2 pi r h + 2 pi r^2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe curved surface of the cylinder can be 'unrolled' to form a rectangle whose dimensions are equal to the circumference and the altitude of the cylinder, so the curved surface has volume Acurved = circumference * altitude = 2 pi r * h, where r is the radius and h the altitude. The top and bottom of the cylinder are both circles of radius r, each with resulting area pi r^2. {]The total surface area is therefore Acylinder = 2 pi r h + 2 pi r^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q012. Part 1 Summary Question 2: What is the formula for the surface area of a sphere? Your solution: The formula of the surface area of a sphere is, from memory, as follows: S.A. = 4 pi r^2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe surface area of a sphere is A = 4 pi r^2, where r is the radius of the sphere. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q013. Part 1 Summary Question 3: What is the meaning of the term 'density'. Your solution: Density is basically how much mass of a substance is allocated to one unit of volume of that substance. This is the best way I understand it. So the formula can be D = M / V (where M is mass and V is volume). There is a difference between average density, and density because some substances can be more dense in certain regions and less dense in other regions. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Given Solution: `aThe average density of an object is its mass per unit of volume, calculated by dividing its total mass by its total volume. If the object is uniform and homogeneous then its density is constant and we can speak of its 'density' as opposed to its 'average density' &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q014. Part 1 Summary Question 4: If we know average density and mass, how can we find volume? Your solution: If we know the average density and mass, we can find the volume by dividing the average density by that mass. D = M/V…… If we have volume then we can write M/V average density * (1/M) mass…..Mass cancels out and you are left with volume. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since mass = ave density * volume, it follows by simple algebra that volume = mass / ave density. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q015. Part 1 Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I have done the same as with the other assignments similar to this. I have attempted every problem with to the best of my abilities, compared my answer with the given solution and finally study any discrepancies in detail. I have also recorded any concepts that were not familiar to me to begin with in my notes. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q016. The hypotenuse of a right triangle is also the diameter of a certain circle. If the legs of the triangle are 4 feet and 9 feet, what is the area of the circle? Optional question (somewhat challenging): The hypotenuse of a right triangle is also the diameter of a certain circle. The legs of the triangle are alsodiameters of circles, and the areas of those circles are respectively 50 pi and 90 pi. Can you find the area of the largest circle without actually calculating the hypotenuse first? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First Question: a^2 + b^2 = c^2 16 ft^2 + 81 ft^2 = c^2 C^2 = 97 ft^2 c = sqrt(97) ft A = pi * r^2 R = sqrt(97)ft / 2 A = pi * (sqrt(97)/2 ft)^2 A = pi * 97/4 ft^2 = approximately 76.2 ft^2 Optional: Yes, the method that I used in order to find the area of the largest circle is to simply add together the areas of the two smaller circles. This gives: 50 pi + 90 pi = 140 pi This is possible from the following relationship. Let a = diameter of the circle with area 50 pi Let b = diameter of the circle with area 90 pi Let c = diameter of the circle with the unknown area. a^2 + b^2 = c^2 (because of the right triangle relationship) We can manipulate the corresponding diameters in this equation so that they are radii instead of diameters without changing the truth of the equation because we are doing it to every term. (a/2)^2 + (b/2)^2 = (c/2)^2 …We can also multiply through by pi. pi * (a/2)^2 + pi * (b/2)^2 = pi * (c/2)^2 now since pi * (a/2)^2 = 50 pi pi * (b/2)^2 = 90 pi and pi * (c/2)^2 = area of largest circle…….all you have to do is add. 50 pi + 90 pi = 140 pi which is, by the above equation, the area of the largest circle. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Self-critique: ??? I thought long and hard on the optional part of this question. Is my solution what you were looking for???