#$&* course Phy 201 documentshort description of contentwhat you'll know when you're doneVolumesreviews the meaning, reasoning and calculation of volumes of some common geometric figuresthe meaning of volume, reasoning about volumes, some important formulas for volumes
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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): Self-critiqueRating: ********************************************* 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: 48 * 2 = 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): Self-critiqueRating: ********************************************* 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: 20 * 40 = 800m^3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aV = A * h applies to uniform cylinders as well as to rectangular solids.We are given the altitude h and the base area A so we conclude that V = A * h = 20 m^2 * 40 m = 800 m^3. The relationship V = A * h applies to any solid object whose cross-sectional area A is constant.This is the case for uniform cylinders and uniform prisms. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critiqueRating: ********************************************* Question:`q004.What is the volume of a uniform cylinder whose base has radius 5 cm and whose altitude is 30 cm? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Pi * 5^2 = 25pi 25pi * 30 = 750pi 750pi = 2356.2cm^3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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( 5cm)^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. STUDENT QUESTION I should have stated that my answer was an approximate. ???? When using pi, should I calculate this out or just leave pi in the solution? INSTRUCTOR RESPONSE I would say to do both when in doubt. If the given dimensions are known to be approximate, and when the numbers aren't simple in the first place, it's appropriate to just multiply everything out and use an appropriate number of significant figures. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critiqueRating: ********************************************* Question:`q005.Estimate the dimensions of a metal can containing food. What is its volume, as indicated by your estimates? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Estimes: Altitude = 6 inches Diameter = 3 inches Pi * r^2 Pi * 1.5^2 = 2.25pi 2.25pi * 6 = 13.5pi 13.5pi = 42.41 inches^3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aPeople will commonly estimate the dimensions of a can of food in centimeters or in inches, though other units of measure are possible (e.g., millimeters, feet, meters, miles, km).Different cans have different dimensions, and your estimate will depend a lot on what can you are using. A typical can might have a circular cross-section with diameter 3 inches and altitude 5 inches.This can would have volume V = A * h, where A is the area of the cross-section.The diameter of the cross-section is 3 inches so its radius will be 3/2 in..The cross-sectional area is therefore A = pi r^2 = pi * (3/2 in)^2 = 9 pi / 4 in^2 and its volume is V = A * h = (9 pi / 4) in^2 * 5 in = 45 pi / 4 in^3. Approximating, this comes out to around 35 in^3. Another can around the same size might have diameter 8 cm and height 14 cm, giving it cross-sectional area A = pi ( 4 cm)^2 = 16 pi cm^2 and volume V = A * h = 16 pi cm^2 * 14 cm = 224 pi cm^3. STUDENT QUESTION Should my in^3 come after the total solution even though it is associated with the 9? As in your example the in^3 is associated with 224 but you have it at the end of the solution. INSTRUCTOR RESPONSE I wouldn't be picky at this point of the course, but the generally used order has the numbers first and the units last. This is what most readers will expect. It's a lot like using good grammar, which makes everything easier to understand. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): My guesses were actually really close Self-critiqueRating: ********************************************* Question:`q006.What is the volume of a pyramid whose base area is 50 square cm and whose altitude is 60 cm? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 50 * 60 * 1/3 = 1,000cm^3 confidence rating #$&*: 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 50 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): Self-critiqueRating: ********************************************* 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: 1/3 * 20 * 9 = 60m^3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aJust as the volume of a pyramid is 1/3 the volume of the 'box' that contains it, the volume of a cone is 1/3 the volume of the cylinder that contains it.Specifically, the cylinder that contains the cone has the base of the cone as its base and matches the altitude of the cone.So the volume of the cone is 1/3 A * h, where A is the area of the base and h is the altitude of the cone. In this case the base area and altitude are given, so the volume of the cone is V = 1/3 A * h = 1/3 * 20 m^2 * 9 m = 60 m^3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critiqueRating: ********************************************* Question:`q008.What is a volume of a sphere whose radius is 4 meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4/3 * pi * r^3 4/3pi * 4^3 4/3pi * 64 = 268.1m^3 confidence rating #$&*: 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.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): Self-critiqueRating: ********************************************* Question:`q009.What is the volume of a planet whose diameter is 14,000 km? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 14,000 / 2 = 7,000 4/3pi * 7000^3 = 1,436,755,040,241km^3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe planet is presumably a sphere, so to the extent that this is so the volume of this planet is V = 4/3 pi r^3, where r is the radius of the planet.The diameter of the planet is 14,000 km so the radius is half this, or 7,000 km.It follows that the volume of the planet is V = 4/3 pi r^3 = 4/3 pi * (7,000 km)^3 = 4/3 pi * 343,000,000,000 km^3 = 1,372,000,000,000 / 3 * pi km^3. This result can be approximated to an appropriate number of significant figures. STUDENT QUESTION How did we go from 343,000,000,000 to 1,372,000,000,000? INSTRUCTOR RESPONSE We go from 4/3 pi * 343,000,000,000 to 1,372,000,000,000 / 3 * pi by multiplying 343 000 000 000 by 4. Like a lot of thing, this is fairly obvious once you see it, hard to see until you do. Let me know if after thinking about it for a few minutes, then if necessary giving it a rest for awhile (say, a day) and coming back to it, you don't see it. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I guess I didn’t leave it in a very simple number times pi Self-critiqueRating: ********************************************* Question:`q010.Summary Question 1:What basic principle do we apply to find the volume of a uniform cylinder of known dimensions? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We first find the area of a circle (the base) and then multiply it by the height. Imagine the areas of all those circles stacked up all the way to the height of the cylinder. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Self-critiqueRating: ********************************************* Question:`q011.Summary Question 2:What basic principle do we apply to find the volume of a pyramid or a cone? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because of their “triangular” shapes, we multiply the volumes of “would be cylinders or boxes” by 1/3. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe volumes of these solids are each 1/3 the volume of the enclosing figure.Each volume can be expressed as V = 1/3 A * h, where A is the area of the base and h the altitude as measured perpendicular to the base. STUDENT QUESTION I thought I had the right idea but I got lost. I’m not sure how to handle the square roots, even after reading the solution, I am confused about this one. INSTRUCTOR RESPONSE Think of a simple example, the equation x^2 = 25. It should be clear that x = 5 is a solution to this equation, as is x = -5. Now 5 is the square root of 25, since 25 is the square of 5. In notation, the same sentence would read 5 = sqrt(25) since 25 = 5^2. So the solutions to this equation are x = sqrt(25) and x = -sqrt(25). We often write that as x = +- sqrt(25), where the '+-' means 'plus or minus'. More generally, if c is any positive number, the equation x^2 = c has solutions x = +- sqrt(c). Now sometimes only one of the two solutions makes sense. In the present problem A radius is a distance, and a distance can't be negative. So after finding the two solutions, we discard the negative solution. However we always find both solutions before discarding everything, in order to make sure we don't throw out something important &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critiqueRating: ********************************************* Question:`q012.Summary Question 3:What is the formula for the volume of a sphere? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: V = 4/3 * pi * r^3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Self-critiqueRating: ********************************************* Question:`q013.Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I honestly didn’t know all these volume formulas right away and I did search for most of them, and I don’t look at the answers before I give my own answer to the questions. These formulas can be written down in the notebook, they will always be readily available. Self-critiqueRating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
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Given Solution: ** Radius 2.8 * 10^4 cm means that the radius is between 2.75 * 10^4 cm and 2.85 * 10^4 cm. We regard 2.75 *10^4 cm as the lower bound and 2.85 *10^4 cm as the upper bound on the radius. 2.75 is .05 less than 2.8, and 2.85 is .05 greater than 2.8, so we say that the actual number is 2.8 +- .05. • Thus we express the actual radius as (2.8 +- .05) * 10^4 cm, and we call .05 * 10^4 cm theuncertaintyin the measurement. The area of a circle is pi r^2, with which you should be familiar (if for no reason other than that you used it and wrote it down in the orientation exercises).. With this uncertainty estimate, we find thatthe area is betweenalower area estimateofpi * (2.75 * 10^4 cm)^2 = 2.376 * 10^9 cm^2 andandupper area estimateofpi * (2.85 * 10^4 cm)^2 = 2.552 * 10^9 cm^2. • Thedifferencebetween the lower and upper estimateis .176 * 10^9 cm^2 = 1.76 * 10^8 cm^2. • Theareawe would get from thegiven radiusis abouthalfway betweenthese estimates, so theuncertaintyin theareais abouthalf of the difference. • We therefore say that theuncertainty in areais about 1/2 * 1.76 * 10^8 cm^2, or about .88 * 10^8 cm^2. Note that the .05* 10^4 cm uncertainty in radius is about2% of the radius, whilethe .88* 10^8cm uncertainty in area is about4% of the area. • Theareaof a circleis proportional to thesquared radius. • Asmall percent uncertaintyin theradiusgives very nearlydoublethepercent uncertaintyin thesquared radius. ** STUDENT COMMENT: I don't recall seeing any problems like this in any of our readings or assignments to this point INSTRUCTOR RESPONSE: The idea of percent uncertainty is presented in Chapter 1 of your text. The formula for the area of a circle should be familiar. Of course it isn't a trivial matter to put these ideas together. STUDENT COMMENT: I don't understand the solution. How does .176 * 10^9 become 1.76 * 10^8? I understand that there is a margin of error because of the significant figure difference, but don't see how this was calculated. INSTRUCTOR RESPONSE: .176 = 1.76 * .1, or 1.76 * 10^-1. So .176 * 10^9 = 1.76 * 10^-1 * 10^9. Since 10^-1 * 10^9 = 10^(9 - 1) =10^8, we have • .176 * 10^9 = 1.76 * 10^8. The key thing to understand is the first statement of the given solution: • Radius 2.8 * 10^4 cm means that the radius is between 2.75 * 10^4 cm and 2.85 * 10^4 cm. This is because any number between 2.75 and 2.85roundsto 2.8. A number whichroundsto 2.8 can therefore lieanywhere between 2.75 and 2.85. The rest of the solution simply calculates the areas corresponding to these lower and upper bounds on the number 2.8, then calculates the percent difference of the results. STUDENT COMMENT: I understand how squaring the problem increases uncertainty and I understand the concept of a range of uncertainty but I am having trouble figuring out how the range of 2.75 * 10^4 and 2.85*10^4 were established for the initial uncertainties in radius. INSTRUCTOR RESPONSE: The key is the first sentence of the given solution: 'Radius 2.8 * 10^4 cm means that the radius is between 2.75 * 10^4 cm and 2.85 * 10^4 cm.' You know this because you know that any number which is at least 2.75, and less than 2.85, rounds to 2.8. Ignoring the 10^4 for the moment, and concentrating only on the 2.8: Since the given number is 2.8, with only two significant figures, all you know is that when rounded to two significant figures the quantity is 2.8. So all you know is that it's between 2.75 and 2.85. STUDENT QUESTION I honestly didn't consider the fact of uncertainty at all. I misread the problem and thought I was simply solving for area. I'm still not really sure how to determine the degree of uncertainty. INSTRUCTOR RESPONSE Response to Physics 121 student: This topic isn't something critical to your success in the course, but the topic will come up. You're doing excellent work so far, and it might be worth a little time for you to try to reconcile this idea. Consider the given solution, the first part of which is repeated below, with some questions (actually the same question repeated too many times). I'm sure you have limited time so don't try to answer the question for every statement in the given solution, but try to answer at least a few. Then submit a copy of this part of the document. Note that a Physics 201 or 231 student should understand this solution very well, and should seriously consider submitting the following if unsure. This is an example of how to break down a solution phrase by phrase and self-critique in the prescribed manner. ##&* ** Radius 2.8 * 10^4 cm means that the radius is between 2.75 * 10^4 cm and 2.85 * 10^4 cm. Do you understand what this is saying, and why it is so? If not, tell me what you think you understand, what you are pretty sure you don't understand, and what you think you might understand but aren't sure. ##&* We regard 2.75 *10^4 cm as the lower bound and 2.85 *10^4 cm as the upper bound on the radius. 2.75 is .05 less than 2.8, and 2.85 is .05 greater than 2.8, so we say that the actual number is 2.8 +- .05. Do you understand what this is saying, and why it is so? If not, tell me what you think you understand, what you are pretty sure you don't understand, and what you think you might understand but aren't sure. ##&* Thus we express the actual radius as (2.8 +- .05) * 10^4 cm, and we call .05 * 10^4 cm the uncertainty in the measurement. Do you understand what this is saying, and why it is so? If not, tell me what you think you understand, what you are pretty sure you don't understand, and what you think you might understand but aren't sure. ##&* The area of a circle is pi r^2, with which you should be familiar (if for no reason other than that you used it and wrote it down in the orientation exercises). With this uncertainty estimate, we find that the area is between a lower area estimate of pi * (2.75 * 10^4 cm)^2 = 2.376 * 10^9 cm^2 and and upper area estimate of pi * (2.85 * 10^4 cm)^2 = 2.552 * 10^9 cm^2. Do you understand what this is saying, and why it is so? If not, tell me what you think you understand, what you are pretty sure you don't understand, and what you think you might understand but aren't sure. ##&* The difference between the lower and upper estimate is .176 * 10^9 cm^2 = 1.76 * 10^8 cm^2. Do you understand what this is saying, and why it is so? If not, tell me what you think you understand, what you are pretty sure you don't understand, and what you think you might understand but aren't sure. ##&* The area we would get from the given radius is about halfway between these estimates, so the uncertainty in the area is about half of the difference. We therefore say that the uncertainty in area is about 1/2 * 1.76 * 10^8 cm^2, or about .88 * 10^8 cm^2. Note that the .05 * 10^4 cm uncertainty in radius is about 2% of the radius, while the .88 * 10^8 cm uncertainty in area is about 4% of the area. Do you understand what this is saying, and why it is so? If not, tell me what you think you understand, what you are pretty sure you don't understand, and what you think you might understand but aren't sure. ##&* The area of a circle is proportional to the squared radius. A small percent uncertainty in the radius gives very nearly double the percent uncertainty in the squared radius. Do you understand what this is saying, and why it is so? If not, tell me what you think you understand, what you are pretty sure you don't understand, and what you think you might understand but aren't sure. ##&* If you wish you can submit the above series of questions in the usual manner. STUDENT QUESTION I said the uncertainty was .1, which gives me .1 / 2.8 = .4. INSTRUCTOR RESPONSE A measurement of 2.8 can be taken to imply a number between 2.75 and 2.85, which means that the number is 2.8 +- .05 and the uncertainty is .05. This is the convention used in the given solution. (The alternative convention is that 2.8 means a number between 2.7 and 2.9; when in doubt the alternative convention is usually the better choice. This is the convention used in the text. It should be easy to adapt the solution given here to the alternative convention, which yields an uncertainty in area of about 8% as opposed to the 4% obtained here). Using the latter convention, where the uncertainty is estimated to be .1: The uncertainty you calculated would indeed be .04 (.1 / 2.8 is .04, not .4), or 4%. However this would be the percent uncertainty in the radius. The question asked for the uncertainty in the area. Since the calculation of the area involves squaring the radius, the percent uncertainty in area is double the percent uncertainty in radius. This gives us a result of .08 or 8%. The reasons are explained in the given solution. NOTE FOR UNIVERSITY PHYSICS STUDENTS (calculus-based answer): Note the following: A = pi r^2, so the derivative of area with respect to radius is dA/dr = 2 pi r. The differential is therefore dA = 2 pi r dr. Thus an uncertainty `dr in r implies uncertainty `dA = 2 pi r `dr, so that `dA / `dr = 2 pi r `dr / (pi r^2) = 2 `dr / r. `dr / r is the proportional uncertainty in r. We conclude that the uncertainty in A is 2 `dr / r, i.e., double the uncertainty in r. STUDENT QUESTION I looked at this, and not sure if I calculated the uncertainty correctly, as the radius squared yields double the uncertainty. I know where this is in the textbook, and do ok with uncertainty, but this one had me confused a bit. INSTRUCTOR RESPONSE: In terms of calculus, since you are also enrolled in a second-semester calculus class: A = pi r^2 The derivative r^2 with respect to r is 2 r, so the derivative of the area with respect to r is dA / dr = pi * (2 r). If you change r by a small amount `dr, the change in the area is dA / dr * `dr, i.e., rate of change of area with respect to r multiplied by the change in r, which is a good commonsense notion. Thus the change in the area is pi * (2 r) `dr. As a proportion of the original area this is pi ( 2 r) `dr / (pi r^2) = 2 `dr / r. The change in the radius itself was just `dr. As a proportion of the initial radius this is `dr / r. The proportional change in area is 2 `dr / r, compared to the proportional change in radius `dr / r. That is the proportional change in area is double the proportional change in radius. STUDENT COMMENT I used +-.1 instead of using +-.05. I understand why your solution used .05 and will use this method in the future. INSTRUCTOR RESPONSE Either way is OK, depending on your assumptions. When it's possible to assume accurate rounding, then the given solution works. If you aren't sure the rounding is accurate, the method you used is appropriate. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I have no idea where the .05 came from, so I didn’t know how to calculate any uncertainty.