Take any large spherical piece of cheese. Then, I cut off a slice of the cheese equal to 1/4 of its circumference. If the weight of the slice is 5 pounds, what is the weight of the cheese? Thanks.
+2440 Well, Mr. Banker, I thought you might like to know I was playing monopoly with my dog and (cheating) cat, when this question popped up. I’ve seen this question before. A student in my Trig study group presented it as a practice question. We all took a stab at it, but none of us knew how to solve it. The student who presented it also had the solution for it, which is near verbatim to what you presented. There is one major problem: the solution is wrong.
I knew it was wrong, not because I am a math whiz –as I said in a previous post, that’s not on my résumé yet, and it certainly wasn’t then. I knew it was wrong because I have an intuitive understanding and a well-developed skill for art – including three-dimensional art. Moreover, and most importantly, I’ve dipped into more than a few cheese bàlls in my life. Cheese bàlls are the only spherical-cheeses I’ve ever seen. I’ve seen many circular cylinders of cheese, including one of six feet in diameter and weighing 1323 lbs., displayed nine years ago in New York City’s Grand Central Market. It was Gouda from Holland.
When I saw your solution, I thought, “Is this the banker? The banker doesn’t show work product, except for simple TVM interest rate problems.” You post as a guest, so it’s not easy to tell. I decided you are the banker for a few reasons.
First, there was your introduction:
I hope you can follow this. You certainly can if you draw up the picture:
Sorry, I couldn't upload the picture.
Then your wrong (and PLAGERIZED) answer kind of gave it away. It was nearly verbatim to the one present by the student in my trig study group. The only difference was the last line gave the approximation of pi as (22/7) which gives the answer as exactly 50lbs.
Finally, you followed your solution with a confirmation of plagiarism:
I think I got it right!!.
It’s always a good idea to make sure the answer you are copying is the correct answer, because you are less likely to get caught (most of us learn that in grade school). In this case, the presentation is too unique to be duplicated by random chance. In case you do not know what plagiarism is, it’s analogous to embezzlement. I’m sure you understand embezzlement. I figured out this is why you haven’t taken my suggestion to choose a user name.
http://web2.0calc.com/questions/hey-guys-i-posted-a-few-questions-and-i-can-t-find-them-anywhere-i-posted-it-yesterday-but-it-s-gone-what-do-i-do
it’s also why you cannot upload images. You have restricted access where you are I understand, it’s a country club by comparison but there are limits.
As I said above, I was playing monopoly with my dog and cat. Normally it takes something important to pull me away from these games. However, it was the cat’s turn to be the banker and the dog always insists on the right to do a bank audit at the end, and, at a time of his choosing, before the end of the game. It’s easy to understand why: the cat not only cheats, he embezzles when it’s his turn to be the banker. He can’t help himself. (I’m thinking, you’d probably like my cat). About the time you posted your reply to Sir Cphill’s answer, the dog called for an audit. This is never pretty, but it’s usually entertaining.
In their younger years they use to “duke-it-out.” They still do, sometimes, but it’s usually a short-lived battle. The dog will snarl at the cat, while the cat hisses and extends his claws. The dog will lunge at the cat but pivot his rump around as the cat swipes with his extended claws. The dog will yelp, as his moderately massive rump plows into the cat, sending him flying across the slippery floor, at which point the cat jumps on top of the china cabinet and waits for the dog to cool off.
Usually, by the time I correct the cash discrepancies and fine the cat 20% of his winnings, the dog will have cooled off enough to let the cat return to the game, so they can trade-in their winnings for Scooby-Snacks and catnip. Not this time, though. He decided to keep the cat on his china cabinet “cell” for a while. Actually, they both just fell asleep -- they are elderly: seventeen for the cat, and fourteen for the dog.
While the dog and cat were napping, I searched for my notes where my mentor explained the correct solution to me. My mentor’s explanation clearly shows why the solution you (and the book) present is wrong.
Hi Ginger,
You are right, the answer is BS. (The author may have smoked a doobie before constructing this solution. In fairness to the author, it would work if it was a circular cylinder.
This solution is incorrect, because the h(eight) is both subtracted from the radius and then squared and multiplied by it. This means the volume is not uniformly proportional over the height. The only time a spherical segment of one base (spherical cap) is proportional to the other is when it’s divided in half.
This formula will give the volume for a spherical cap
\(V_{cap} = \dfrac{3}{4}*(\dfrac{h}{r})^{2} - \dfrac{1}{4}*(\dfrac{h} {r})^{3}*\dfrac{4}{3}(\pi r^{3}) \)
This is derived from an integration of Gauss’s Divergence Theorem
\(V_{cap}=\int_{cap} V = \int_{cap}{1 \over 3}\,\nabla\cdot\vec{r}\, V \int_{sur} \vec{r} \cdot \vec{s} \)
The easiest method to use for solving the spherical cap volume is the formula using the radius and height.
\(\displaystyle V_{cap} = \dfrac{1}{3}\pi h^2 (3r - h)\\ \)
The height can be found using
\(\text {Cap height (h)} = (r) \left( 1- (\cos\frac{\alpha}{2}) \right)\)
The radius is 1 (unit circle) and it’s 90 degrees for the angle corresponding to the chord length of sqrt(2). Divide the volume of the sphere (v=4/3pi*r^3 or 4.1888 for unit radius), by the volume of the cap.
\((1)* \left (1 - \cos \left ( \dfrac {90}{2} \right ) \right ) = 0.29289 \text{ (Cap height)} \\\)
\(\dfrac {1}{3} \pi * (0.29289)^2 * \left ( 3(1) -0.29289 \right ) = 0.24319 \text { (Cap volume)} \\ \)
Sv/Capv = ratio
4.1888/0.2432= 17.2237
Multiply this ratio by the 5 lbs.
5*17.223 = 86.12 lbs
One monster cheese ball, for sure! :)
----------------------------------
Sir CPhill’s method is the same, but it looks like he took a 1/4 of the radius instead of the circumference.
Well Mr. Banker, if you’ve not fallen asleep on (or in) your china cabinet --or even if you have, have yourself a Merry little Christmas.
GA
I hope you can follow this. You certainly can if you draw up the picture:
Sorry, I couldn't upload the picture. Draw a circle, with a square inscribed within the circle. Label the 4 corner of the square, from the top right-hand corner, clockwise: LPSR. And the centre of both as:O, then: Assume the circle containing the diameter of the spherical cheese with the radius OL = 1. Then the area of the circle is:Pi xr^2 = Pi. Let LP represent the straight line cut made, cutting arc LP = 1/4 of the circumference. Then PLRS is an inscribed square: OL = OP = OR = 1. In the right triangle PLR, RP^2 = RL^2 = 2 [LP]^2, and LP = sqrt(2). But the diameter RP = 2, hence 2^2= 2 [LP]^2, or 2 = [LP]^2, and LP = sqrt(2). It follows that the area of the shaded segment, LP = 1/4 [area of the circle - area of square) = 1/4 [Pi -2). Thus, if W equals the weight of cheese, then 1/4 [Pi -2): Pi = 5: W and W = 5 Pi ÷1/4 [Pi -2). Assuming Pi = 3.141592, then W = ~55 pounds.
I think I got it right!!.
+128408 I think that the answer is much larger.......
See the pic below....it represents a cross-section of the sphere.....for our puposes....we'll set the radius of the sphere at R
The cut is made at DE with the radius of the cut = CD = .25R thus ...the circumference of the slice is 1/4 that of the circumference of the whole sphere
The "formula" for the volme of a spherical cap is [ pi * H^2 * (3R - H)] / 3
Where H is the height of the cap = the distance measured along the x axis between "C" and the right edge of the circle = R (1 - √15/4) .........
And the voliume of the cap [ with a little computational help from WolframAlpha ] is :
[((2 π)/3 - (11 sqrt(15) π)/64) R^3]
And the volume of the entire sphere is just (4/3) * pi * (R)^3
So......taking the ratio of weights to volumes......the weight of the whole cheese, W, is given by
W/ 5 = [ (4/3)piR^3] / [((2 π)/3 - (11 sqrt(15) π)/64) R^3] solve for W
W = 5 * [ (4/3)piR^3] / [((2 π)/3 - (11 sqrt(15) π)/64) R^3] ≈ 6682.34 lbs.
As Melody often says ; "That is what I think....."
CPhill: I was of the opinion that the cheese is more of circle, like pizza, rather than a sphere!! No?. They obviously have more thickness to them than a pizza. Besides, as a practical matter, I have never seen or heard of piece of cheese weighing in at more than 3 1/2 tons!!. I think my answer is closer to the true answer in this case!.
+2440 Well, Mr. Banker, I thought you might like to know I was playing monopoly with my dog and (cheating) cat, when this question popped up. I’ve seen this question before. A student in my Trig study group presented it as a practice question. We all took a stab at it, but none of us knew how to solve it. The student who presented it also had the solution for it, which is near verbatim to what you presented. There is one major problem: the solution is wrong.
I knew it was wrong, not because I am a math whiz –as I said in a previous post, that’s not on my résumé yet, and it certainly wasn’t then. I knew it was wrong because I have an intuitive understanding and a well-developed skill for art – including three-dimensional art. Moreover, and most importantly, I’ve dipped into more than a few cheese bàlls in my life. Cheese bàlls are the only spherical-cheeses I’ve ever seen. I’ve seen many circular cylinders of cheese, including one of six feet in diameter and weighing 1323 lbs., displayed nine years ago in New York City’s Grand Central Market. It was Gouda from Holland.
When I saw your solution, I thought, “Is this the banker? The banker doesn’t show work product, except for simple TVM interest rate problems.” You post as a guest, so it’s not easy to tell. I decided you are the banker for a few reasons.
First, there was your introduction:
I hope you can follow this. You certainly can if you draw up the picture:
Sorry, I couldn't upload the picture.
Then your wrong (and PLAGERIZED) answer kind of gave it away. It was nearly verbatim to the one present by the student in my trig study group. The only difference was the last line gave the approximation of pi as (22/7) which gives the answer as exactly 50lbs.
Finally, you followed your solution with a confirmation of plagiarism:
I think I got it right!!.
It’s always a good idea to make sure the answer you are copying is the correct answer, because you are less likely to get caught (most of us learn that in grade school). In this case, the presentation is too unique to be duplicated by random chance. In case you do not know what plagiarism is, it’s analogous to embezzlement. I’m sure you understand embezzlement. I figured out this is why you haven’t taken my suggestion to choose a user name.
http://web2.0calc.com/questions/hey-guys-i-posted-a-few-questions-and-i-can-t-find-them-anywhere-i-posted-it-yesterday-but-it-s-gone-what-do-i-do
it’s also why you cannot upload images. You have restricted access where you are I understand, it’s a country club by comparison but there are limits.
As I said above, I was playing monopoly with my dog and cat. Normally it takes something important to pull me away from these games. However, it was the cat’s turn to be the banker and the dog always insists on the right to do a bank audit at the end, and, at a time of his choosing, before the end of the game. It’s easy to understand why: the cat not only cheats, he embezzles when it’s his turn to be the banker. He can’t help himself. (I’m thinking, you’d probably like my cat). About the time you posted your reply to Sir Cphill’s answer, the dog called for an audit. This is never pretty, but it’s usually entertaining.
In their younger years they use to “duke-it-out.” They still do, sometimes, but it’s usually a short-lived battle. The dog will snarl at the cat, while the cat hisses and extends his claws. The dog will lunge at the cat but pivot his rump around as the cat swipes with his extended claws. The dog will yelp, as his moderately massive rump plows into the cat, sending him flying across the slippery floor, at which point the cat jumps on top of the china cabinet and waits for the dog to cool off.
Usually, by the time I correct the cash discrepancies and fine the cat 20% of his winnings, the dog will have cooled off enough to let the cat return to the game, so they can trade-in their winnings for Scooby-Snacks and catnip. Not this time, though. He decided to keep the cat on his china cabinet “cell” for a while. Actually, they both just fell asleep -- they are elderly: seventeen for the cat, and fourteen for the dog.
While the dog and cat were napping, I searched for my notes where my mentor explained the correct solution to me. My mentor’s explanation clearly shows why the solution you (and the book) present is wrong.
Hi Ginger,
You are right, the answer is BS. (The author may have smoked a doobie before constructing this solution. In fairness to the author, it would work if it was a circular cylinder.
This solution is incorrect, because the h(eight) is both subtracted from the radius and then squared and multiplied by it. This means the volume is not uniformly proportional over the height. The only time a spherical segment of one base (spherical cap) is proportional to the other is when it’s divided in half.
This formula will give the volume for a spherical cap
\(V_{cap} = \dfrac{3}{4}*(\dfrac{h}{r})^{2} - \dfrac{1}{4}*(\dfrac{h} {r})^{3}*\dfrac{4}{3}(\pi r^{3}) \)
This is derived from an integration of Gauss’s Divergence Theorem
\(V_{cap}=\int_{cap} V = \int_{cap}{1 \over 3}\,\nabla\cdot\vec{r}\, V \int_{sur} \vec{r} \cdot \vec{s} \)
The easiest method to use for solving the spherical cap volume is the formula using the radius and height.
\(\displaystyle V_{cap} = \dfrac{1}{3}\pi h^2 (3r - h)\\ \)
The height can be found using
\(\text {Cap height (h)} = (r) \left( 1- (\cos\frac{\alpha}{2}) \right)\)
The radius is 1 (unit circle) and it’s 90 degrees for the angle corresponding to the chord length of sqrt(2). Divide the volume of the sphere (v=4/3pi*r^3 or 4.1888 for unit radius), by the volume of the cap.
\((1)* \left (1 - \cos \left ( \dfrac {90}{2} \right ) \right ) = 0.29289 \text{ (Cap height)} \\\)
\(\dfrac {1}{3} \pi * (0.29289)^2 * \left ( 3(1) -0.29289 \right ) = 0.24319 \text { (Cap volume)} \\ \)
Sv/Capv = ratio
4.1888/0.2432= 17.2237
Multiply this ratio by the 5 lbs.
5*17.223 = 86.12 lbs
One monster cheese ball, for sure! :)
----------------------------------
Sir CPhill’s method is the same, but it looks like he took a 1/4 of the radius instead of the circumference.
Well Mr. Banker, if you’ve not fallen asleep on (or in) your china cabinet --or even if you have, have yourself a Merry little Christmas.
GA
