kedaman
Oct 19th, 2000, 04:40 PM
A sphere with radius R is inside a cone with radius r (small r) and height h.
What's the smallest volume the cone can have?
What's the smallest volume the cone can have?
Click to See Complete Forum and Search --> : For the Math pros!
Oct 19th, 2000, 05:46 PM
Do you mean the volume w/ or w/out the sphere?
w/o
.33333333 * (r ^ 2) * h
w/
(.33333333 * (r ^ 2) * h) - (1.33333 * 3.14 * R^3 )
Oh crap!!! I knew I made a mistake
I hit the wrong button (* instead of ^, and I copied and pasted... so it was wrong on both)
[Edited by denniswrenn on 10-21-2000 at 08:30 AM]
w/o
.33333333 * (r ^ 2) * h
w/
(.33333333 * (r ^ 2) * h) - (1.33333 * 3.14 * R^3 )
Oh crap!!! I knew I made a mistake
I hit the wrong button (* instead of ^, and I copied and pasted... so it was wrong on both)
[Edited by denniswrenn on 10-21-2000 at 08:30 AM]
kedaman
Oct 19th, 2000, 06:34 PM
V=1/3*pi*r^2*h
one third of a cylinder you know
one third of a cylinder you know
kedaman
Oct 19th, 2000, 07:26 PM
Nobody knows? I'm really disappointed. I've worked out a huge complicated derivative and I got the answer, but I had to use my calculators equationsolver to reach that value and i don't think my math teacher likes such things...
HarryW
Oct 19th, 2000, 07:45 PM
Oh, I thought it was just another test. I'll give it some thought.
Oct 21st, 2000, 08:32 AM
Kedaman,
*** do you mean by smallest?
with, or without the frigin sphere in it?
if not... then what?
*** do you mean by smallest?
with, or without the frigin sphere in it?
if not... then what?
kedaman
Oct 21st, 2000, 09:12 AM
The sphere have to fit into the cone :) And if you have a too small or too large radius it will get huge, so the point is to find the optimal radius to have the smallest size of the cone.
Guv
Dec 26th, 2000, 11:00 PM
After fooling around a bit, I could not get any equation in a proper form to determine an analytical solution. The equations I developed were not difficult, merely messy to deal with analytically. They were quite suitable for numerical analysis.
After some computations, I became convinced that the volume of the minimum cone is twice the volume of the embedded sphere. This seems to occur when the ratio (h/r) is equal to approximately 2.828293.
After some computations, I became convinced that the volume of the minimum cone is twice the volume of the embedded sphere. This seems to occur when the ratio (h/r) is equal to approximately 2.828293.
[Digital-X-Treme]
Dec 27th, 2000, 02:08 PM
Guv, that was my problem as well. It seems that you need a value to play with, such as the volume or radius of the cone or sphere to start with. Did u just come up with an expression to show the smallest volume that the cone can have? Will you post the solutions that you worked out?
Later
Later
HarryW
Dec 27th, 2000, 07:31 PM
Gosh, forgot all about this thread. I played around with some of it when it was first posted, but soon found that any expression I wanted to differentiate was pretty complicated, and I wasn't masochistic enough to try and solve it any further :) I don't remember any of it now though. What was the answer then Ked?
Guv
Dec 27th, 2000, 11:24 PM
Digital-X, analysis based on a rough sketch suggested considering an angle (say A). The sketch indicated the following.r = h * Tan(A)
R = (h - R) * sin(A)For a given value of A and R, you can compute h from the second relation, and then r from the first. With values for r and h, you can obviously compute the volume of the cone.
If you assume the largest possible sphere tangent to the base of the cone and entirely inside the cone (Id est: Sphere is tangent to cone at some circular cross-section), the above relations are required. If you consider a plane through the vertex of the cone and perpendicular to the base, the plane and cone intersect in an isosceles triangle. A is the interior angle of this triangle at the vertex of the cone. I hope this describes my diagram to you.
After spending an hour so, I could not come up with a suitable analytical equation.
Using MathCad 7.0 software, I experimented with various values of R and the angle A. It was fairly easy to numerically determine an approximate minimum volume for the cone.
For every combination of R that I tried, A seemed to be a constant, and the volume of the minimum cone seemed to be twice the volume of the sphere. At this point, I got interrupted by higher priority activities. I quickly posted the data about the double volume result.
Further analysis (assuming the above volume relationship) indicated the following.
sin(A) = 1/3
h = 4 * R
r = R * sqr(2) or r^2 = 2* R^2I have not had time to do further analysis.
It looks to me that a proof of the above could be achieved by the following approach. For very large values of h, the cone approaches a cylinder with r = R and a very large volume. For very large values of r, the cone approaches a cylinder with h = 2 * R and a very large volume. Starting with either of the above two extremes, you can vary either r or h continuously to reach the other extreme. Starting at either extreme, the volume initially decreases. Consideration of these transitions is a convincing argument (perhaps not a proof) that a minimum volume exists between these extremes. Now prove that the formulae for h and r as functions of angle A are independent of the value of R, and necessary for the sphere to be inclosed in the cone. It looks easy to formalize this. Now show that for sin(A) = 1/3, the volume of the cone is twice the volume of the sphere. This is easy to formalize. Call this angle "X" Now comes the first tuff part: Show that for angle A = (X + d), the volume of the cone is larger than twice the volume of the sphere. Now prove that for angle A = (X - d), the volume of the cone is also larger than twice the volume of the cone. If you can manage the proof for A = (X + d), this proof should be a minor variation on it. If the last two proofs can be done, you have proved that the minimum volume is twice the volume of the sphere.The above at least proves the existence of a local minium.
If nobody posts a suitable proof, I will try to find time to work on it.
R = (h - R) * sin(A)For a given value of A and R, you can compute h from the second relation, and then r from the first. With values for r and h, you can obviously compute the volume of the cone.
If you assume the largest possible sphere tangent to the base of the cone and entirely inside the cone (Id est: Sphere is tangent to cone at some circular cross-section), the above relations are required. If you consider a plane through the vertex of the cone and perpendicular to the base, the plane and cone intersect in an isosceles triangle. A is the interior angle of this triangle at the vertex of the cone. I hope this describes my diagram to you.
After spending an hour so, I could not come up with a suitable analytical equation.
Using MathCad 7.0 software, I experimented with various values of R and the angle A. It was fairly easy to numerically determine an approximate minimum volume for the cone.
For every combination of R that I tried, A seemed to be a constant, and the volume of the minimum cone seemed to be twice the volume of the sphere. At this point, I got interrupted by higher priority activities. I quickly posted the data about the double volume result.
Further analysis (assuming the above volume relationship) indicated the following.
sin(A) = 1/3
h = 4 * R
r = R * sqr(2) or r^2 = 2* R^2I have not had time to do further analysis.
It looks to me that a proof of the above could be achieved by the following approach. For very large values of h, the cone approaches a cylinder with r = R and a very large volume. For very large values of r, the cone approaches a cylinder with h = 2 * R and a very large volume. Starting with either of the above two extremes, you can vary either r or h continuously to reach the other extreme. Starting at either extreme, the volume initially decreases. Consideration of these transitions is a convincing argument (perhaps not a proof) that a minimum volume exists between these extremes. Now prove that the formulae for h and r as functions of angle A are independent of the value of R, and necessary for the sphere to be inclosed in the cone. It looks easy to formalize this. Now show that for sin(A) = 1/3, the volume of the cone is twice the volume of the sphere. This is easy to formalize. Call this angle "X" Now comes the first tuff part: Show that for angle A = (X + d), the volume of the cone is larger than twice the volume of the sphere. Now prove that for angle A = (X - d), the volume of the cone is also larger than twice the volume of the cone. If you can manage the proof for A = (X + d), this proof should be a minor variation on it. If the last two proofs can be done, you have proved that the minimum volume is twice the volume of the sphere.The above at least proves the existence of a local minium.
If nobody posts a suitable proof, I will try to find time to work on it.
Guv
Dec 28th, 2000, 10:49 PM
I spent some time on the problem of a proof. My original idea was only partially correct. I thought that I would work with slightly different values for an angle. It turns out that working with slightly different values for the height of the cone works. The proof is not formalized, but I believe in it.
The general idea is to pick a value for the height of the cone that results in the volume of the cone being exactly twice the volume of the embedded sphere. Then show that for a slightly larger or smaller value of the height, the volume of the cone is greater than twice the volume of the sphere.
It goes as follows. h = 4 * R leads to r^2 = 2 * R^2. With just a little effort, this can be worked out from h = ( R + R / sin(A) ) and r = h * tan(A). See previous post for these relations. The first leads to sin(A) = 1/3, from which tan(A) can be determined. The above values for h & r result in volume of cone = 8 * Pi * R^3 / 3, twice volume of sphere of radius R. Start with h = ( 4 * R + d ), where d is some small value. This leads to r^2 equal to a messy quotient of two polynomials in R and d. Volume of cone is an even messier quotient of polynomials. Some algebraic effort yields Volume of cone = ( 8 * Pi * R^3 / 3 + a messy quotient). The messy quotient is obviously a positive quantity. Hence for h slightly greater than 4 * R, the volume of the cone increases. Starting with h = ( 4 * R - d ) leads to a similar result.
I do not know how to format all the messy algebra for posting here. If I did know how, it would probably be more effort than I want to spend on this.
Those who are interested should be able to work out the dirty details from the above. If you do not want to work it out, you have to trust me or come up with another proof or counter-proof.
The general idea is to pick a value for the height of the cone that results in the volume of the cone being exactly twice the volume of the embedded sphere. Then show that for a slightly larger or smaller value of the height, the volume of the cone is greater than twice the volume of the sphere.
It goes as follows. h = 4 * R leads to r^2 = 2 * R^2. With just a little effort, this can be worked out from h = ( R + R / sin(A) ) and r = h * tan(A). See previous post for these relations. The first leads to sin(A) = 1/3, from which tan(A) can be determined. The above values for h & r result in volume of cone = 8 * Pi * R^3 / 3, twice volume of sphere of radius R. Start with h = ( 4 * R + d ), where d is some small value. This leads to r^2 equal to a messy quotient of two polynomials in R and d. Volume of cone is an even messier quotient of polynomials. Some algebraic effort yields Volume of cone = ( 8 * Pi * R^3 / 3 + a messy quotient). The messy quotient is obviously a positive quantity. Hence for h slightly greater than 4 * R, the volume of the cone increases. Starting with h = ( 4 * R - d ) leads to a similar result.
I do not know how to format all the messy algebra for posting here. If I did know how, it would probably be more effort than I want to spend on this.
Those who are interested should be able to work out the dirty details from the above. If you do not want to work it out, you have to trust me or come up with another proof or counter-proof.
paulw
Dec 29th, 2000, 06:28 AM
Surely the approach to this should be to define the volume of the cone surrounding the sphere in terms of r (the radius of the sphere) and Ø (theta). The angle at the vertex of the cone. You can then set r to be 1 to eliminate it and differentiate the volume formulae with respect to Ø.
To make it easier, you can reduce this to a 2-D problem by finding the smallest triangle which will contain a circle. Since the triangle will be tangential to the circle at three places, you can create a diagram consisting of three quadrangles which can be further decomposed into six right-angled triangles. Using symmetry, it is only necessary to consider three of these versus the semi-circle.
I will follow this up later on. Any comments?
Cheers,
P.
To make it easier, you can reduce this to a 2-D problem by finding the smallest triangle which will contain a circle. Since the triangle will be tangential to the circle at three places, you can create a diagram consisting of three quadrangles which can be further decomposed into six right-angled triangles. Using symmetry, it is only necessary to consider three of these versus the semi-circle.
I will follow this up later on. Any comments?
Cheers,
P.
HarryW
Dec 29th, 2000, 02:04 PM
I was doing it Paul's way, until I got bored and gave up :)
[Digital-X-Treme]
Dec 29th, 2000, 02:04 PM
Yeah, i see what ya mean with the 2d model. Didn't think of doing things that way... Lets see what else you comes up on this subject... ;)
Laterz
Laterz
Guv
Dec 29th, 2000, 10:45 PM
PaulW, the derivative approach should always (I think) be applicable to this type of problem, but there are other approaches which work. The proof I posted earlier could be formalized into a valid proof and does not use a derivative.
PaulW & Digital-Z, It was not intuitively obvious to me that the 2D & 3D solutions would be the same. After all the volumes are 3rd order functions, while the areas are 2nd order functions. If you get equations for both problems and differentiate, it does not seem guaranteed that the derivatives will have zeros at the same values. Also the volume of a cone is a transcendental number, while the area of a triangle is not.
The above having been said, it seems that the problems are equivalent. I did some numerical calculations which convinced me that if the height of the cone (or triangle) is four times the Radius of the sphere (or circle), the volume of the cone (area of the triangle) is a minimum. In this case, the volume of the cone is exactly twice the volume of the sphere. As might be expected for the 2D problem, the minimum area is not an integer multiple of the area of the circle.
After seeing some recent posts, I decided to try the derivative approach again. The equation I originally obtained using an angle (A) as the variable looked too messy. Angle A is the angle at the vertex of an isosceles triangle which is the intersection of the cone and a plane through the vertex and perpendicular to the base.
Instead of working with angle A as the variable, I introduced a variable X. h = X * R, where h is height of cone, R is radius of sphere, and X is a variable > 2 (required by conditions of the problem). sin(A) = R / (h - R) or sin(A) = R / (X * R - R) or sin(A) = 1 / (X - 1) r = h * tan(A) or r = X * R * tan(A) A little work shows that tan(A) * tan(A) = 1 / (X ^ 2 - 2 * X) r ^ 2 = [ X * R * tan(A) ] ^ 2 or r ^ 2 = (X * R) ^ 2 / (X ^ 2 - 2 * X) Volume of cone, V = Pi * r ^ 2 * h / 3 V = Pi * (X * R) ^ 2 * X * R / 3 * (X ^ 2 - 2 * X) V = Pi * X ^ 2 * R ^ 3 / 3 * (X - 2) derivative D = (Pi * R ^ 3 / 3)[ 2 * X / (X - 2) - X ^ 2 / (X - 2) ^ 2 ] D = 0 implies that 2 * X / (X - 2 ) - X ^ 2 / (X - 2) ^ 2 = 0 Note that X > 2 [ 2 * X * (X - 2) - X ^ 2 / (X - 2) ^ 2 ] = 0 2 * X * (X - 2 ) - X ^ 2 = 0, but X > 2, so X = 0 not allowed. 2 * (X - 2 ) - X = 0 X - 4 = 0 or X = 4 QEDMinimum occurs when height of cone is 4 times the radius of inscribed sphere, and minimum volume of cone is exactly twice the volume of the sphere. This is the same as the result I posted earlier, but the proof is a bit more convincing only because this post includes most of the steps involved.
PaulW & Digital-Z, It was not intuitively obvious to me that the 2D & 3D solutions would be the same. After all the volumes are 3rd order functions, while the areas are 2nd order functions. If you get equations for both problems and differentiate, it does not seem guaranteed that the derivatives will have zeros at the same values. Also the volume of a cone is a transcendental number, while the area of a triangle is not.
The above having been said, it seems that the problems are equivalent. I did some numerical calculations which convinced me that if the height of the cone (or triangle) is four times the Radius of the sphere (or circle), the volume of the cone (area of the triangle) is a minimum. In this case, the volume of the cone is exactly twice the volume of the sphere. As might be expected for the 2D problem, the minimum area is not an integer multiple of the area of the circle.
After seeing some recent posts, I decided to try the derivative approach again. The equation I originally obtained using an angle (A) as the variable looked too messy. Angle A is the angle at the vertex of an isosceles triangle which is the intersection of the cone and a plane through the vertex and perpendicular to the base.
Instead of working with angle A as the variable, I introduced a variable X. h = X * R, where h is height of cone, R is radius of sphere, and X is a variable > 2 (required by conditions of the problem). sin(A) = R / (h - R) or sin(A) = R / (X * R - R) or sin(A) = 1 / (X - 1) r = h * tan(A) or r = X * R * tan(A) A little work shows that tan(A) * tan(A) = 1 / (X ^ 2 - 2 * X) r ^ 2 = [ X * R * tan(A) ] ^ 2 or r ^ 2 = (X * R) ^ 2 / (X ^ 2 - 2 * X) Volume of cone, V = Pi * r ^ 2 * h / 3 V = Pi * (X * R) ^ 2 * X * R / 3 * (X ^ 2 - 2 * X) V = Pi * X ^ 2 * R ^ 3 / 3 * (X - 2) derivative D = (Pi * R ^ 3 / 3)[ 2 * X / (X - 2) - X ^ 2 / (X - 2) ^ 2 ] D = 0 implies that 2 * X / (X - 2 ) - X ^ 2 / (X - 2) ^ 2 = 0 Note that X > 2 [ 2 * X * (X - 2) - X ^ 2 / (X - 2) ^ 2 ] = 0 2 * X * (X - 2 ) - X ^ 2 = 0, but X > 2, so X = 0 not allowed. 2 * (X - 2 ) - X = 0 X - 4 = 0 or X = 4 QEDMinimum occurs when height of cone is 4 times the radius of inscribed sphere, and minimum volume of cone is exactly twice the volume of the sphere. This is the same as the result I posted earlier, but the proof is a bit more convincing only because this post includes most of the steps involved.
[Digital-X-Treme]
Dec 30th, 2000, 07:30 AM
Some very nice work there GUV. Well done.
We should have a mathematics forum on this site, where people could post their maths problems. This would be an excellent idea! What do you think?
Later
We should have a mathematics forum on this site, where people could post their maths problems. This would be an excellent idea! What do you think?
Later
paulw
Jan 2nd, 2001, 07:30 AM
Derivatives (and all calculus) are simply short handmethods for simplifying complicated numerical problems. There are ALWAYS at least two approaches to any computed problem.
The 2D/3D approach is not intuitive, 3d regions are produced by integrating the 2D area through a sweep of 360 degrees - that is how you formally prove that the volume of a sphere is 4/3 pi * r^3 (damn, wish I had formulae available).
The derivative approach should be simpler to compute but I cannot remember the formula for differentiating sin and cos when they are anything but simple.
Cheers,
P.
The 2D/3D approach is not intuitive, 3d regions are produced by integrating the 2D area through a sweep of 360 degrees - that is how you formally prove that the volume of a sphere is 4/3 pi * r^3 (damn, wish I had formulae available).
The derivative approach should be simpler to compute but I cannot remember the formula for differentiating sin and cos when they are anything but simple.
Cheers,
P.
parksie
Jan 2nd, 2001, 09:14 AM
I think for sin and cos you need integration by parts when they get complicated.
kedaman
Jan 2nd, 2001, 03:03 PM
LOL, i've been away a week and what happens to a thread you posted some three months ago hehe. I don't know where my answers ended up, probably at the last place i search, if i start now.
kedaman
Jan 2nd, 2001, 03:23 PM
damn, i found it at the first place. Hmm, i didn't recall i ever read this after the teacher commented it with "How do you know it's a minimum?" and gave me ~1p of 1p.
My derivations are lost though, or if i search enough i might find them.
a= angle between cone base and cone side
R=Circle Radius
r=Cone Base radius
h=Cone height
tan(a)=h/r
tan(a/2)=R/r
h=r*tan(2*atn(R/r))
V=1/3*pi*r^3*tan(2*atn(R/r))=0
V'=pi*r^2*tan(2atn(R/r)-(1+tan^2(2*atn(R/r))*R/(r^2+R*r)=0
R/r=k=sqr(2) <-this is probably some conclusion i don't know how i got, it's probably on the other papers.
-> V=1/3*pi*R^2*4R=8/3*pi*R^3 <-and the answer is definitely correct.
The derivation is just not shown on this paper.
My derivations are lost though, or if i search enough i might find them.
a= angle between cone base and cone side
R=Circle Radius
r=Cone Base radius
h=Cone height
tan(a)=h/r
tan(a/2)=R/r
h=r*tan(2*atn(R/r))
V=1/3*pi*r^3*tan(2*atn(R/r))=0
V'=pi*r^2*tan(2atn(R/r)-(1+tan^2(2*atn(R/r))*R/(r^2+R*r)=0
R/r=k=sqr(2) <-this is probably some conclusion i don't know how i got, it's probably on the other papers.
-> V=1/3*pi*R^2*4R=8/3*pi*R^3 <-and the answer is definitely correct.
The derivation is just not shown on this paper.
kedaman
Jan 2nd, 2001, 03:37 PM
Woohoo! when did this forum show up? Math forum!
parksie
Jan 2nd, 2001, 03:42 PM
People asked John, and Dennis rammed the point home over ICQ :)
So here's our Maths forum! (All we need now is for John to put LaTeX on, and we're done!)
So here's our Maths forum! (All we need now is for John to put LaTeX on, and we're done!)
Guv
Jan 2nd, 2001, 09:06 PM
Kedamen, your derivative looks incorrect. Using "D(...)" to indicate a differential, the following looks correct to me.D(U * V ) = D(U) * V + U * D(V)
Applying that to V = (1 / 3) * Pi * r^3 * tan[ 2 * atan(R / r) ]
Results in D(V) = Pi * r^2 * tan[ 2 * atan(R / r) ] + (1 / 3) * Pi * r^3 * [1 + tan^2(2 * atan(R / r)] * D[2 * atan(R / r) ]When I got to this point, I decided I did not want to try to figure out what the zeros of that mess were.
In fact, I was not even sure that I had the mess correct.
If I expanded the last differential, I felt sure I would make a mistake.
When you set the mess equal to zero, Pi and r^2 can be factored out,
leaving a slightly simpler mess.
That is why I took the approaches posted previously.
If you can fight your way through the above, more power to you.
With the derivative you posted, I do not think you can get the correct result.
Applying that to V = (1 / 3) * Pi * r^3 * tan[ 2 * atan(R / r) ]
Results in D(V) = Pi * r^2 * tan[ 2 * atan(R / r) ] + (1 / 3) * Pi * r^3 * [1 + tan^2(2 * atan(R / r)] * D[2 * atan(R / r) ]When I got to this point, I decided I did not want to try to figure out what the zeros of that mess were.
In fact, I was not even sure that I had the mess correct.
If I expanded the last differential, I felt sure I would make a mistake.
When you set the mess equal to zero, Pi and r^2 can be factored out,
leaving a slightly simpler mess.
That is why I took the approaches posted previously.
If you can fight your way through the above, more power to you.
With the derivative you posted, I do not think you can get the correct result.
kedaman
Jan 3rd, 2001, 09:53 AM
Actually i recall i used my calculator (damn, why do i have to that everytime) to estimate k, my teacher didn't commment the derivate at all, i'm not sure if it's correct though, however using it i got sqr(2) and that was correct. I'll look further into this if i only kind find my papers
VBTZH
Feb 28th, 2002, 01:49 PM
I don't understand this thread at all :rolleyes:
sail3005
Mar 1st, 2002, 12:48 AM
*** you dig up a year old thread to say that?
No one gives a rats ass if you understand it.
No one gives a rats ass if you understand it.
vbforums.com
Copyright 2007 Jupitermedia Corporation All Rights Reserved.
Copyright 2007 Jupitermedia Corporation All Rights Reserved.