A/S Level Pure Maths Help please
06 February 2003, 09:36 PM
#1
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Anyway, this is plane (
-- sic) old two-dimensional geometry. When I did A level pure maths we did stuff like group theory, conic sections, polar coordinates, permutations and combinations, integration and differentiation (recognizing the substitutions is always the trick in the former).
Surely this is GCSE stuff you're talking about here
[Edited by carl - 6/2/2003 9:40:38 PM]
-- sic) old two-dimensional geometry. When I did A level pure maths we did stuff like group theory, conic sections, polar coordinates, permutations and combinations, integration and differentiation (recognizing the substitutions is always the trick in the former). Surely this is GCSE stuff you're talking about here
[Edited by carl - 6/2/2003 9:40:38 PM]
06 February 2003, 09:54 PM
#2
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Solid ball...
Stop waffling...
To simplify further... how much 1mm diameter wire can you get from a solid ball of 1metre diameter?
No tricks or traps...
[Edited by Mycroft - 6/2/2003 9:57:37 PM]
Stop waffling...
To simplify further... how much 1mm diameter wire can you get from a solid ball of 1metre diameter?
No tricks or traps...
[Edited by Mycroft - 6/2/2003 9:57:37 PM]
06 February 2003, 10:29 PM
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OK. It's like this:
(4/3)*pi*R^3=pi*r^2*h
cancel the pis and note that R=1000r
(4/3)*1000^3*r=h
2/3*(1*10^9)=h [in mm]
h=666,666.7m
Odd how the errors mount up when you do it in decimal with some rounding.
[Edited by carl - 6/2/2003 10:31:54 PM]
(4/3)*pi*R^3=pi*r^2*h
cancel the pis and note that R=1000r
(4/3)*1000^3*r=h
2/3*(1*10^9)=h [in mm]
h=666,666.7m
Odd how the errors mount up when you do it in decimal with some rounding.
[Edited by carl - 6/2/2003 10:31:54 PM]
06 February 2003, 10:53 PM
#4
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It matters not at all... speed, grinding rate any 'variable matters not a jot...
And remember we are in the 17th Century... your solution must be suitable for use then...
There is a simple arithmetical answer and a beautiful answer that can be drawn on the wheel...
[Edited by Mycroft - 6/2/2003 10:59:07 PM]
And remember we are in the 17th Century... your solution must be suitable for use then...
There is a simple arithmetical answer and a beautiful answer that can be drawn on the wheel...
[Edited by Mycroft - 6/2/2003 10:59:07 PM]
06 March 2003, 08:30 PM
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Well it's even easier than the previous problem as it's only two dimensional.
I'm assuming here that the goal is to divide the wheel so that they both use the same amount of stone, i.e. the same volume of stone. Assuming the wheel is of uniform thickness, the goal is then to find the radius s at which half the useable area will have been used up.
Assume total radius of the stone is R and the radius of the hole in the middle is r.
Then the goal is:
pi(s^2-r^2)=(pi/2)(R^2-r^2)
s^2=0.5(R^2-r^2)+r^2
s^2=0.5(R^2+r^2)
s^2=0.5(225+(25/4))
The simplest I can get it is s=(5/4)*sqrt(74)
[Edited by carl - 6/3/2003 8:36:13 PM]
I'm assuming here that the goal is to divide the wheel so that they both use the same amount of stone, i.e. the same volume of stone. Assuming the wheel is of uniform thickness, the goal is then to find the radius s at which half the useable area will have been used up.
Assume total radius of the stone is R and the radius of the hole in the middle is r.
Then the goal is:
pi(s^2-r^2)=(pi/2)(R^2-r^2)
s^2=0.5(R^2-r^2)+r^2
s^2=0.5(R^2+r^2)
s^2=0.5(225+(25/4))
The simplest I can get it is s=(5/4)*sqrt(74)
[Edited by carl - 6/3/2003 8:36:13 PM]
02 June 2003, 03:56 PM
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Question is:
The Square ABCD has sides of length 10cm. Circular arcs of radius 10cm are drawn from centres A, B, C, D, as shown.
Find a) the perimeter of the shaded region
Find b) the area of the shaded region
Here is a simple diagram similar to the one of the paper.
Formula that may help:
area of a sector = 1/2 * r^2 * [T]
length of arc or s = r * [T]
Note: r^2 is r squared and [T] is theta (sp) in radians.
Cheers
The Square ABCD has sides of length 10cm. Circular arcs of radius 10cm are drawn from centres A, B, C, D, as shown.
Find a) the perimeter of the shaded region
Find b) the area of the shaded region
Here is a simple diagram similar to the one of the paper.
Formula that may help:
area of a sector = 1/2 * r^2 * [T]
length of arc or s = r * [T]
Note: r^2 is r squared and [T] is theta (sp) in radians.
Cheers
02 June 2003, 04:23 PM
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Now this may be a test to see if you really know your stuff...
This is easy to answer if you know your Euclidean maths...
twin bi-sectors of any equal radii always create a 30deg or equal division...
so the 4 side will equal 120deg of the circumference...
so pi×D/3 will give you the perimeter length... about 209.44mm
The area is also as simple... but will let you see, now that you have been given a simple lever into the 'thinking', if you can work it out...
This is easy to answer if you know your Euclidean maths...
twin bi-sectors of any equal radii always create a 30deg or equal division...
so the 4 side will equal 120deg of the circumference...
so pi×D/3 will give you the perimeter length... about 209.44mm
The area is also as simple... but will let you see, now that you have been given a simple lever into the 'thinking', if you can work it out...
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02 June 2003, 05:16 PM
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Good Lord... I have already given one answer and a clue as to the 'thinking' behind it...
I can work out the area in my head for crying out loud!
Another clue...
if you 'squared up' that bloated 'square' shape it would I reckon make a squre just over 5.6 cm square... what area would that give you...
I can work out the area in my head for crying out loud!
Another clue...
if you 'squared up' that bloated 'square' shape it would I reckon make a squre just over 5.6 cm square... what area would that give you...
02 June 2003, 05:45 PM
#13
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I have got the answers, but unfortunatly without working and the perimeter is indeed 20.9 cm, the area is 31.5cm^2 which by the look of the 5.6cm bloated square edge, Mycroft has already got. But I personally have not heard of this method although it does work, it must be possible to work it out just with the maths I have been taught so far?
Any clues or alternative methods?
Cheers
Any clues or alternative methods?
Cheers
02 June 2003, 05:56 PM
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That puffed up square shape formed by the use of existing radii will always equal 1/10th the area of the whole circle... 31.4159265358979... sq.cm. depending on the extent of your memory of'Pi'
02 June 2003, 07:44 PM
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As we all know our Euclidean maths so well... not!...
Try this one... the answer should take you 1 minute and you should need NO pocket calculator or computer...
A man wants to buy enough 1mm Diameter wire to make a ball 1metre in diameter... he is very clever and he has found a cunning way to wind this wire so as to have no voids or spaces in the finished 1 metre ball.
How much wire did he buy...
Try this one... the answer should take you 1 minute and you should need NO pocket calculator or computer...
A man wants to buy enough 1mm Diameter wire to make a ball 1metre in diameter... he is very clever and he has found a cunning way to wind this wire so as to have no voids or spaces in the finished 1 metre ball.
How much wire did he buy...
02 June 2003, 09:31 PM
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That puffed up square shape formed by the use of existing radii will always equal 1/10th the area of the whole circle... 31.4159265358979... sq.cm. depending on the extent of your memory of'Pi'
]Actually it can only be 32 cm^2 as your input variables are only to two significant figures
02 June 2003, 09:33 PM
#18
Guess I don't remember as much of my maths as I thought I did!
Maybe I should have a quick browse through some of my old notes to get the brain ticking over again!
02 June 2003, 10:00 PM
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Surely this is GCSE stuff you're talking about here
02 June 2003, 10:07 PM
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To follow in your footsteps Mycroft, I will say that I know how to do it but I'm not going to tell you....
However, volume of a sphere is (4/3)*pi*r^3
volume of a cylinder is pi*r^2*h where h is the height/length/whatever you want to call it.
Substitute in the values and bob's your uncle. Either do it in units of a metre (1mm = 1x10^-3m) or a millimetre (1m =1x1-^3mm) to get your answer.
However, volume of a sphere is (4/3)*pi*r^3
volume of a cylinder is pi*r^2*h where h is the height/length/whatever you want to call it.
Substitute in the values and bob's your uncle. Either do it in units of a metre (1mm = 1x10^-3m) or a millimetre (1m =1x1-^3mm) to get your answer.
02 June 2003, 10:16 PM
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So what is the answer?
How many metres?
NO CALCULATOR/PC
Exact amount... to the final decimal point!
Additional clue... you DO NOT need to know the value of 'pi'... it is not required!!!
[Edited by Mycroft - 6/2/2003 10:21:13 PM]
How many metres?
NO CALCULATOR/PC
Exact amount... to the final decimal point!
Additional clue... you DO NOT need to know the value of 'pi'... it is not required!!!
[Edited by Mycroft - 6/2/2003 10:21:13 PM]
02 June 2003, 10:23 PM
#23
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Well I didn't think it was essential to plug the numbers in 
I'll do the lot in metres.
Volume of sphere is (4/3)*pi*0.5^3=0.524 m^3
To get a bit of 1mm wire the same volume you need
pi*(.5*10^-3)^2*h=0.5
h=636,620m
or about 634 km.
FWIW mass would be a bit over 4 tonnes.
I'll do the lot in metres.
Volume of sphere is (4/3)*pi*0.5^3=0.524 m^3
To get a bit of 1mm wire the same volume you need
pi*(.5*10^-3)^2*h=0.5
h=636,620m
or about 634 km.
FWIW mass would be a bit over 4 tonnes.
02 June 2003, 10:30 PM
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Right!
Simple 'Euclidean' explanation...
Area of circles... as the diameter/Radus increase the area increases at its square, so 1metre is 1,000× the diameter of 1mm so 1,000 sqared is 1,000,000....
The volume of a sphere of a given diameter is 2/3s' that of the same diameter cylinder... so we have 2/3s' of a metre × 1,000,000...
666.666*kms...
[Edited by Mycroft - 6/2/2003 10:36:17 PM]
Simple 'Euclidean' explanation...
Area of circles... as the diameter/Radus increase the area increases at its square, so 1metre is 1,000× the diameter of 1mm so 1,000 sqared is 1,000,000....
The volume of a sphere of a given diameter is 2/3s' that of the same diameter cylinder... so we have 2/3s' of a metre × 1,000,000...
666.666*kms...
[Edited by Mycroft - 6/2/2003 10:36:17 PM]
02 June 2003, 10:45 PM
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That was too easy...
Let's really have ahard one...
2 Smithies in the 17th C clubbed together to buy a grind stone... they would share this wheel in turn... the wheel was 30inches in diameter and it was agreed that when down to 5inches in diameter it would be useless.... at what diameter did the first Smithy have to hand the wheel over to the second?
Let's really have ahard one...
2 Smithies in the 17th C clubbed together to buy a grind stone... they would share this wheel in turn... the wheel was 30inches in diameter and it was agreed that when down to 5inches in diameter it would be useless.... at what diameter did the first Smithy have to hand the wheel over to the second?
02 June 2003, 10:50 PM
#27
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Don't know much about grinding stones. Does the outside turn at a constant angular speed or a constant linear speed? I assume the amount of "grindingness" that it does is a function of the linear speed of the outside of the wheel against the thing to be ground?
