Q. I wonder, how people invented the constant e which often called Euler's number or base of natural logrithm and e is also an irrational number.
For instance, I know pi which is ratio of the circumference of a circle to its diameter and it is also a constant and of course an irrational number.
And Golden mean is also a constant which is ratio of line to its larger segment is equal to ratio of the larger segment to its smaller segment. And you can Golden mean almost everywhere in the nature, i.e. your own body.
WHAT ABOUT e?
I would give the 10 points to whom, who explain it better.
Thanks for answering.
For instance, I know pi which is ratio of the circumference of a circle to its diameter and it is also a constant and of course an irrational number.
And Golden mean is also a constant which is ratio of line to its larger segment is equal to ratio of the larger segment to its smaller segment. And you can Golden mean almost everywhere in the nature, i.e. your own body.
WHAT ABOUT e?
I would give the 10 points to whom, who explain it better.
Thanks for answering.
A. e first came about during John Napier's studies of logarithms, although the first calculations of it appear to have been in the work of Bernoulli:
lim (as n -> inf) of (1 + 1/n)^n
In fact, outside the sciences, the place where e most often occurs naturally is in the financial world.
When interest on an investment is compounded, we talk about the "compounding period": interest can be compounded annually, monthly, daily. The formulas for the change in value of an investment is:
A = P·(1 + r/n)^(nt)
where A = the final amount, P = the principal, r = the annual interest rate, n = the number of compounding periods per year, and t = the number of years.
When money is compounded annually, n = 1; when it's compounded monthly, n = 12; and when it's compounded daily, n = 365 (though sometimes banks use 360, to simplify calculations).
See how n is going up? So what happens if we compound every hour? every minute? every second? In fact, if we "compound continuously," then we're letting n approach infinity, and the function actually becomes
A = P·e^(rt)
There are a couple articles linked below that give more info. Hope that answers your question!
lim (as n -> inf) of (1 + 1/n)^n
In fact, outside the sciences, the place where e most often occurs naturally is in the financial world.
When interest on an investment is compounded, we talk about the "compounding period": interest can be compounded annually, monthly, daily. The formulas for the change in value of an investment is:
A = P·(1 + r/n)^(nt)
where A = the final amount, P = the principal, r = the annual interest rate, n = the number of compounding periods per year, and t = the number of years.
When money is compounded annually, n = 1; when it's compounded monthly, n = 12; and when it's compounded daily, n = 365 (though sometimes banks use 360, to simplify calculations).
See how n is going up? So what happens if we compound every hour? every minute? every second? In fact, if we "compound continuously," then we're letting n approach infinity, and the function actually becomes
A = P·e^(rt)
There are a couple articles linked below that give more info. Hope that answers your question!
It is mathematically possible to calculate the area of a penis?
Q. It is not a rectangle or a cylinder. Using purely mathematical formulas, it is possible to calculate the area?
A. Hello :P
I suppose you mean areal surface area of the penis.
Imagining develop a penis on a plane, with "area of the penis" there refers to the area occupied by the development of the penis on the floor.
Suppose we develop an erect penis so as to prevent the variety that represents the surface is twisted around itself or other problems occur topological [in this case is really the development of an area].
From the mathematical point of life is a penis? We can assume that a subset Ω â IR ³ or a solid.
How can we describe this solid Ω which we calculate the surface area? Given the particularly bizarre form which unfortunately [or fortunately] is not attributable to any known primary function must use one or more functions that approximate the shape.
At the application layer after performing the steps trying to determine "the equation of the penis, or recourse to methods provided by the theory of functions tries to find a function or equation that best approximates the values obtained from the measures.
We hypothesize that the penis has a uniform cylindrical shape, at least for the trunk. Under this hypothesis, intersects the cylinder forming the trunk with a plan will get a circle of radius R. A better approximation can be obtained by considering an elliptical rather than circular cylinder with a director [eccentricity may be more or less pronounced].
We now give some definitions that will be useful later
DEFINITION 1
We define the total length â of the penis the following size
â: = â â + â â
where
â â penis length is measured from the bottom to the base of the glans
â â is the length of the glans
DEFINITION 2
We define the penis in this way
Ω: = Ω â U Ω â
where
Ω â: = ((x, y, z) â IR ³ x ² + y ² = R ², 0 ⤠z ⤠â â) is the trunk of the penis [R is the radius of the circle described previously]
Ω â = ((x, y, z) â IR ³ ...) is the glans
I left because of ellipses describe the glans from the mathematical point of view is a real problem.
As I mentioned previously, it is necessary to find the equation of a surface or a function that approximates the shape.
The first function that comes to mind is the Gauss bell function described by equation
Æ (x, y) = exp (- y ² - x ²)
http://img147.imageshack.us/i/oeb2d8tmp.jpg/
or even the paraboloid of equation
Æ (x, y) = â â - y ² - x ²
http://img147.imageshack.us/i/immaginext.jpg/
If we use this second function to approximate the glans will
Ω â = ((x, y, z) â IR ³: â â ⤠z ⤠â â - y ² - x ²)
Now we have to do some 'healthy crafts.
Regarding the surface area of the trunk will
A_Ω â = 2ÏR â â
We now calculate the surface area of the glans by calculating the surface area underlying the paraboloid
Æ (x, y) = â â - y ² - x ²
To do this we will use a formula that represents the same two variables calcoalo of the length of a curve subtended a function of a variable.
In a variable length of a curve subtended a function Æ (x) is given by the following equation:
L: = INTEGRAL between α & β â (1 + Æ '(x)) dx
where α & β are the endpoints of the curve
In two variables underlying the surface area is a function Æ (x, y) given by the following equation:
A double integral over T = (â (1 + [â / â x [Æ (x, y)]] ² + [â / â y [Æ (x, y) dxdy )]]²)
where T is the set bounded by the edge of the area.
In our case we
T: = ((x, y) â IR ²: x ² + y ² ⤠â â)
Æ (x, y) = â â - y ² - x ²
â / â x [Æ (x, y)] = â / â x [â â - y ² - x ²] = - 2x
â / â y [Æ (x, y)] = â / â y [â â - y ² - x ²] = - 2y
and therefore
A_Ω â = double integral over T (â (1 + [â / â x [Æ (x, y)]] ² + [â / â y [Æ (x, y) dxdy = )]]²)
Double integral over T = (â (1 + (-2x) ² + (-2y) ²)) dxdy =
Double integral over T = (â (1 + 4x ² + 4y ²)) dxdy =
At this point we switch to polar coordinates to simplify the calculations
(X: = Ïcos (θ)
(Y = Ïsen (θ)
T: = ((Ï, θ) â IR ²: 0 â¤ Ï â¤ â (â â), 0 ⤠θ ⤠2Ï)
Recalling that by applying the generic change of coordinates
(X: = Ï (u, v)
(Y = Ï (u, v)
is called "Jacobian of transformation" for the following
J: = |. . . â / â u [Ï (u, v)]. . â / â v [Ï (u, v)]. . . |
. . . . |. . . â / â u [Ï (u, v)]. . â / â v [Ï (u, v)]. . . |
and recalling that by applying the above change of coordinates is the following relation
Double integral on T Æ (x, y) dxdy =
Double integral over T = Æ (Ï (Ï, θ), Ï (Ï, θ)) * det | J | dÏdθ
in our case we
(X = Ï (Ï, θ) = Ïcos (θ)
(Y = Ï (Ï, θ) = Ïsen (θ)
â / â θ [Ïcos (θ)] =-Ïsen (θ)
â / â Ï [Ïcos (θ)] = cos (θ)
â / â θ [Ïsen (θ)] = Ïcos (θ)
â / â Ï [Ïsen (θ)] = sin (θ)
J: = |. . . â / â u [Ï (Ï, θ)]. . â / â v [Ï (Ï, θ)]. . . |
. . . . |. . . â / â u [Ï (Ï, θ)]. . â / â v [Ï (Ï, θ)]. . . |
J = |. . . â / â Ï [Ïcos (θ)]. . â / â θ [Ïcos (θ)]. . |
. . . |. . . â / â Ï [Ïsen (θ)]. . â / â θ [Ïsen (θ)]. . |
J = |. . cos (θ). . -Ρsen (θ). . |
. . . |. . sin (θ). . . Ρcos (θ). . |
det | J | = Ïcos ² (θ) + Ïsen ² (θ) = Ï [cos ² (θ) + sen ² (θ)] = Ï
and therefore
Double integral over T = (â (1 + 4x ² + 4y ²)) dxdy =
Double integral over T = (â (1 + 4 (Ïcos (θ)) ² + 4 (Ïsen (θ)) ²)) = ÏdÏdθ
Double integral over T = (â (1 + 4Ï Â² cos ² (θ) + 4Ï sen ² ² (θ))) = ÏdÏdθ
Double integral over T = (â (1 + 4Ï Â² (cos ² (θ) + sen ² (θ)))) = ÏdÏdθ
Double integral over T = (â (1 + 4Ï Â²)) = ÏdÏdθ
= (INTEGRAL between 0 & 2Ï dθ) * (INTEGRAL between 0 & â (â â) (Ï â (1 + 4Ï Â²))) = dÏ
= ([Î] _calcolato between 0 & 2Ï) * ((1 / 12) â ((1 + 4Ï Â²)³)]_ calculated 0 & â (â â)) =
= (2Ï) * (1 / 12) [â ((1 + 4 (â (â â ))²)³) - â ((1 + 0) ³)] =
= (Î / 6) [â ((1 + 4 â â) ³) - 1]
We then discovered how good the surface area of the glans:
A_Ω â = (Ï / 6) [â ((1 + 4 â â) ³) - 1]
The total area of the surface of the penis will therefore be the sum of the surface area of the trunk and the surface area of the glans
A: = + A_Ω A_Ω â â = 2ÏR â â + (Ï / 6) [â ((1 + 4 â â) ³) - 1]
This final formula which we have just obtained or
A = 2ÏR â â + (Ï / 6) [â ((1 + 4 â â) ³) - 1]
is the formula to calculate the area of a penis
____________________________
Now I do have a question: when do we go from theory to practice? : P
I suppose you mean areal surface area of the penis.
Imagining develop a penis on a plane, with "area of the penis" there refers to the area occupied by the development of the penis on the floor.
Suppose we develop an erect penis so as to prevent the variety that represents the surface is twisted around itself or other problems occur topological [in this case is really the development of an area].
From the mathematical point of life is a penis? We can assume that a subset Ω â IR ³ or a solid.
How can we describe this solid Ω which we calculate the surface area? Given the particularly bizarre form which unfortunately [or fortunately] is not attributable to any known primary function must use one or more functions that approximate the shape.
At the application layer after performing the steps trying to determine "the equation of the penis, or recourse to methods provided by the theory of functions tries to find a function or equation that best approximates the values obtained from the measures.
We hypothesize that the penis has a uniform cylindrical shape, at least for the trunk. Under this hypothesis, intersects the cylinder forming the trunk with a plan will get a circle of radius R. A better approximation can be obtained by considering an elliptical rather than circular cylinder with a director [eccentricity may be more or less pronounced].
We now give some definitions that will be useful later
DEFINITION 1
We define the total length â of the penis the following size
â: = â â + â â
where
â â penis length is measured from the bottom to the base of the glans
â â is the length of the glans
DEFINITION 2
We define the penis in this way
Ω: = Ω â U Ω â
where
Ω â: = ((x, y, z) â IR ³ x ² + y ² = R ², 0 ⤠z ⤠â â) is the trunk of the penis [R is the radius of the circle described previously]
Ω â = ((x, y, z) â IR ³ ...) is the glans
I left because of ellipses describe the glans from the mathematical point of view is a real problem.
As I mentioned previously, it is necessary to find the equation of a surface or a function that approximates the shape.
The first function that comes to mind is the Gauss bell function described by equation
Æ (x, y) = exp (- y ² - x ²)
http://img147.imageshack.us/i/oeb2d8tmp.jpg/
or even the paraboloid of equation
Æ (x, y) = â â - y ² - x ²
http://img147.imageshack.us/i/immaginext.jpg/
If we use this second function to approximate the glans will
Ω â = ((x, y, z) â IR ³: â â ⤠z ⤠â â - y ² - x ²)
Now we have to do some 'healthy crafts.
Regarding the surface area of the trunk will
A_Ω â = 2ÏR â â
We now calculate the surface area of the glans by calculating the surface area underlying the paraboloid
Æ (x, y) = â â - y ² - x ²
To do this we will use a formula that represents the same two variables calcoalo of the length of a curve subtended a function of a variable.
In a variable length of a curve subtended a function Æ (x) is given by the following equation:
L: = INTEGRAL between α & β â (1 + Æ '(x)) dx
where α & β are the endpoints of the curve
In two variables underlying the surface area is a function Æ (x, y) given by the following equation:
A double integral over T = (â (1 + [â / â x [Æ (x, y)]] ² + [â / â y [Æ (x, y) dxdy )]]²)
where T is the set bounded by the edge of the area.
In our case we
T: = ((x, y) â IR ²: x ² + y ² ⤠â â)
Æ (x, y) = â â - y ² - x ²
â / â x [Æ (x, y)] = â / â x [â â - y ² - x ²] = - 2x
â / â y [Æ (x, y)] = â / â y [â â - y ² - x ²] = - 2y
and therefore
A_Ω â = double integral over T (â (1 + [â / â x [Æ (x, y)]] ² + [â / â y [Æ (x, y) dxdy = )]]²)
Double integral over T = (â (1 + (-2x) ² + (-2y) ²)) dxdy =
Double integral over T = (â (1 + 4x ² + 4y ²)) dxdy =
At this point we switch to polar coordinates to simplify the calculations
(X: = Ïcos (θ)
(Y = Ïsen (θ)
T: = ((Ï, θ) â IR ²: 0 â¤ Ï â¤ â (â â), 0 ⤠θ ⤠2Ï)
Recalling that by applying the generic change of coordinates
(X: = Ï (u, v)
(Y = Ï (u, v)
is called "Jacobian of transformation" for the following
J: = |. . . â / â u [Ï (u, v)]. . â / â v [Ï (u, v)]. . . |
. . . . |. . . â / â u [Ï (u, v)]. . â / â v [Ï (u, v)]. . . |
and recalling that by applying the above change of coordinates is the following relation
Double integral on T Æ (x, y) dxdy =
Double integral over T = Æ (Ï (Ï, θ), Ï (Ï, θ)) * det | J | dÏdθ
in our case we
(X = Ï (Ï, θ) = Ïcos (θ)
(Y = Ï (Ï, θ) = Ïsen (θ)
â / â θ [Ïcos (θ)] =-Ïsen (θ)
â / â Ï [Ïcos (θ)] = cos (θ)
â / â θ [Ïsen (θ)] = Ïcos (θ)
â / â Ï [Ïsen (θ)] = sin (θ)
J: = |. . . â / â u [Ï (Ï, θ)]. . â / â v [Ï (Ï, θ)]. . . |
. . . . |. . . â / â u [Ï (Ï, θ)]. . â / â v [Ï (Ï, θ)]. . . |
J = |. . . â / â Ï [Ïcos (θ)]. . â / â θ [Ïcos (θ)]. . |
. . . |. . . â / â Ï [Ïsen (θ)]. . â / â θ [Ïsen (θ)]. . |
J = |. . cos (θ). . -Ρsen (θ). . |
. . . |. . sin (θ). . . Ρcos (θ). . |
det | J | = Ïcos ² (θ) + Ïsen ² (θ) = Ï [cos ² (θ) + sen ² (θ)] = Ï
and therefore
Double integral over T = (â (1 + 4x ² + 4y ²)) dxdy =
Double integral over T = (â (1 + 4 (Ïcos (θ)) ² + 4 (Ïsen (θ)) ²)) = ÏdÏdθ
Double integral over T = (â (1 + 4Ï Â² cos ² (θ) + 4Ï sen ² ² (θ))) = ÏdÏdθ
Double integral over T = (â (1 + 4Ï Â² (cos ² (θ) + sen ² (θ)))) = ÏdÏdθ
Double integral over T = (â (1 + 4Ï Â²)) = ÏdÏdθ
= (INTEGRAL between 0 & 2Ï dθ) * (INTEGRAL between 0 & â (â â) (Ï â (1 + 4Ï Â²))) = dÏ
= ([Î] _calcolato between 0 & 2Ï) * ((1 / 12) â ((1 + 4Ï Â²)³)]_ calculated 0 & â (â â)) =
= (2Ï) * (1 / 12) [â ((1 + 4 (â (â â ))²)³) - â ((1 + 0) ³)] =
= (Î / 6) [â ((1 + 4 â â) ³) - 1]
We then discovered how good the surface area of the glans:
A_Ω â = (Ï / 6) [â ((1 + 4 â â) ³) - 1]
The total area of the surface of the penis will therefore be the sum of the surface area of the trunk and the surface area of the glans
A: = + A_Ω A_Ω â â = 2ÏR â â + (Ï / 6) [â ((1 + 4 â â) ³) - 1]
This final formula which we have just obtained or
A = 2ÏR â â + (Ï / 6) [â ((1 + 4 â â) ³) - 1]
is the formula to calculate the area of a penis
____________________________
Now I do have a question: when do we go from theory to practice? : P
Please doublecheck my physics work on this problem?
Q. I think I got this one right, just want to doublecheck is all. Thanks in advance!
A plane is flying in a loop to simulate weightlessness for a brief instant. If the speed is 215m/s, what is the radius of the circle?
I have since v = sqrt rg where v is velocity, r is the radius, and g is gravity, that it comes out to (215^2)/9.98 = 4631.7635m
Did I do this correctly? If not, please explain what I did wrong and get me on the right track. Thanks!
A plane is flying in a loop to simulate weightlessness for a brief instant. If the speed is 215m/s, what is the radius of the circle?
I have since v = sqrt rg where v is velocity, r is the radius, and g is gravity, that it comes out to (215^2)/9.98 = 4631.7635m
Did I do this correctly? If not, please explain what I did wrong and get me on the right track. Thanks!
A. If the formula is correct, your calculation is correct but you may consider taking the value of g as 9.8 m/sec^2 instead of 9.98
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