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  • derivation of heat equation | planetmath.org
    Reader in sparql request line 92 of home jcorneli beta sites all modules sparql sparql module Primary tabs View active tab Coauthors PDF Source Edit derivation of heat equation Let us consider the heat conduction in a homogeneous matter with density ϱ ϱ varrho and specific heat capacity c c c Denote by u x y z t u x y z t u x y z t the temperature in the point x y z x y z x y z at the time t t t Let a a a be a simple closed surface in the matter and v v v the spatial region restricted by it When the growth of the temperature of a volume element d v d v dv in the time d t d t dt is d u d u du the element releases the amount d u c ϱ d v u t d t c ϱ d v d u c ϱ d v subscript superscript u normal t d t c ϱ d v du c varrho dv u prime t dt c varrho dv of heat which is the heat flux through the surface of d v d v dv Thus if there are no sources and sinks of heat in v v v the heat flux through the surface a a a in d t d t dt is d t v c ϱ u t d v d t subscript v c ϱ subscript superscript u normal t d v displaystyle dt int v c varrho u prime t dv 1 On the other hand the flux through d a d a da in the time d t d t dt must be proportional to a a a to d t d t dt and to the derivative of the temperature in the direction of the normal line of the surface element d a d a da i e the flux is k u d a d t k normal normal u d normal a d t k nabla u cdot d vec a dt where k k k is a positive constant because the heat flows always from higher temperature to lower one Consequently the heat flux through the whole surface a a a is d t a k u d a d t subscript contour integral a k normal normal u d normal a dt oint a k nabla u cdot d vec a which is by the Gauss s theorem same as d t v k u d v d t v k 2 u d v d t subscript v normal k normal normal u d v d t subscript v k superscript normal 2 u d v displaystyle dt int v k nabla cdot nabla u dv dt int v k nabla 2 u dv 2 Equating the expressions 1 and 2 and dividing by d t d t dt one obtains v k 2 u d v v c ϱ

    Original URL path: http://www.planetmath.org/derivationofheatequation (2016-04-25)
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  • derivation of Pappus's centroid theorem | planetmath.org
    org 8890 sparql query 0APREFIX msc 3A 3Chttp 3A 2F 2Fmsc2010 org 2Fresources 2FMSC 2F2010 2F 3E PREFIX skos 3A 3Chttp 3A 2F 2Fwww w3 org 2F2004 2F02 2Fskos 2Fcore 23 3E PREFIX dct 3A 3Chttp 3A 2F 2Fpurl org 2Fdc 2Fterms 2F 3E PREFIX local 3A 3Chttp 3A 2F 2Flocal virt 2F 3E SELECT 3Flabel WHERE 7B GRAPH 3Chttp 3A 2F 2Flocalhost 3A8890 2FDAV 2Fhome 2Fpm 2Frdf sink 23this 3E 7B msc 3A35Q99 skos 3AprefLabel 3Flabel FILTER langMatches 28 lang 28 3Flabel 29 2C 22en 22 29 7D 7D via ARC2 Reader in sparql request line 92 of home jcorneli beta sites all modules sparql sparql module Primary tabs View active tab Coauthors PDF Source Edit derivation of Pappus s centroid theorem I Let s s s denote the arc rotating about the x x x axis and its length and R R R be the y y y coordinate of the centroid of the arc If the arc may be given by the equation y y x y y x y y x where a x b a x b a leq x leq b the area of the formed surface of revolution is A 2 π a b y x 1 y x 2 d x A 2 π superscript subscript a b y x 1 superscript superscript y normal x 2 d x A 2 pi int a b y x sqrt 1 y prime x 2 dx This can be concisely written A 2 π s y d s A 2 π subscript s y d s displaystyle A 2 pi int s y ds 1 since differential geometrically the product 1 y x 2 d x 1 superscript superscript y normal x 2 d x sqrt 1 y prime x 2 dx is the arc element We rewrite 1 as A s 2 π 1 s s y d s A normal normal s 2 π 1 s subscript s y d s A s cdot 2 pi cdot frac 1 s int s y ds Here the last factor is the ordinate of the centroid of the rotating arc whence we have the result A s 2 π R A normal s 2 π R A s cdot 2 pi R which states the first Pappus s centroid theorem II For deriving the second Pappus s centroid theorem we suppose that the region defined by a x b 0 y 1 x y y 2 x formulae sequence a x b 0 subscript y 1 x y subscript y 2 x a leq x leq b quad 0 leq y 1 x leq y leq y 2 x having the area A A A and the centroid with the ordinate R R R rotates about the x x x axis and forms the solid of revolution with the volume V V V The centroid of the area element between the arcs y y 1 x y subscript y 1 x y y 1

    Original URL path: http://www.planetmath.org/derivationofpappusscentroidtheorem (2016-04-25)
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  • derivation of plastic number | planetmath.org
    Edit derivation of plastic number The plastic number may be defined to be the limit of the ratio of two successive members of the Padovan sequence or the Perrin sequence both of which obey the recurrence relation a n a n 3 a n 2 subscript a n subscript a n 3 subscript a n 2 displaystyle a n a n 3 a n 2 1 Supposing that such a limit P lim n a n 1 a n assign P subscript normal n subscript a n 1 subscript a n displaystyle P lim n to infty frac a n 1 a n 2 exists and is 0 absent 0 neq 0 we first write 1 as a n a n 1 a n 1 a n 2 a n 3 a n 2 1 normal subscript a n subscript a n 1 subscript a n 1 subscript a n 2 subscript a n 3 subscript a n 2 1 displaystyle frac a n a n 1 cdot frac a n 1 a n 2 frac a n 3 a n 2 1 3 and then let n normal n n to infty It follows the limit equation P P 1 P 1 normal P P 1 P 1 P cdot P frac 1 P 1 which is same as P 3 P 1 superscript P 3 P 1 displaystyle P 3 P 1 4 Thinking the graphs of the equations y x 3 y superscript x 3 y x 3 and y x 1 y x 1 y x 1 it is clear that the cubic equation x 3 x 1 0 superscript x 3 x 1 0 displaystyle x 3 x 1 0 5 has only one real root which is P P P For solving the plastic number from the cubic substitute by Cardano into 5 the sum x u v assign x u v x u v of two auxiliary unknowns when the equation may be written u 3 v 3 1 3 u v 1 u v 0 superscript u 3 superscript v 3 1 3 u v 1 u v 0 u 3 v 3 1 3uv 1 u v 0 Then as in the example of solving a cubic equation u u u and v v v are determined such that the first two parentheses vanish u 2 v 3 1 u v 1 3 or u 3 v 3 1 27 u2v3 1 uv13or u3v3127 displaystyle begin cases u 2 v 3 1 uv frac 1 3 quad mbox or quad u 3 v 3 frac 1 27 end cases Thus u 3 superscript u 3 u 3 and v 3 superscript v 3 v 3 are the roots of the resolvent equation z 2 z 1 27 0 superscript z 2 z 1 27 0 z 2 z frac 1 27 0 i e z 9 69 18 z plus or minus 9 69 18 z frac 9 pm sqrt 69

    Original URL path: http://www.planetmath.org/derivationofplasticnumber (2016-04-25)
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  • derivation of Pythagorean triples | planetmath.org
    in sparql request line 92 of home jcorneli beta sites all modules sparql sparql module Primary tabs View active tab Coauthors PDF Source Edit derivation of Pythagorean triples For finding all positive solutions of the Diophantine equation x 2 y 2 z 2 superscript x 2 superscript y 2 superscript z 2 displaystyle x 2 y 2 z 2 1 we first can determine such triples x y z x y z x y z which are coprime When these are then multiplied by all positive integers one obtains all positive solutions Let x y z x y z x y z be a solution of the mentioned kind Then the numbers are pairwise coprime since by 1 a common divisor of two of them is also a common divisor of the third Especially x x x and y y y cannot both be even Neither can they both be odd since because the square of any odd number is 1 mod 4 absent annotated 1 pmod 4 equiv 1 mathop rm mod 4 the equation 1 would imply an impossible congruence 2 z 2 mod 4 2 annotated superscript z 2 pmod 4 2 equiv z 2 mathop rm mod 4 Accordingly one of the numbers e g x x x is even and the other y y y odd Write 1 to the form x 2 z y z y superscript x 2 z y z y displaystyle x 2 z y z y 2 Now both factors on the right hand side are even whence one may denote z y 2 u z y 2 v fragments z y normal 2 u normal z y normal 2 v displaystyle z y 2u quad z y 2v 3 giving z u v y u v formulae sequence z u v y u v displaystyle z u v quad y u v 4 and thus 2 reads x 2 4 u v superscript x 2 4 u v displaystyle x 2 4uv 5 Because z z z and y y y are coprime and z y 0 z y 0 z y 0 one can infer from 4 and 3 that also u u u and v v v must be coprime and u v 0 u v 0 u v 0 Therefore it follows from 5 that u m 2 v n 2 formulae sequence u superscript m 2 v superscript n 2 u m 2 quad v n 2 where m m m and n n n are coprime and m n 0 m n 0 m n 0 Thus 5 and 4 yield x 2 m n y m 2 n 2 z m 2 n 2 formulae sequence x 2 m n formulae sequence y superscript m 2 superscript n 2 z superscript m 2 superscript n 2 displaystyle x 2mn quad y m 2 n 2 quad z m 2 n 2 6 Here one of m m m and n n n is

    Original URL path: http://www.planetmath.org/derivationofpythagoreantriples (2016-04-25)
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  • derivation of wave equation | planetmath.org
    derivation of wave equation Let a string of homogeneous matter be tightened between the points x 0 x 0 x 0 and x p x p x p of the x x x axis and let the string be made vibrate in the x y x y xy plane Let the line density of mass of the string be the constant σ σ sigma We suppose that the amplitude of the vibration is so small that the tension T normal T vec T of the string can be regarded to be constant The position of the string may be represented as a function y y x t y y x t y y x t where t t t is the time We consider an element d m d m dm of the string situated on a tiny interval x x d x x x d x x x dx thus its mass is σ d x σ d x sigma dx If the angles the vector T normal T vec T at the ends x x x and x d x x d x x dx of the element forms with the direction of the x x x axis are α α alpha and β β beta then the scalar components of the resultant force F normal F vec F of all forces on d m d m dm the gravitation omitted are F x T cos α T cos β F y T sin α T sin β formulae sequence subscript F x T α T β subscript F y T α T β F x T cos alpha T cos beta quad F y T sin alpha T sin beta Since the angles α α alpha and β β beta are very small the ratio F x F y cos β cos α sin β sin α 2 sin β α 2 sin β α 2 2 sin β α 2 cos β α 2 subscript F x subscript F y β α β α 2 β α 2 β α 2 2 β α 2 β α 2 frac F x F y frac cos beta cos alpha sin beta sin alpha frac 2 sin frac beta alpha 2 sin frac beta alpha 2 2 sin frac beta alpha 2 cos frac beta alpha 2 having the expression tan β α 2 β α 2 tan frac beta alpha 2 also is very small Therefore we can omit the horizontal component F x subscript F x F x and think that the vibration of all elements is strictly vertical Because of the smallness of the angles α α alpha and β β beta their sines in the expression of F y subscript F y F y may be replaced with their tangents and accordingly F y T tan β tan α T y x x d x t y x x t T y x x x t d x subscript F y normal T β

    Original URL path: http://www.planetmath.org/derivationofwaveequation (2016-04-25)
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  • derivative and differentiability of complex function | planetmath.org
    of home jcorneli beta sites all modules sparql sparql module Primary tabs View active tab Coauthors PDF Source Edit derivative and differentiability of complex function Let f z f z f z be given uniquely in a neighborhood of the point z z z in ℂ ℂ mathbb C If the difference quotient Δ f Δ z f z Δ z f z Δ z normal Δ f normal Δ z f z normal Δ z f z normal Δ z frac Delta f Delta z frac f z Delta z f z Delta z tends to a finite limit A A A as Δ z 0 normal normal Δ z 0 Delta z to 0 then A A A is the derivative of f f f at the point z z z and is denoted by f z A lim Δ z 0 Δ f Δ z superscript f normal z A subscript normal normal Δ z 0 normal Δ f normal Δ z displaystyle f prime z A lim Delta z to 0 frac Delta f Delta z 1 Thus the difference λ Δ f Δ z A λ normal Δ f normal Δ z A lambda frac Delta f Delta z A tends to zero simultaneously with Δ z normal Δ z Delta z and Δ f normal Δ f Delta f has the expansion Δ f A Δ z λ Δ z normal Δ f A normal Δ z λ normal Δ z Delta f A Delta z lambda Delta z If we denote Δ z ϱ fragments normal Δ z normal normal ϱ Delta z varrho we have λ Δ z λ Δ z ϱ ϱ ϱ ϱ λ normal Δ z normal λ normal Δ z ϱ ϱ ϱ ϱ lambda Delta z frac

    Original URL path: http://www.planetmath.org/derivativeanddifferentiabilityofcomplexfunction (2016-04-25)
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  • derivative as parameter for solving differential equations | planetmath.org
    p mapsto f prime p are continuous and f p 0 superscript f normal p 0 f prime p neq 0 on an interval p 1 p 2 subscript p 1 subscript p 2 p 1 p 2 It follows that on the interval the function p f p maps to p f p p mapsto f p changes monotonically from f p 1 x 1 assign f subscript p 1 subscript x 1 f p 1 x 1 to f p 2 x 2 assign f subscript p 2 subscript x 2 f p 2 x 2 whence conversely the equation x f p x f p displaystyle x f p 3 defines from p 1 p 2 subscript p 1 subscript p 2 p 1 p 2 onto x 1 x 2 subscript x 1 subscript x 2 x 1 x 2 a bijection p g x p g x displaystyle p g x 4 which is continuously differentiable Thus on the interval x 1 x 2 subscript x 1 subscript x 2 x 1 x 2 the differential equation 2 can be replaced by the equation d y d x g x d y d x g x displaystyle frac dy dx g x 5 and therefore the solution of 2 is y g x d x C y g x d x C displaystyle y int g x dx C 6 If we cannot express g x g x g x in a closed form we take p p p as an independent variable through the substitution 3 which maps x 1 x 2 subscript x 1 subscript x 2 x 1 x 2 bijectively onto p 1 p 2 subscript p 1 subscript p 2 p 1 p 2 Then 6 becomes a function of p p p and by the chain rule d y d p g f p f p p f p d y d p g f p superscript f normal p p superscript f normal p frac dy dp g f p f prime p pf prime p Accordingly the solution of the given differential equation may be presented on p 1 p 2 subscript p 1 subscript p 2 p 1 p 2 as x f p y p f p d p C x fp y pf pdpC displaystyle begin cases displaystyle x f p displaystyle y int p f prime p dp C end cases 7 II With corresponding considerations one can write the solution of the differential equation y f d y d x f p y f d y d x assign f p displaystyle y f frac dy dx f p 8 where p p p changes on some interval p 1 p 2 subscript p 1 subscript p 2 p 1 p 2 where f p f p f p and f p superscript f normal p f prime p are continuous and p f p 0 normal p superscript f normal p

    Original URL path: http://www.planetmath.org/derivativeasparameterforsolvingdifferentialequations (2016-04-25)
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  • derivative for parametric form | planetmath.org
    28 lang 28 3Flabel 29 2C 22en 22 29 7D 7D via ARC2 Reader in sparql request line 92 of home jcorneli beta sites all modules sparql sparql module Primary tabs View active tab Coauthors PDF Source Edit derivative for parametric form Instead of the usual way y f x y f x y f x to present plane curves it is in many cases more comfortable to express both coordinates x x x and y y y by means of a suitable auxiliary variable the parametre It is true e g for the cycloid curve Suppose we have the parametric form x x t y y t formulae sequence x x t y y t displaystyle x x t quad y y t 1 For getting now the derivative d y d x d y d x displaystyle frac dy dx in a point P 0 subscript P 0 P 0 of the curve we chose another point P P P of the curve If the values of the parametre t t t corresponding these points are t 0 subscript t 0 t 0 and t t t we thus have the points x t 0 y t 0 x subscript t 0 y subscript t 0 x t 0 y t 0 and x t y t x t y t x t y t and the slope of the secant line through the points is the difference quotient y t y t 0 x t x t 0 y t y t 0 t t 0 x t x t 0 t t 0 y t y subscript t 0 x t x subscript t 0 y t y subscript t 0 t subscript t 0 x t x subscript t 0 t subscript t 0 displaystyle frac y t y t 0 x t x t 0 frac frac y t y t 0 t t 0 frac x t x t 0 t t 0 2 Let us assume that the functions 1 are differentiable when t t 0 t subscript t 0 t t 0 and that x t 0 0 superscript x normal subscript t 0 0 x prime t 0 neq 0 As we let t t 0 normal t subscript t 0 t to t 0 the left side of 2 tends to the derivative d y d x d y d x frac dy dx and the right side to the quotient y t 0 x t 0 superscript y normal subscript t 0 superscript x normal subscript t 0 frac y prime t 0 x prime t 0 Accordingly we have the result d y d x t t 0 y t 0 x t 0 subscript d y d x t subscript t 0 superscript y normal subscript t 0 superscript x normal subscript t 0 displaystyle left frac dy dx right t t 0 frac y prime t 0 x prime t 0 3 Note that the

    Original URL path: http://www.planetmath.org/derivativeforparametricform (2016-04-25)
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