# Multivariable derivative rule

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 When we consider multivariable derivatives related to same variable with different values, there can be a simplification of the manner to calculate a derivative that gives rise to the possibility to calculate by other means the la grangian point at which the total volume mass time expression between two celestial bodies equals , a point at which the distance gives rise to a certain value that express the limit of the derivative to zero and that is equivalent to a distance between the bodies where the value of gravity time multiple is equal meaning that the bodie is pulled in equal terms of proportion. r1= RandomReal[6471000,{200}] radius of earth in meters r2=RandomReal[695510000,{200}] radius of the sun in meters s=RandomReal[149600000000,{200}] distance from the sun in meters g1=RandomReal[9.86,{200}] gravity of earth g2=RandomReal[274,{200}] gravity of sun s2=(g2*r1*s)/(2*r2*g1) interrelated equation to obtain an intermediate distance from earth to the sun s1=s-s2 a=(r1*s1)/g1 total space mass time relation to earth b=(r2*s2)/g2 total space mass time relation to sun; both equals each other at a given distance y=(g2*r1*s1)/(g1*r2*s2) third variable obtained from a rule of proportion that equals the aplication of the theorem for the limit of a derivative ListPlot[a] ListPlot[b] ListPolarPlot[s2] ListPolarPlot[y] data=Table[Mod[s2,y],{s2,0,Pi,Pi},{y,0,Pi,Pi}] ListPlot[data,InterpolationOrder->3, ColorFunction->{s2->Yellow,y->Blue,s1->Green}] ContourPlot3D[Sin[s2*s1/y] ,{s2,-Pi,Pi},{s1,-Pi,Pi},{y,-Pi,Pi}] Run the program and notice that the simple proportion between two analogus functions alows to obtain a third variable that is related to the applied principle of limit of a derivative. Notice that a and b have equal values for a certain distance from the sun and from earth expressed in space per mass per second. the distance at this point makes the bodie there located to be in the same relative position. r1= RandomReal[6471000,{200}] r2=RandomReal[695510000,{200}] s=RandomReal[149600000000,{200}] g1=RandomReal[9.86,{200}] g2=RandomReal[274,{200}] s2=(g2*r1*s)/(2*r2*g1) s1=s-s2 a=(r1*s1)/g1 b=(r2*s2)/g2 y=(g2*r1*s1)/(g1*r2*s2) ListPlot[a] ListPlot[b] ListPolarPlot[s2] ListPolarPlot[y] data=Table[Mod[s2,y],{s2,0,Pi,Pi},{y,0,Pi,Pi}] ListPlot[data,InterpolationOrder->3, ColorFunction->{s2->Yellow,y->Blue,s1->Green}] ContourPlot3D[Sin[s2*s1/y] ,{s2,-Pi,Pi},{s1,-Pi,Pi},{y,-Pi,Pi}] tahnk you , feel free to aply the rule to any other object distance gravity related condition: black holes for instance. Attachments:
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Posted 1 month ago
 In shorter terms at a distance of aproximate 130 millions kilometers from the Sun the total volumetric knetic energy flow from sun to earth or earth to the sun is exactly the same.
 Further exploring the relation of the radius to the distance and gravity of two distinct points or bodies with mass , relating it to the hydrogen atom considering the values for the gravity of the proton and electron to correct when expressed in terms of energy the value for the limit or derivative between two distinc function qith equal variables and different values the value is the same as for the relation of earth to the moon. Just an amazing discovery that leads to quantize the value of gravity. Think of gravity bewtween two points as being the force of the wieght of two bodies ina seesaw whcih distances from the point of equilibrium compensates for the value of the force exerting downward orientation so that theere can be an equiblirium bewteen the two bodies.If you run the program you will see that this relation is equal to 2 for the value of y for both systems and probably aplies else where. r1= RandomReal[1*10^-15,{200}] radius of the proton r2= RandomReal[1*10^-18,{200}] radius of the electron s=RandomReal[5.29*10^-11,{200}] electron distance to the nucleus c=RandomReal[300000000,{200}] speed of light g1= 1.610^-7/(c^4r1) gravity of the proton expressed as energy g2= 1.510^-19/(c^4 r2) gravity of the electron expressed as energy s2=(g2r1s)/(2r2g1) s1=s-s2 distance of equilibrium considering the pendulum of gravity between two points a=(r1*s1)/g1 total volumetric energy flow of proton b=(r2*s2)/g2 total volumetric energy flow of electron y=(g2r1s1)/(g1r2s2) derivative or limit of x to zero for the relation between gravity distance and radius ListPlot[a] ListPlot[b] ListPolarPlot[s2] ListPolarPlot[y] data=Table[Mod[s2,y],{s2,0,Pi,Pi},{y,0,Pi,Pi}] ListPlot[data,InterpolationOrder->3, ColorFunction->{s2->Yellow,y->Blue,s1->Green}] ContourPlot3D[Mod[a*b,y] ,{a,-Pi,Pi},{b,-Pi,Pi},{y,-Pi,Pi}]Run these lines and compare to the previous post for the values of the sun and earth...this lines refer to the hydrogen atom. r1= RandomReal[1*10^-15,{200}] r2= RandomReal[1*10^-18,{200}] s=RandomReal[5.29*10^-11,{200}] c=RandomReal[300000000,{200}] g1= 1.6*10^-7/(c^4*r1) g2= 1.5*10^-19/(c^4* r2) s2=(g2*r1*s)/(2*r2*g1) s1=s-s2 a=(r1*s1)/g1 b=(r2*s2)/g2 y=(g2*r1*s1)/(g1*r2*s2) ListPlot[a] ListPlot[b] ListPolarPlot[s2] ListPolarPlot[y] data=Table[Mod[s2,y],{s2,0,Pi,Pi},{y,0,Pi,Pi}] ListPlot[data,InterpolationOrder->3, ColorFunction->{s2->Yellow,y->Blue,s1->Green}] ContourPlot3D[Mod[a*b,y] ,{a,-Pi,Pi},{b,-Pi,Pi},{y,-Pi,Pi}] Thank you Attachments:
 Corrected gravity for electron and proton considering G r1= RandomReal[1*10^-15,{200}] r2= RandomReal[1*10^-18,{200}] s=RandomReal[5.29*10^-11,{200}] c=RandomReal[300000000,{200}] G=6.67*10^-11 g1= (1.6*10^-7)*G/(c^4*r1) g2= (1.5*10^-19)*G/(c^4* r2) s2=(g2*r1*s)/(2*r2*g1) s1=s-s2 a=(r1*s1)/g1 b=(r2*s2)/g2 y=(g2*r1*s1)/(g1*r2*s2) ListPlot[a] ListPlot[b] ListPolarPlot[s2] ListPolarPlot[y] data=Table[Mod[s2,y],{s2,0,Pi,Pi},{y,0,Pi,Pi}] ListPlot[data,InterpolationOrder->3, ColorFunction->{s2->Yellow,y->Blue,s1->Green}] ContourPlot3D[Mod[a*b,y] ,{a,-Pi,Pi},{b,-Pi,Pi},{y,-Pi,Pi}]