Solutions of the Cambridge Problems: From 1800 to 1820, Volum 2 |
Inni boken
Resultat 1-5 av 100
Side 1
... equal to either of these . .. the whole area A MB M ' Ambm'A = 4a2 3 N. B. Unless the branches of an oval intersect , it is not per- fectly quadrable . See Waring , Vince , & c . ( 2 ) Let E F ( Fig . 3 ) = y , AE = x AD , the diameter ...
... equal to either of these . .. the whole area A MB M ' Ambm'A = 4a2 3 N. B. Unless the branches of an oval intersect , it is not per- fectly quadrable . See Waring , Vince , & c . ( 2 ) Let E F ( Fig . 3 ) = y , AE = x AD , the diameter ...
Side 7
... equal para- meter with , the generating one Q E.D. If in other applications it should be necessary to ascertain the angle at which PM is inclined to Cc , in order to transform the co - ordinates of the section to rectangular or ...
... equal para- meter with , the generating one Q E.D. If in other applications it should be necessary to ascertain the angle at which PM is inclined to Cc , in order to transform the co - ordinates of the section to rectangular or ...
Side 14
... equal and similar teeth made by the union of the equal and similar spirals EF , E'F ; E′F ′ , E ′′ F ′ ; E ′′ F " , EF " , similarly posited around the circumference of the wheel ; then if the piston AB be constrained to move in the ver ...
... equal and similar teeth made by the union of the equal and similar spirals EF , E'F ; E′F ′ , E ′′ F ′ ; E ′′ F " , EF " , similarly posited around the circumference of the wheel ; then if the piston AB be constrained to move in the ver ...
Side 16
... equal negative values of x , the curve will con- sist of two equal and similar ovals AMBA , AmbA , intersecting , ( the curve is therefore quadrable ) in A. Again dy dx = √2 . 2x - x8x2 + 1 √ 8x2 + 1 × ( √ 8x2 + 1 − 2x2 − 1 ) 1⁄2 ...
... equal negative values of x , the curve will con- sist of two equal and similar ovals AMBA , AmbA , intersecting , ( the curve is therefore quadrable ) in A. Again dy dx = √2 . 2x - x8x2 + 1 √ 8x2 + 1 × ( √ 8x2 + 1 − 2x2 − 1 ) 1⁄2 ...
Side 21
... √x2 + y2 : m . x Hence y = ± √ m2 - ( n + x ) 2 which is the equation of the curve . diameter . n + x It is an oval similar and equal on each side the 25 . Given two radii vectores of a logarithmic spiral CONSTRUCTION OF CURVES . 21.
... √x2 + y2 : m . x Hence y = ± √ m2 - ( n + x ) 2 which is the equation of the curve . diameter . n + x It is an oval similar and equal on each side the 25 . Given two radii vectores of a logarithmic spiral CONSTRUCTION OF CURVES . 21.
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Solutions of the Cambridge Problems, from 1800 to 1820, Volum 2 John Martin Frederick WRIGHT Uten tilgangsbegrensning - 1836 |
Solutions of the Cambridge Problems: From 1800 to 1820, Volum 2 John Martin Frederick Wright Uten tilgangsbegrensning - 1825 |
Solutions of the Cambridge Problems, from 1800 to 1820, Volum 2 John Martin Frederick Wright Uten tilgangsbegrensning - 1836 |
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