Solutions of the Cambridge Problems: From 1800 to 1820, Volum 2 |
Inni boken
Resultat 1-5 av 100
Side 20
... question Sydx area AGC AGCB - ACB = fy'dx y'r 2 = Sdx √2x - x2 . ydxdx 2x - x2 - x √2x - x2 2 - dx √ 2x - x2 - xdr ( 1 - 2√2x - x2 x Hence y = 9 2√2x - x2 2 the equation of the curve APF . The curve cuts the semicircle when y ' = y ...
... question Sydx area AGC AGCB - ACB = fy'dx y'r 2 = Sdx √2x - x2 . ydxdx 2x - x2 - x √2x - x2 2 - dx √ 2x - x2 - xdr ( 1 - 2√2x - x2 x Hence y = 9 2√2x - x2 2 the equation of the curve APF . The curve cuts the semicircle when y ' = y ...
Side 37
... question . n Hence , we get ( n · G : 1 ) xydz = dxf ydz ydz Tydz dx n- -1 I : . l . fydz 1 = l ‚ x + C = l.xn2i + l . c ± l.c x : fydz = ca n -- 1 = с :: y√ dx2 + dy2 = xdx n- 1 ( a ) the general differential equation , which is ...
... question . n Hence , we get ( n · G : 1 ) xydz = dxf ydz ydz Tydz dx n- -1 I : . l . fydz 1 = l ‚ x + C = l.xn2i + l . c ± l.c x : fydz = ca n -- 1 = с :: y√ dx2 + dy2 = xdx n- 1 ( a ) the general differential equation , which is ...
Side 58
... question - dy TT ' $ = PP ' - since they move uniformly . n ydx : . ns = x dy Hence , making dy constant , we have n . √ dx2 + dy2 = dx = - GORD y . dxy d2x dy d2x dy p . dp dx Let ] ] dy Then - ny dy n 1 У dy = · √1 + p2 dp √1 + p2 ...
... question - dy TT ' $ = PP ' - since they move uniformly . n ydx : . ns = x dy Hence , making dy constant , we have n . √ dx2 + dy2 = dx = - GORD y . dxy d2x dy d2x dy p . dp dx Let ] ] dy Then - ny dy n 1 У dy = · √1 + p2 dp √1 + p2 ...
Side 66
... in P , and draw PM 1 CA and make PM y , MV = x . Now , by similar A , we have y : MA :: CB ' : CA ' :: b + BB ' : a + AA ' . — But , by the question , BB ' AA ' AV ' -b VM - MA ' - b . x - b - MA ' , and MA ' 66 THE NATURE AND.
... in P , and draw PM 1 CA and make PM y , MV = x . Now , by similar A , we have y : MA :: CB ' : CA ' :: b + BB ' : a + AA ' . — But , by the question , BB ' AA ' AV ' -b VM - MA ' - b . x - b - MA ' , and MA ' 66 THE NATURE AND.
Side 72
... question . ydy dx Hence the common parabola does not satisfy the conditions of the problem . Mr. Herschel's method , however holds good in other cases ; for instance , ( we suppose the reader possessed of the book ) when ƒ ( y . qy ) ...
... question . ydy dx Hence the common parabola does not satisfy the conditions of the problem . Mr. Herschel's method , however holds good in other cases ; for instance , ( we suppose the reader possessed of the book ) when ƒ ( y . qy ) ...
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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 |
Vanlige uttrykk og setninger
abscissa accelerating force altitude angular axes axis base bisected body centre of gravity chord circle co-declination co-ordinates cone curve cycloid cylinder denote density descending diameter distance earth ecliptic ellipse equal equation fluid focus given point gives Hence horizon hyperbola inclination intersection latitude latus rectum length locus logarithmic spiral moving force orbit ordinate orifice oscillation parabola paraboloid parallel perpendicular plane position problem projection Prop question radius ratio right angles right ascension shew sides specific gravity sphere spherical straight line substituting subtangent supposing surface tangent triangle values velocity vers vertex vertical Vince weight whence whole