Examples of the Processes of the Differential and Integral CalculusJ. and J.J. Deighton, 1846 - 529 стор. |
З цієї книги
Результати 1-5 із 100
Сторінка 17
... ( − ) } } ' + ' = ( a2 + x2 ) * { cos ( r + 1 ) 9 − ( − ) 1 sin ( r + 1 ) 0 } , 2 { x + a ( − ) } } r + 1 = ( a2 + x2 ) * { cos ( r + 1 ) 8 + ( − ) 1 sin ( r + 1 ) 0 } . Hence we have d \ " dx ( 1 = 2 SUCCESSIVE DIFFERENTIATION . 17.
... ( − ) } } ' + ' = ( a2 + x2 ) * { cos ( r + 1 ) 9 − ( − ) 1 sin ( r + 1 ) 0 } , 2 { x + a ( − ) } } r + 1 = ( a2 + x2 ) * { cos ( r + 1 ) 8 + ( − ) 1 sin ( r + 1 ) 0 } . Hence we have d \ " dx ( 1 = 2 SUCCESSIVE DIFFERENTIATION . 17.
Сторінка 21
... Hence multiplying by ( e * + 1 ) * + 1 we must have ( e * + 1 ) ' + ' d'u dx = a ‚ € TM ” + ɑ‚_16 ( ̃ − 1 ) 2 + & c . + α1e * ( 1 ) . Now as u = € ' - 2x - E -32 + € - & c . d'u = dx ( − ) ' { 1 ′ e ̄ I - 2 ' € -2x + 3 ′ € -3x ...
... Hence multiplying by ( e * + 1 ) * + 1 we must have ( e * + 1 ) ' + ' d'u dx = a ‚ € TM ” + ɑ‚_16 ( ̃ − 1 ) 2 + & c . + α1e * ( 1 ) . Now as u = € ' - 2x - E -32 + € - & c . d'u = dx ( − ) ' { 1 ′ e ̄ I - 2 ' € -2x + 3 ′ € -3x ...
Сторінка 31
... Hence , substituting these binomial factors , we find d " u y ” = dyn [ { a - ( n - d dx - ( n - ... ( 7 ) Change the independent variable in d dx - d U. dx . - ( 1 − y2 ) d2 u dy - y dy du + n2 u = 0 The result is from y tox , having ...
... Hence , substituting these binomial factors , we find d " u y ” = dyn [ { a - ( n - d dx - ( n - ... ( 7 ) Change the independent variable in d dx - d U. dx . - ( 1 − y2 ) d2 u dy - y dy du + n2 u = 0 The result is from y tox , having ...
Сторінка 36
... Hence we have dx == dx de . d Ꮎ Substituting these values in the double integral it becomes SSV dx dy da dy - d Ꮎ dr dr de dr de . If we had three variables x , y , ≈ to be transformed into three others p , q , r , we should have ...
... Hence we have dx == dx de . d Ꮎ Substituting these values in the double integral it becomes SSV dx dy da dy - d Ꮎ dr dr de dr de . If we had three variables x , y , ≈ to be transformed into three others p , q , r , we should have ...
Сторінка 42
... Hence dx dy • - dx dy độ đẹp đẹp do dz dy dx dy = аф ab sin cos 0 , = bc ( sin 0 ) 2 cos , do do do do dx dx dx dx ᏧᎾ do dp do = ac ( sin 0 ) 2 sin p . Substituting these values in the general expressions for dz dz > dx ' dy and do dy ...
... Hence dx dy • - dx dy độ đẹp đẹp do dz dy dx dy = аф ab sin cos 0 , = bc ( sin 0 ) 2 cos , do do do do dx dx dx dx ᏧᎾ do dp do = ac ( sin 0 ) 2 sin p . Substituting these values in the general expressions for dz dz > dx ' dy and do dy ...
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a² b2 a²x² angle arbitrary constant asymptote becomes C₁ c²x² Cambridge circle co-ordinates condition Crelle's Journal curvature curve cycloid determine differential coefficients differential equation dx dx dx dy dx dy dx dx² dy dx dy dy dy dy dz dz dz eliminate ellipse equal Euler factor formula fraction function Geometry gives Hence hypocycloid infinite intersection John Bernoulli Let the equation lines of curvature locus logarithmic logarithmic spiral Multiply negative origin parabola perpendicular plane of reference radius SECT singular solution spiral Substituting subtangent surface tangent plane theorem triangle vanish whence x²)³