Examples of the Processes of the Differential and Integral CalculusJ. and J.J. Deighton, 1846 - 529 стор. |
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Сторінка 17
... gives it under a form which is more convenient in practice ; 1 a2 + x 2 1 = 2a ( - ) x + a ( - ) x - a - Differentiating r times , 1 d − ( − ) ' + 1 ̧ † ( r − 1 ) ... ..2 . - { = ( ~ ) " + 1 " T 1 a2 + x2 1 2a ( - ) { { x + a ...
... gives it under a form which is more convenient in practice ; 1 a2 + x 2 1 = 2a ( - ) x + a ( - ) x - a - Differentiating r times , 1 d − ( − ) ' + 1 ̧ † ( r − 1 ) ... ..2 . - { = ( ~ ) " + 1 " T 1 a2 + x2 1 2a ( - ) { { x + a ...
Сторінка 20
... ( r + & c . 1 ( 27 ) Let u = € + 1 We might in this case expand the function and differen- tiater times each term in the development , but as this would give d'u dx expressed in an infinite series , 20 SUCCESSIVE DIFFERENTIATION .
... ( r + & c . 1 ( 27 ) Let u = € + 1 We might in this case expand the function and differen- tiater times each term in the development , but as this would give d'u dx expressed in an infinite series , 20 SUCCESSIVE DIFFERENTIATION .
Сторінка 29
... gives us the means of simplification . ( 1 ) Change the formula dy + ( d ) " } dy dy + ( y - a ) = 0 dx da into one where y is the independent variable . The result is 1 + dy ( da ) * - dx ( y - a ) = 0 . dy ( 2 ) The expression for the ...
... gives us the means of simplification . ( 1 ) Change the formula dy + ( d ) " } dy dy + ( y - a ) = 0 dx da into one where y is the independent variable . The result is 1 + dy ( da ) * - dx ( y - a ) = 0 . dy ( 2 ) The expression for the ...
Сторінка 37
... give it here , as the consideration of the particular conditions of any given transformation will usually give us its value more readily than a substitution in the general formula . * ( 1 ) Transform x d R dy - dR y dx having given x ...
... give it here , as the consideration of the particular conditions of any given transformation will usually give us its value more readily than a substitution in the general formula . * ( 1 ) Transform x d R dy - dR y dx having given x ...
Сторінка 61
... gives = = 3 u = 0 , u = ± a . Taking the first of these values , we find the series x3 25 u = - - & c . a2 a3 a1 Taking the positive value of a , u = a - x x2 2 3x3 + " & c . 8 a 16a2 Taking the negative value of a , v u = a + + x2 5.23 ...
... gives = = 3 u = 0 , u = ± a . Taking the first of these values , we find the series x3 25 u = - - & c . a2 a3 a1 Taking the positive value of a , u = a - x x2 2 3x3 + " & c . 8 a 16a2 Taking the negative value of a , v u = a + + x2 5.23 ...
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a² b2 a²x² angle arbitrary constant assume asymptote becomes branches C₁ Cambridge circle co-ordinates condition Crelle's Journal curvature curve cycloid determine differential coefficients differential equation dx dx 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 radius SECT singular points singular solution spiral Substituting subtangent surface tangent plane theorem triangle University of Cambridge vanish whence x²)³