Examples of the Processes of the Differential and Integral Calculus |
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Page 36
... condition we proceed as follows . Let for example there be a double integral ffVda dy , and let x = $ ( r , 0 ) , so that y = ¥ ( r , 0 ) , dx dx dx = dr + d Ꮎ , dr d Ꮎ dy dy dy = dr + do . dr do Since is to vary when y is constant ...
... condition we proceed as follows . Let for example there be a double integral ffVda dy , and let x = $ ( r , 0 ) , so that y = ¥ ( r , 0 ) , dx dx dx = dr + d Ꮎ , dr d Ꮎ dy dy dy = dr + do . dr do Since is to vary when y is constant ...
Page 37
... 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 = r cos 0 , and therefore x2 + y2 = r2 , dR dx dR dy ...
... 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 = r cos 0 , and therefore x2 + y2 = r2 , dR dx dR dy ...
Page 40
... condition being similar , we find dev d2 V + = d'V 1 + d2 V + d p2 do dx2 dr r2 1 dv - r dr Also , as in the first part of the last example , 1 dv 1 dv = p dp r dr cot 0 dv + do Adding these three expressions , d'V d'V de V + + dy d dx ...
... condition being similar , we find dev d2 V + = d'V 1 + d2 V + d p2 do dx2 dr r2 1 dv - r dr Also , as in the first part of the last example , 1 dv 1 dv = p dp r dr cot 0 dv + do Adding these three expressions , d'V d'V de V + + dy d dx ...
Page 45
... condition that a curve in three dimensions should be a plane curve . ( 11 ) Eliminate the exponentials from € + € y = - Multiply numerator and denominator by e " , then +1 y = - whence e2 = y + 1 and 2x = log y + 1 ' y - 1 y - 1 and ...
... condition that a curve in three dimensions should be a plane curve . ( 11 ) Eliminate the exponentials from € + € y = - Multiply numerator and denominator by e " , then +1 y = - whence e2 = y + 1 and 2x = log y + 1 ' y - 1 y - 1 and ...
Page 50
... condition ( 2 ) is reduced to dz f ( a ) + ( a ) 1 = 1 0 . dx 0 ; In the same way , differentiating with respect to y , we have dz $ ( a ) + 4 ( a ) = 0 . dy dz dz dy Since from these two equations it appears that and dx are both ...
... condition ( 2 ) is reduced to dz f ( a ) + ( a ) 1 = 1 0 . dx 0 ; In the same way , differentiating with respect to y , we have dz $ ( a ) + 4 ( a ) = 0 . dy dz dz dy Since from these two equations it appears that and dx are both ...
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