Examples of the Processes of the Differential and Integral CalculusJ. and J.J. Deighton, 1846 - 529 стор. |
З цієї книги
Результати 1-5 із 17
Сторінка 102
... line PQ so that the triangle PAQ shall be a mini- mum . Draw MN parallel to AQ , and let AN = a , MN = b , AP . Then 2a makes PAQ a minimum , and PQ is bisected in M. ( 25 ) Given the length of the arc of 102 MAXIMA AND MINIMA .
... line PQ so that the triangle PAQ shall be a mini- mum . Draw MN parallel to AQ , and let AN = a , MN = b , AP . Then 2a makes PAQ a minimum , and PQ is bisected in M. ( 25 ) Given the length of the arc of 102 MAXIMA AND MINIMA .
Сторінка 103
... triangles ACX and BXY shall be a minimum . If AC = a , AB = b , AX = ∞ , it is easily seen that the area of the triangle ACX is proportional to ax , and that of a ( b − x ) 2 BXY to - 00 9 so that we have - a { x + ( b = x ) ...
... triangles ACX and BXY shall be a minimum . If AC = a , AB = b , AX = ∞ , it is easily seen that the area of the triangle ACX is proportional to ax , and that of a ( b − x ) 2 BXY to - 00 9 so that we have - a { x + ( b = x ) ...
Сторінка 104
... triangles , tan 0 = cos a tan Ø . Differentiating with respect to 0 , we have do 1 = 0 , de аф 1 + tan20 = cos a ( 1 + tan ' p ) ; d Ꮎ therefore or Whence 1 + tan30 = cos a ( 1 + tan3Ø ) , 1 + cos3 a tan3 = cos a ( 1 + tan2 ( ) . tan ...
... triangles , tan 0 = cos a tan Ø . Differentiating with respect to 0 , we have do 1 = 0 , de аф 1 + tan20 = cos a ( 1 + tan ' p ) ; d Ꮎ therefore or Whence 1 + tan30 = cos a ( 1 + tan3Ø ) , 1 + cos3 a tan3 = cos a ( 1 + tan2 ( ) . tan ...
Сторінка 105
... triangles POR , PMR are equal , the pyramids PSRO and PMRQ are equal , so that , whatever be the in- clination of SQ to OM , the part cut off from the prism is equal to the part included in the pyramid SPR , and the con- tent of the ...
... triangles POR , PMR are equal , the pyramids PSRO and PMRQ are equal , so that , whatever be the in- clination of SQ to OM , the part cut off from the prism is equal to the part included in the pyramid SPR , and the con- tent of the ...
Сторінка 117
... , x , y , z being subject to the condition find x2 y2 & 2 + + = 1 . a2 2 By the method of indeterminate multipliers , we easily a b c 8abc ľ = " Y = % = u = ( 13 ) To find the triangle of least perimeter 33 MAXIMA AND MINIMA . 117.
... , x , y , z being subject to the condition find x2 y2 & 2 + + = 1 . a2 2 By the method of indeterminate multipliers , we easily a b c 8abc ľ = " Y = % = u = ( 13 ) To find the triangle of least perimeter 33 MAXIMA AND MINIMA . 117.
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