Examples of the Processes of the Differential and Integral CalculusJ. and J.J. Deighton, 1846 - 529 стор. |
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Сторінка 45
... 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 differentiating , dy ...
... 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 differentiating , dy ...
Сторінка 124
... curve shall pass through the point aa , y = 0 , gives A ( a − a ) 2 - 2 B ( a − a ) ß + CB2 + 1 = 0 . - - Subtracting ( 1 ) from ( 2 ) we have ( 2 ) A ( 2a - a ) + 2BB = 0 . ( 3 ) The condition that the curve shall pass through the ...
... curve shall pass through the point aa , y = 0 , gives A ( a − a ) 2 - 2 B ( a − a ) ß + CB2 + 1 = 0 . - - Subtracting ( 1 ) from ( 2 ) we have ( 2 ) A ( 2a - a ) + 2BB = 0 . ( 3 ) The condition that the curve shall pass through the ...
Сторінка 129
... curve , named after Diocles , a Greek mathema- tician , who is supposed to have lived about the sixth century of our era , was invented by him for the purpose of constructing the solution of the problem of finding two mean proportionals ...
... curve , named after Diocles , a Greek mathema- tician , who is supposed to have lived about the sixth century of our era , was invented by him for the purpose of constructing the solution of the problem of finding two mean proportionals ...
Сторінка 130
... curve in a point P. Joining AP , Joining AP , and pro- ducing it to meet CS produced , we determine the line CT which is the first of the two mean proportionals required . According to the geometrical ideas of the ancients a problem was ...
... curve in a point P. Joining AP , Joining AP , and pro- ducing it to meet CS produced , we determine the line CT which is the first of the two mean proportionals required . According to the geometrical ideas of the ancients a problem was ...
Сторінка 131
... curve , including both the su- perior and the inferior conchoid . It is evident from the construction of the curve that the line KH is an asymptote to both branches . When a > b there is a loop in the inferior conchoid at O as in the ...
... curve , including both the su- perior and the inferior conchoid . It is evident from the construction of the curve that the line KH is an asymptote to both branches . When a > b there is a loop in the inferior conchoid at O as in the ...
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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