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 ix
... Curves and the Investigation of their Equations from their Geometrical Properties 129 IX . X. On the Tangents , Normals , and Asymptotes to Curves ...... 144 Singular Points in Curves 162 XI . On the Tracing of Curves from their ...
... Curves and the Investigation of their Equations from their Geometrical Properties 129 IX . X. On the Tangents , Normals , and Asymptotes to Curves ...... 144 Singular Points in Curves 162 XI . On the Tracing of Curves from their ...
Сторінка 45
... curve in three dimensions should be a plane curve . ( 11 ) Eliminate the exponentials from € + € y = - E Multiply numerator and denominator by e , then y = y + 1 whence € 2 " = y and differentiating , - dy dx € 2 + 1 1 9 = - and 2x = 1 ...
... curve in three dimensions should be a plane curve . ( 11 ) Eliminate the exponentials from € + € y = - E Multiply numerator and denominator by e , then y = y + 1 whence € 2 " = y and differentiating , - dy dx € 2 + 1 1 9 = - and 2x = 1 ...
Сторінка 124
... curve shall pass through the point a = a , y = 0 , gives - - Subtracting ( 1 ) from ( 2 ) we have A ( a − a ) 2 - 2 B ( a − a ) ẞ + C ẞ2 + 1 = 0 . ( 2 ) A ( 2a - a ) + 2BB = 0 . ( 3 ) The condition that the curve shall pass through ...
... curve shall pass through the point a = a , y = 0 , gives - - Subtracting ( 1 ) from ( 2 ) we have A ( a − a ) 2 - 2 B ( a − a ) ẞ + C ẞ2 + 1 = 0 . ( 2 ) A ( 2a - a ) + 2BB = 0 . ( 3 ) The condition that the curve shall pass through ...
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 , 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 not thought ...
... curve in a point P. 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 not thought ...
Зміст
1 | |
9 | |
28 | |
43 | |
52 | |
77 | |
79 | |
94 | |
224 | |
237 | |
249 | |
271 | |
282 | |
291 | |
340 | |
351 | |
129 | |
132 | |
144 | |
162 | |
175 | |
188 | |
200 | |
386 | |
400 | |
412 | |
440 | |
464 | |
506 | |
Інші видання - Показати все
Загальні терміни та фрази
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