Examples of the Processes of the Differential and Integral CalculusJ and J. J. Deighton, 1846 - 529 стор. |
З цієї книги
Результати 1-5 із 33
Сторінка 106
... ellipse which can be cut from a given cone . Let AC ( fig . 8 ) = a , major axis of the ellipse . CD = b , CN = x , BP being the Then the condition that the area of the ellipse shall be a maximum gives 2b ( a2 — b2 ) ± b ( a1 − 14a2b2 ...
... ellipse which can be cut from a given cone . Let AC ( fig . 8 ) = a , major axis of the ellipse . CD = b , CN = x , BP being the Then the condition that the area of the ellipse shall be a maximum gives 2b ( a2 — b2 ) ± b ( a1 − 14a2b2 ...
Сторінка 110
... ellipse which revolves round an axis parallel to the major axis . In these cases we have d'u d2 u dx2 dy2 - u ( dvdy ) 2 = = 0 , an equation which is usually excluded from Lagrange's con- dition . It is to be observed , however , that ...
... ellipse which revolves round an axis parallel to the major axis . In these cases we have d'u d2 u dx2 dy2 - u ( dvdy ) 2 = = 0 , an equation which is usually excluded from Lagrange's con- dition . It is to be observed , however , that ...
Сторінка 123
... ellipse is Ax2 + Bxy + Cy2 + Dx + Ey + 1 = 0 , which involves five arbitrary constants ; three of these may be determined by the conditions that the ellipse shall pass through the three points A , B , C. Instead however of directly ...
... ellipse is Ax2 + Bxy + Cy2 + Dx + Ey + 1 = 0 , which involves five arbitrary constants ; three of these may be determined by the conditions that the ellipse shall pass through the three points A , B , C. Instead however of directly ...
Сторінка 124
... ellipse may be put under the form - A ( x − a ) 2 + 2B ( x − a ) ( y − ẞ ) + C ( y − ẞ ) 2 + 1 = 0 , - - - where A , B , C are to be determined . Now the condition that the ellipse shall pass through the origin gives Aa2 + 2Baß + C ...
... ellipse may be put under the form - A ( x − a ) 2 + 2B ( x − a ) ( y − ẞ ) + C ( y − ẞ ) 2 + 1 = 0 , - - - where A , B , C are to be determined . Now the condition that the ellipse shall pass through the origin gives Aa2 + 2Baß + C ...
Сторінка 125
... ellipse is therefore 2 п ab sin 0 . 31 It appears then that the area of the ellipse is to that of the triangle as 47 : 3 , and that its centre coincides with the centre of gravity of the triangle . This problem is given by Euler in the ...
... ellipse is therefore 2 п ab sin 0 . 31 It appears then that the area of the ellipse is to that of the triangle as 47 : 3 , and that its centre coincides with the centre of gravity of the triangle . This problem is given by Euler in the ...
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Загальні терміни та фрази
a² b2 a²x² angle arbitrary constant asymptote axis becomes C₁ c²x² Cambridge circle co-ordinates condition curvature curve cycloid cylinder determine differential coefficients differential equation dx dx dx dy dx dx² dy dx dy dy dy dy dz eliminate ellipse equal Euler find the value formula function Geometry gives Hence hypocycloid infinite Integrating with respect intersection John Bernoulli Let the equation lines of curvature locus logarithmic logarithmic spiral maximum minimum Multiply negative origin parabola perpendicular radius radius of curvature singular solution spiral Substituting subtangent surface tangent plane theorem tractory triangle vanish whence x²)³