Examples of the Processes of the Differential and Integral Calculus |
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Page 103
D. F. Gregory. ( 25 ) Given the length of the arc of a circle , find the angle which it must subtend at the centre in order that the corresponding segment may be a maximum . Let a be the half - length of the arc , the radius of the α circle ...
D. F. Gregory. ( 25 ) Given the length of the arc of a circle , find the angle which it must subtend at the centre in order that the corresponding segment may be a maximum . Let a be the half - length of the arc , the radius of the α circle ...
Page 104
... circles on a sphere , and the arc PM is drawn perpendicular to OM ; find when the difference between OP and OM is the greatest . Let POM = α , ОР - ф - Ө ф , ом ; then we have = = a maximum ; while , by Napier's rules for the solution ...
... circles on a sphere , and the arc PM is drawn perpendicular to OM ; find when the difference between OP and OM is the greatest . Let POM = α , ОР - ф - Ө ф , ом ; then we have = = a maximum ; while , by Napier's rules for the solution ...
Page 127
... circles containing angles of 120 ° , their intersection will determine the point 0. The actual length of the lines u , v , w , and the value of the minimum sum may be found . For from the geometry of the figure we have the equations v2 ...
... circles containing angles of 120 ° , their intersection will determine the point 0. The actual length of the lines u , v , w , and the value of the minimum sum may be found . For from the geometry of the figure we have the equations v2 ...
Page 129
... circle ADBE take BM = AN , erect the ordinates QM , RN and , join AQ ; the locus of the point P where the line AQ cuts the ordinate RN is the cissoid of Diocles . To find its equation , put AN , PN - y , AC - a : then as PN QM y AN = AM ...
... circle ADBE take BM = AN , erect the ordinates QM , RN and , join AQ ; the locus of the point P where the line AQ cuts the ordinate RN is the cissoid of Diocles . To find its equation , put AN , PN - y , AC - a : then as PN QM y AN = AM ...
Page 130
... circle ADB ( fig . 13 ) erect the perpendicular CDE , of indefinite length . Take a point 0 in CA produced such that AO = AC ; then if the rectangular ruler NLM , of which the leg LM is equal to the diameter of the circle , be moved so ...
... circle ADB ( fig . 13 ) erect the perpendicular CDE , of indefinite length . Take a point 0 in CA produced such that AO = AC ; then if the rectangular ruler NLM , of which the leg LM is equal to the diameter of the circle , be moved so ...
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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