## The Mathematical Questions Proposed in the Ladies' Diary: And Their Original Answers, Together with Some New Solutions, from Its Commencement in the Year 1704 to 1816, Volume 4 |

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Acaster Malbis Alfred Fox altitude Answered axis ball base bisect centre of gravity circumference cone consequently construction cosine curve cycloid denote descent described diameter difference distance draw drawn earth's ellipse equal equation expression feet fluent fluxion force formula frustum given circle given point gives greatest height hence Henry Atkinson horizon Hutton's Course Hutton's Mensuration hyperbola inches inscribed John John Davey John Surtees join latitude length Let ABC maximum parabola paraboloid parallel perpendicular plane pole Ponteland problem quantity question radius ratio rectangle right-angled triangle segment semicircle similar triangles sine slant side solidity solution square number sun's Suppose tang h tangent theor theorem trigonometry velocity vertex vertical angle whence whole zenith

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Page 271 - C, and whose principal axis is the difference between AZ and CZ; and QS a perpendicular on AC may be drawn, to which (QS) if from any point Z of this hyperbola a perpendicular ZS is let fall (this ZS), shall be to AZ as the difference between AZ and CZ is to AC. Wherefore the ratios of ZR and ZS to AZ are given, and...

Page 271 - PR perpendicular to AB, and let fall ZR perpendicular to PR; then from the nature of the hyperbola, ZR will be to AZ as MN is to AB. And by the like argument, the locus of the point Z will be another hyperbola, whose foci are A, C, and...

Page 271 - TA are drawn, the figure TRZS will be given in specie, and the right line TZ, in which the point Z is somewhere placed, will be given in position. There will be given also the right line TA, and the angle ATZ; and because the ratios of AZ and TZ to ZS are given, their ratio to each other is given also; and thence will be given likewise the triangle ATZ, whose vertex is the point ZQEI CASE 2.

Page 165 - Given the vertical angle, the difference of the two sides containing it, and the difference of the segments of the base made by a perpendicular from the vertex ; construct the triangle.

Page 154 - IN a Right-angled Triangle, having given the Perimeter or Sum of all the Sides, and the Perpendicular let fall from the Right Angle on the Hypothenuse ; to determine the Triangle, that is, its Sides. PROBLEM XVIII.

Page 391 - ... 0. The diameter, therefore, and the radius of curvature at A are infinite. In other words, no circle, having its centre in AB produced, and passing through A, can be described with so great a radius, but that, at the point A, it will be within the curve of equal attraction. The solid of greatest attraction, then, at the extremity of its axis, where the attracted particle is placed, is exceedingly flat, approaching more nearly to a plane than the superficies of any sphere can do, however great...

Page 394 - I, от m + l =2, this equation becomes y* — ax я* being that of a circle of which the diameter is AB. If, therefore, the attracting force were inversely as the distance, the solid of greatest attraction would be a sphere. If the force be inversely as the cube of the distance, or m = 3, and m + 1 = 4, the equation is y = о?

Page 225 - In any triangle, the centre of the circumscribed circle, the intersection of the medial lines, and the intersection of the perpendiculars from the angles upon the opposite sides, are in the...

Page 388 - ACS, therefore, is the locus of all the points in which a body being placed, will attract the particle A in the direction AB, with the same force. This condition is sufficient to determine the nature of the curve ACB.

Page 389 - AG, from the nature of the semicircle, and therefore AC= v'AB x AG, which has been shewn to be a property of the curve. In this way, any number of points of the curve may be determined ; and the solid of greatest attraction will be described, as already explained, by the revolution of this curve about the axis AB.