2012, ISBN: 1231890150

Paperback

[EAN: 9781231890158], Neubuch, [PU: Rarebooksclub.com, United States], Language: English Brand New Book ***** Print on Demand *****. This historic book may have numerous typos and missing… More...

[EAN: 9781231890158], Neubuch, [PU: Rarebooksclub.com, United States], Language: English Brand New Book ***** Print on Demand *****. This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1814 Excerpt: .--Fe X Fr is to be a maximum, that is, the rectangle Feqr inscribed in the semi-circle Aeqp is to be a maximum. Now, it is well known that a square is the greatest rectangle that can be inscribed in a circle, therefore when the rectangle FF.O.R is a maximum, it will be half a square inscribed in the whole circle, that is 2fe = Eq, or 2 arc Ae =2 arc Pq, = arc Eq, . Consequently /. Aqd = Z. Dae = 22, and Z. Daq, = 671-. Second Solution, by Mr. John Cavill, Brighton. It is well known that if two projectiles are delivered with equal velocities, and the angle of elevation of the one is the complement of the angle of elevation of the other, then these two projectiles will have equal horizontal ranges. Having made this remark the solution of the question may be as follows. Let d, denote the velocity per second with which the two projectile s are delivered, and put g--i6T 7 feet. Also, let denote the sine of the angle of elevation of the lower projectile radius 1, then /(i--Xs) will express the sine of the elevation i the other. And by the doctrine of projectiles in a non Dalby s Count of Mathematics, vol. s, page 326, art. 325, resisting medium (dx- -g) /--x1) will express the horizontal range of both projectiles, or the common double ordinate of the two paiabolic arcs described by the projectiles. Again by the doctrine of projectiles, (d1--4g ( --J) will express the altitude or abscissa of the greater arc; and (dr-j-qg) that of the less or lower; and therefore (d2-- g) (1--a) will express the difference of the two heights. And hence, by the known principles of mensuration, 14 14-X--rW---JX(l--7.X1) =---2X4X/(1--)x(i--8X) will express the area included between the two curves. Now let A, denote a circular arc radius 1, and sine x; then from the .<

2012, ISBN: 1231890150

Paperback

[EAN: 9781231890158], Neubuch, [PU: Rarebooksclub.com, United States], Language: English Brand New Book ***** Print on Demand *****.This historic book may have numerous typos and missing … More...

[EAN: 9781231890158], Neubuch, [PU: Rarebooksclub.com, United States], Language: English Brand New Book ***** Print on Demand *****.This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1814 Excerpt: .--Fe X Fr is to be a maximum, that is, the rectangle Feqr inscribed in the semi-circle Aeqp is to be a maximum. Now, it is well known that a square is the greatest rectangle that can be inscribed in a circle, therefore when the rectangle FF.O.R is a maximum, it will be half a square inscribed in the whole circle, that is 2fe = Eq, or 2 arc Ae =2 arc Pq, = arc Eq, . Consequently /. Aqd = Z. Dae = 22, and Z. Daq, = 671-. Second Solution, by Mr. John Cavill, Brighton. It is well known that if two projectiles are delivered with equal velocities, and the angle of elevation of the one is the complement of the angle of elevation of the other, then these two projectiles will have equal horizontal ranges. Having made this remark the solution of the question may be as follows. Let d, denote the velocity per second with which the two projectile s are delivered, and put g--i6T 7 feet. Also, let denote the sine of the angle of elevation of the lower projectile radius 1, then /(i--Xs) will express the sine of the elevation i the other. And by the doctrine of projectiles in a non Dalby s Count of Mathematics, vol. s, page 326, art. 325, resisting medium (dx- -g) /--x1) will express the horizontal range of both projectiles, or the common double ordinate of the two paiabolic arcs described by the projectiles. Again by the doctrine of projectiles, (d1--4g ( --J) will express the altitude or abscissa of the greater arc; and (dr-j-qg) that of the less or lower; and therefore (d2-- g) (1--a) will express the difference of the two heights. And hence, by the known principles of mensuration, 14 14-X--rW---JX(l--7.X1) =---2X4X/(1--)x(i--8X) will express the area included between the two curves. Now let A, denote a circular arc radius 1, and sine x; then from the .<