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Now, according to the problem we have a = r' = R, P = 90°.

.. equations (1) and (2) become

a

× {2R ×

...

.. (2).

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226.

We generalize this problem by stating it, Let P (Fig. 76,) be the pole of the great ○ ABM of a given sphere whose centre is C; then supposing the quadrant AP to revolve uniformly about the axis PC with a given velocity, whilst a point from P moves along the quadrant also with a given uniform velocity, it is required to find the surface APQM traced out by the point P in moving through the whole quadrant.

Let v and nv be the velocities of P and of A (which measures the angular velocity of the quadrant), and let PQP, Pqb be any two successive positions of the quadrant, and Q, q of the point P. Then S denoting the surface APB,

dS=QqbB.

Now QBbq Zone corresponding to 'x (CN) :: Bb : ABMA = 2′′r.

But the zone = 2xrx, and Bb = nd. PQ =

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nrdx

and SC + nr √ r2 = x2 = nr √ r2 — x2

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nr2 = (2x — n)r3.

In the problem n = 4, when M coincides with A, and the surface of the hemisphere is divided into the two parts 47a = (2r)1 and 2. (x-2).

The curve of double curvature AQM is called the Spiral of Pappus.

227.

Let a, b, c, be the sides of the spherical ▲, and a, ß, y, their chords, and suppose A, A', the included by b, c, and B, y. Then it is well known that

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Hence, by substitution and reduction, we get

cos. A' cos. A + sin. sin.

b 2

C

2

b

whence cos. A' is always > cos. A (for sin.

sin. can never be

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negative), and consequently A is always > A' whether the spherical A be isosceles or not.

228.

Measuring the co-ordinates of the semicircle S, and

the sector S', we have

dS = ydz, and dS' = — √ dx3 + dy3

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and when z is indefinitely diminished, cos. z = 1

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sin. z=z

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z) sin. x z cos. x

sin. x.

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a

and altitude a

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x, is

max. and putting

dV = 0, we get a =, which determines the cone required.

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231.

Let C (Fig. 77,) be the centre of the earth, AB = 9 feet, the height of the person's eyes from the ground, and CA = radius of the earth 4000 miles. Then if from B the tangent BD be drawn, the arc AD will be the distance required.

Now CB (which is evidently a straight line) = 4000 miles +

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But since C is very small compared with 1, we may neglect the

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may easily be converted into miles, furlongs, poles, yards, &c.

232. Let A (Fig. 78,) be the given, and with AD, the given line drawn from the vertex A bisecting the opposite side of the A, describe a circular arc MN. Then ABC being any A of which BC is bisected in D by AD, and Abc that whose side bc is bisected by Ad at right angles, and .. touches MN, we have ▲ Abc > ABC. For rDs being drawn parallel to bc meeting the from C and B in r, s, it is evident that A Cec> CDs > DrB and à fortiori > bBe, and the part AbeC is common to the two ▲ Abc, ABC; ... Abc is > ABC. But ABC is any A whatever, .. Abc is the greatest A required.

233.

Let AB. (Fig. 79,) the given linea, and suppose the velocity of the point A moving along AC from A = 4 × vel. of B moving from B along BA; then B', A', being any contemporaneous positions of A, B, required the curve to which A'B' is perpetually a tangent.

Let ba be the position of A' B' immediately successive to A'B',

and intersecting it in P. Then P is evidently the point at which B'A' touches the curve.

Now let the equation to the line B' A' referred to origin A by co-ordinates x, y, parallel to AC, AB be

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Now at the point P, y and x belong to the curve as well as to the tangent; and since for the next position ab, they may be considered constant, we have

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the equation to the curve required. See pp. 47, 48.

234. Let a be the given line, and Q' the given square, then a being one of the required parts, and .. a - x the other, we have

(a− x) x = Q2

a ± √ a2

4Q, which determines the rectangle re

2

quired. Also since 4Qa cannot exceed a2, (

( - )

2

α

is the greatest

square the rectangle can equal.

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