line and at right angles to it,

P cos 0μR coso,

=

P sin0 W sina = μR sinḍ.

C. Todhunter's Anal. Stat. Cap. XIII.

and Steele's Part. Dyn. Cap. v.

Pure Mathematics.

D. Tait

6. e(1+æ+æ3+ 5xo); −2± √(6) √(x) F

6

7. A minimum when x=0, neither when x=-1; if a3 be the given volume, r the radius of the base, h,

3

the heights of the cones, h=h' = r √√/ (2) = a

h'

+ [(a2 + x)dx.

n+1

13. (i) A right cone, vertex (a, b, c), axis parallel to

(l, m, n), cos (semi-vertical angle). (2) Distance

K

of point on the locus from (a, b, c) = projection of

distance from origin on (l, m, n). By subtraction the surfaces intersect in the two planes

-

lx+my+nz = ± κ {(x − a) l + (y − b) m + (z − c) n}. C. Figures 2 and 3.

WOOLWICH, SEPTEMBER, 1873. Pp. 28-41.

Arithmetic and Logarithms.

1. If AE, BE be the bisectors, the angle AEB_together with half the sum of the angles A and B

=

two right angles, and half the sum of the angles A and B together with half the sum of the two remaining angles = two right angles.

3. Let the straight line AB be divided at H so that the rect. AB, BH = sq. on AH. Take AK=HB; therefore AH- KB. Then the rect. AH, HK together with the rect. AH, AK=sq. on AH (Euc. II. 2) = rect. AB, AK (Euc. II. 11) = sq. on AK together with the rect. AK, KB (Euc. II. 3). Therefore the rect. AH, HK=sq. on AK. 5. All the triangles on the base AB, having the same vertical angle, have their vertices on the circumference of a segment, of which AB is the base. Let ACB be the isosceles triangle, and CE the tangent at C; CE is parallel to AB. Let ADB be any other of the triangles; produce AD to

meet CE at E, and join BE. Then the triangle ACB AEB>ADB.

=

6. See For. p. 16, Prop. v. Circles will Circles will go round BFEC, FDBP, and PDCE; thus the angle EDA the angle FCE= the angle FBE= the angle FDA. 8. See question 4 on p. 13.

=

10. Let ACB be the given arc; divide AB at D so that AD is to DB in the given ratio, and produce AB to E so that AE is to EB in the given ratio. On DE describe the semicircle DCE cutting the given arc at C, the point required.

11. With the figure of Todhunter's Euclid, VI. C., we have rect. AD, EA = rect. BA, AC; therefore

therefore

EA: 1: BA.AC: AD;

EA: BC:: BA.AC: AD.BC.

Algebra.

1

1. 26; 0; 4.

2. (1) x2 - 2 (a2 + b2) x2 + (a2 − b2)2; (2) x2 — 2xy + y2. 3. (1) Take 2n (2n − 1) out of the two first terms, and 2n (2n+1) out of the two last; (2) use For. p. 35, 11. 9, 10. 4. √√(5) −√√(2); 1.

7. y3 + y2 (18 − p3) + y (81 − 2pr) — r2 = 0.

8. a+ (2n−5) b; n {a+b (n − 4)}; 6, -9. 9. 1-x; (1+x*) (1 + x), (1 + x2) (1 − x)". 11. 10(1+b)3.

write x for the value of the continued fraction; thus

9. 4, 4, -3.

4. 1.25893. 5.2. 10. x+3y=0.

2. 35, 0; 24, 7; 13, 14; 2, 21. 7. 3, 3, 1. 11. yy=2a (x+x'); to determine the points of intersection of this tangent with the parabola, y=4a'x,

Y1, y, be the roots of this equation, we then have

2

1

B. ±1, ±√(−1), ± √(2) {1+√(−1)}, ±

C. Use For. p. 98, 1. 2, and p. 99, last line.

1

{1−√(−1)}.

√(2)

and b = 1⁄2[A+C − √ {B2 + (A −C)3}].

1. Diminished.

Mixed Mathematics.

3. The tension of the string, that has its point of attachment shifted, is double that of each of the

others.

7. 2g.

μW

th.

6. √(2ga). 10. 11ths. 11. The action of the pump is assisted. A. With the same notation as in A, p. 25, (R − Q) P− R Q = 0. B. Tait and Steele's Part. Dyn. Cap. v., Art. 144. C. Besant's Hydromech., Cap. II.

2

n (n + 1) x2 ̧

8. Radius of base

7. (1) 1; (2) -1.

12. (1) log(); (2) sin (a"-");

6

83

MARINES, AUGUST, 1873. PP. 42-51.
Arithmetic and Logarithms.

1. 22. 4. 2. 7. 1.0368.

10. 1618. 13. §.

88.

16. 00423.

19. 75081.6.

22. £13.

2. Solving the equation, (2

an impossible quantity.

2

2

n

51) = 4, we get n

3. The coeff. of the 6th term; the first term.

4. 250° 31′ 44′′. 5. nπ+77.

4

6. 89° 59′ 35′′.