Problems in Mathematical Analysis.pdf - pwp.net.ipl.pt

410_Answers_4
where R is the radius of the given sphere. 871. Altitude of the cone, -^ R,
where R is the radius of the given sphere. 872. Radius of the base of the
cone
3
-jr-r, where r is the radius of the base of the given cylinder. 873. That
whose altitude is twice the diameter of the sphere. 874. q> = jr, that is, the crosssection
of the channel is a semicircle. 875. The central angle of the sector
/~2 .
876. The altitude of the cylindrical part must be zero; that
is, the vessel should be in the shape of a hemisphere. 877. h = \l* d 8
878. ^- + -^- = 1. 879. The sides of the rectangle are a^Tand 6J/~2, where
*X Q *yo
a and b are the respective semiaxes of the ellipse. 880. The coordinates of
the vertices of the rectangle which lie on the parabola
881.
/ 1 3 \
( ~Y=., -j J.
(-~-a;
2 I/ ^M-
882. The angle is equal to the greatest of the numbers
arc cos 4- and arc tan ^ 883. AM=a ^L P
885. a) t-y-. b),= =*. =; If 886.
. ^_ . 884. -4r m
,=
P min = y%aqQ. 887. \f~Mm. Hint. For a completely elastic impact of two
spheres, the velocity imparted to the stationary sphere of mass m l after
impact with a
sphere_of
mass m 2 moving with velocity v is equal to
m * V
this number is not an integer or is not a divisor of
/HI + w 2 v r
7V,we take the closest integer which is a divisor of TV). Since the internal resistance
. 888. n= "I/ - (if
of the battery is ^
,
the physical meaning of the solution obtained is as
follows: the internal resistance of the battery must be as close as possible to the
2
external resistance. 889. y = h. 891. ( oo, 2), concave down; (2, oo),
o
concave up; M (2, 12), point of inflection. 892. (00, oo), concave up.
893. ( oo, 3), concave down, ( 3, oo), concave up; no points of inflection.
894. ( oo, 6) and (0, 6), concave up; ( 6, 0) and (6, oo), concave down;
points of inflection M, (6, --|Vo(0, 0), Af 2 ( 6, |-V
895. (00,
^"3") and (0, ^3), concave up; ( 1/~3, 0) and (1^3, oo), concave down;
points of inflection M lj2 (|/* 0) and 0(0, 0). 896.
-jr- J,
concave up; ( (4/e +3) -^, (4 + 5)~ , concave down (& =
J
1, 2, ...); points of inflection, f(2fe+l)y, oV 897. (2/m,
concave up; ((2k l)it, 2&Ji), concave down(fe=0, 1, i2, ...); the abscissas
of the points of inflection are equal to xkn. 898. [0, ^ ), concave
\ V&J