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

-406 Answers
640. Hint. The equation of the tangent is ^-{-^-=1. Hence, the tangent
ZXQ 2yQ
crosses the x-axis at the point A (2x , 0) and the y-axis at B (0, 2f/ ). Finding
the midpoint of AB, we get the point (* , yQ ). 643. 40 36'. 644. The para-
bolas are tangent at the point (0, 0) and intersect at an angle
arctan-i=^88' at the point (1,1). 647. S,
648 ' ' 652 ' T
653. arc tan ~. 654.
-y +2(p ' 655 ' S = f
* = 2jia /1+4JT 2
rt = 2
a ;
}/~l-|-4ji ; tan^i = 2jt. 656. S, = -- >
t n
a; 5,,= -^;
I
-~ ~ fl2_|_p2. . tan|i= q> 657. 3 cm /sec; 0; 9 cm /sec
658. 15 cm/sec. 659.
g
--x 2 2
K uj
. The
2
-^-in/sec. 660. The equation of the trajectory \sy=x tan a
u * sin 2a
velocity,
range is -. The
2u g/sina + g 2 / 8 ; the slope of the velocity
vector is
v cosa .
Hint. To determine the trajectory, eliminate the parameter t from the given
system. The range is the abscissa of the point A (Fig. 17). The projections
of velocity on the axes are ~ and j-
~
The magnitude of the velocity is
' the velocit y vector is directed along the tangent to the
Q
~ ,
CO8
2 f
663. 1 he diagonal increases at a rale ol - '6. 8 cm/sec, the area, at a rate
of 40 cm 2
/sec 664. The surface area increases at a rate of 2ji m 2
/sec,
the volume, at a rate of 0.05 JT m'/'sec. 665. y cm/sec
666. The mass of the rod
is 360 g, the density at M is 5* g/crn, the
density at A is 0, the density
at B is 60g/cm. 667. 56x 6 + 210,Y4 * x
. 668. e (4^ + 2). 669. 2 cos 2x
2(1 ~"* ~~
2)
X
670.
. 671 .
672. 2 arc tan x
2
+
3(1
673. _ + 674 cosh 6 79. y"' = 6. 680. /'"(3)
1 x 2
(1 x*) i* a a
681. y v = ^ .
682.
vl = // ~64sin2x 684. 0; 1; 2; 2. 685. The velocity
is u = 5; 4997; 4.7. The acceleration, a = 0; 0.006; 0.6. 686. The law
of motion of the point M, is x = acoscof; the velocity
at time / is
awsincof; the acceleration at time / is acD 2 cos(of. Initial velocity, 0;
initial acceleration: aw2 ; velocity when * = is -aco; acceleration when
* = is 0. The maximum absolute value of velocity is aa>; the maximum
absolute value of acceleration is aw2 . 687. y (n) = n\a n . 688. a) n\ (n+l
(1 x)~ \
b) ( _ ir + i ~
. 689 .