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

440 Answers
1983. 3* + 4j/-H2z 169 = 0. 1985. x + 4y + 6z = 21 1986.
= / a 2
-f & 2 2
-f c
1987 Ai the points (1, 1. 0), the tangent planes are
parallel to the xz-plane; at the points (0, 0, 0) and (2, 0, 0), to the j/z-plane.
There are no points on the surface at which the tangent plane is parallel to
l _ _ n
the x#-plane. 1991. -^
. 1994. Projection on the xi/-plane: 1.
* =
~ ._w_ )=0
(
Projection on the /z-plane: < 3y* 2 Projection on the xz-plane:
I^Q
=
|
\ 3x 2
, A Hint. The line of tangency of the surface with the cylin-
4-z 2 1=0.
|
der projecting this surface on some plane is a locus at which the tangent
plane to the given surface is perpendicular to the plane of the projection
1996. /(* + /t, y + k) = ax z
+ 2bxy + cy*-{- 2(ax + by)h + 2(b* + cy) k i-ah 2 +
-f 26/ifc-K/z 2
1997. /(*, /)=! (xH-2) 2 + 2(jc-f2)(t/~l)-f3((/ 1)2,
1998. A/(x, i/) = 2/i + fc-M 2
+ 2/i/e-f/i 2 *. 1999. / (x, y, z) = 2
2
(x I) 4- (// I) +
+ (2 -_1)H2(*-1) (*/-!)--( 2005 . a) Un?^
2003.
b)
+ (3 nz_4M)p2j 2006. a) 1.0081; b)J).902. Hint. Apply Taylor's
formula for
r
the functions: a) f(x,y)=\ x yy in the neighbourhood of the point (1,1);
b) f (x, y) = y* in the neighbourhood of the point (2,1). 2007. z= 1 -}-2(x~-l)
-(t/-l)~-8(x~l) 2
+ 10 ( X-l)(y-l)-3 (y- !)'+.. 2008. zmm = when x=l,
i/=0 2009. NocxVcmum. 2010. z -=
min 1 whenx=l, 0=0. 2011. zmax=108 = 3,t/ = 2.20l2. zmin = 8 when x = Y^ y = 1^2 and when x =
There is no extremum for x=j/ = 0. 2013. 2 max ^ at
3 V 3
. . a b , a b ab
the points X^-TTL, t/---^ and*=
at the points ^ = y== a " d * ==
T^ ""Ff '~'Vf' y== F?' 2 14 ' ;?Tnax " =1
when jc = = i/ 0. 2015. zm i n = wne" x=0 = 0; nonrigorous maximum
= 1"\ at points of the circle * 2
2017. min = 3-
+ y 2 = 1. 2016. zmax = 1^3 whenx= 1, y = 1.
4 21
when x = -j , / = ,
z=l. 2018. w min = 4 when
JC = Y J/1' z== ^ 2019. The equation defines two functions, of which one
has a maximum (zmax 8) when *=1, f/=2; the other has a minimum
(Zimn = 2jwhenx 1,0 = 2, at points of the circle (x 1) 2
2 + (t/ -f 2 -^
; 25,
eacn of these functions has a boundary ext-emum = (z 3). Hint. The functions
mentioned in the answer are explicitly defined by the equalities