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(MATH100)midtermFall01.pdf
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Math100,L2,MidtermExam,Solutions,Fall 2001
1.Finethearclength:
x.3t2+2.y.2t3;1.1.t.3:

Solution.Notethat
~~
~r(t).(3t2+2)i+(2t3;1)j: 0(t~~~
~r).6ti+6t2~j.6t(i+tj):
pjj~r0(t)jj.6t1+t2:
Thusthearclengthis
ZZZ
33 10
ppp
s.jj~r0(t)jjdt.6t1+t2dt.3u 12du.2u 32j10.2010;42:
2
112

2.Showthatz.ln(x2+y2)+2tan;1y.xsatis.estheLaplaceequation
@2@2

zz
+.0: @x2 @y2
Proof.Firstwehave
@z2x;2y.x2 x;y

.+.2.
@xx2+y2 1+(y.x)2 x2+y2
@z2y2.xy+x
.+.2:
@yx2+y2 1+(y.x)2 x2+y2
Moreover, @2zx2+y2;2x(x;y)y2;x2+2xy
.2.2.
@x2 (x2+y2)2 (x2+y2)2
@2zx2+y2;2y(y+x)x2;y2;2yx
.2.2: @y2 (x2+y2)2 (x2+y2)2
Thus @2z@2zy2;x2+2xyx2;y2;2yx
+.2+2.0: @x2 @y2 (x2+y2)2 (x2+y2)2
3.Findparametricequationsforthetangentlinetothecurveofintersectionofthe
222222
ellipsoidx+2y+z.4andthehyperboloid5x;2y;2z.1atthepoint(1.1.1). Solution.Let
22222
f(x.y.z).x+2y+z 2.g(x.y.z).5x;2y;2z:
Thentheellipsoidisthelevelsurfacef(x.y.x).4andthehyperboloidisthelevelsurface
g(x.y.x).1.Moreover, fx.2x.fy.4y.fz.2z.
1
gx.10x.gy.;4y.gz.;4z:
Thus
~
~~~~
rf(1.1.1).2i+4j+2k.rg(1.1.1).10i;4j;4~k:
So
. . ~i ~j ~k . .
rf(1.1.1) . r g(1.1.1). . . . 2 4 2 . . . .;8~i + 2 8 ~j;48~k
. 10 ;4 ;4 .

isadirectionvectorofthetangentline.Theequationofthetangentlineis x.1;8t.y.1+28t.z.1;48t: 4.Locateallrelativemaxima,relativeminimaandsaddlepointsofthefunction f(x.y).xy2;6x2;3y2 . Solution.Firstwehave
fx(x.y).y 2;12x.fy(x.y).2xy;6y:
To.ndcriticalpoints,wesolve
y 2;12x.0.2xy;6y.0.)2y(x;3).0.)y.0orx.3:
f
Ify.0,then12x.y2.0,thatisx.0.Ifx.3,theny2.12x.36,andy..6. Thuswehavethreecriticalpoints:(0.0).(3.6).(3.;6). Now xx.;12.fxy.2y.fyy.2x;6:
At(0.0),fxy.0andfyy.;6.SoD.(;12).(;6);02.72.0.Hence(0.0)isa pointofrelativemaximum. At(3..6),fxy..12andfyy.0.SoD.12.0;(.12)2.;144.0.Thus(3.6) and(3.;6)aresaddlepoints.
5.Findtheabsolutemaximumvalueofthefunctionf(x.y).xysubjecttothe constraint(x+1)2+y2.1andthepointatwhichfobtainsthismaximumvalue.
Solution.Letg.(x+1)2+y2;1.Note
fx.y.fy.x.gx.2(x+1).gy.2y: So
Solve
~~~
rf(x.y).yi+xj.rg(x.y).2(x+1)i+2y~j:
yx
rf(x.y)..rg(x.y).g(x.y).0.)...
2(x+1)2y
.)y 2.x(x+1).)(x+1)2+(x+1)x;1.0 .)2x 2+3x.0.)x(2x+3).0.)x.0.x.; 3 :
2
2

Whehx.0,y2.x(x+1).0.Whenx.;3.2,
..p
3333
y 2.;;+1..)y..:
2242
Weevaluate
.!.!
pppp
33333333
f(0.0).0.f;..;.f;.;.:
224224
ppTherefore,f(x.y)obtainsthemaximumvalue33.4at(;3.2.;3.2).
3