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(MATH100)midterm_2000_ans.pdf
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Question1. Evaluatethede.niteintegral
Z
2
kti+t2jkdt:
0
Solution: 100,
Z
2 Z2
p
kti+t2jkdt.t2+t4dt 00
. Z 2 tp1 + t2dt
0
. 1Z 2p1 +t2dt2
2 0
. 1 2 2 3(1+t2)3.2j2 0

p
1
.(125;1)
3
Question2. Theplanethroughthepoint(;1.2.;5)thatisperpendiculartotheplanes 2x;y+z.1andtheplanex+y;2z.3.
Solution: Thenormaldirectionofplane2x;y+z.1is n1.(2.;1.1).2i;j+k: Similarly,forplanex+y;2z.3,thenormaldirectionis n2.(1.1.;2).i+j;2k: Then,thenormaldirectionfortheplanewhichisperpendiculartotheabovetwoplanesis n.n1.n2.i+5j+3k: Withtheabovenandthroughthepoint(;1.2.;5),theequationoftheplaneis (x+1)+5(y;2)+3(z+5).0. or x+5y+3z+6.0:
Solution:
Taking@ .@ x ontheequation
@ @x (x 2+z s in (xyz)).0
gives
2x+sin(xyz)@@ z x +y z 2 cos(xyz) +xyzcos(xyz)@@ z x . 0 :
Then,
.;2x;y z 2cos(xyz)
@ x sin(xyz) +xyzcos(xyz):
Similarly,
@@ z y .; xz2cos(xyz) sin(xyz)+xyzcos(xyz):

Question Calculate@z.@xand@[email protected] andz.
Question4. Findapointonthecirclex2+y2.45thatisclosesttothepoint(1.2)onthexyplane.
Solution:
Sinceminimizingthedistanceandthesquareofthedistanceisidenticalinthiscase.We cande.netheminimizingfunctionFas
F.(x;1)2+(y;2)2.
andtheconstraintfunctionGis
G.x 2+y 2;45.0:
FromtheLagrangemultiplier,wehave5F..5G,whichgives 2(x;1)i+2(y;2)j..(2xi+2yj): Therefore,wehave x;1..xandy;2..y. fromwhichweget y.2x: Takingtheaboverelationintotheconstraint,weget
x 2+4x 2.45.
whichgives
x..3andy..6: Thetwoextremepointsare(3.6)and(;3.;6).But,onlythepoint(3.6)givestheclosest distancewhichis qp
d.(3;1)2+(6;2)2.25: