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SimonFraserUniversity
Departmentof Mathematics
BurnabyCampus
MATH 25l, Summer 2006
Midterm
July 5e, 2006,8:30- 9:
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Instructor: Dr. G. Tanoh
Instructions:
I. DO NOT OPENTHIS BOOKLET LINTIL
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- This examcontains7 pageswith a total of 6 questions.Oncethe exambeginsplease checkto makesureyour exam is complete.
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If you run (^) out of spacein a problem,use the spaceon the backofthe previouspage and clearlyindicatewherethe solution continues. Only scientific, non-programmable calculatorswith no differentiationand integrationcapabilitiesareallowed. No book, paper,or device,other thanthe usualwriting instruments,this bookletand an acceptablecalculator,shallbe within reachof a studentdurine the examination.
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Question Marks
I /
(^2) /
aJ /
5 IJ
6 t t
Total (^) /
\IATH 251- Summer2006 Nlidterm 2
Question 1. The trajectory'of^ a particle^ in spaceis givenby the position^ vector^ r(t):^ (2+3t+3t2)i+ ( 4 t + 4 t 2 ) i - ( 6 c o s f ) k.
(u) (^) [+j Find the normal and tangent componentsof the accelerationvector.
,oI@= (tt c{)i+
(qr Ef)jt(64";t)Lt (^) lrtttl | = [,, (a+zt)'*^ ?6nn'b|?'
)t'k) = 6) + si.,^ U^:rL,^
:rr.rTp)
= 5oQ+et)+48*'nz,b
Iti*
q+igr-e,J"t
I
I e^ E^ 6cill
1nt6e)
^ r"1e)l= r?, ((r-, 2L)c-t- ?'siut)",I (a14t)^ b, t- ?ci' f)"fn/z-/J
Jr'i((4Bt)bt - La;'a0J
a
_ L r
tWn t
, \
2t',f I
,l t+l' 't'(t)^ Eo (4+zb)t4n4"'"zt
I'tt G)
prc)1 l-f t'l+2t)'+ ]6 u'" t J'/"
[2] What is the speed^ and accelerationvector of the particle at the point (2.0, -6)?
zb)z+ (^) Z(
ntVot-
fzt
L-tu
'D4a
q /Y
(b)
1?, o,- ,
-f/*
L,*L
Lr,
u, 6 ) L>rrc f
*' (^) /--" t =^ o ('t(o) =
+1,-a:^ til@l= 5-
l-----/
\2r :
ft caQ-lor,t/,; ur^ rh^.^ =^ d/(o) 6Lts;/@b,)/
6i+ 8d+ 6 /
, 1 1
).IATH 251- Summer2006 \lidterm 2
Question 3. Supposethat we substitutepolar^ coordinatesr:^ r cos0and g:^ rsin 0 in a differentiable f " n n f i n n n ' , -^ f ( n \ " ' v i '^ , ' \
.( u ) \ , n , n. E u '^ 1 0 u ' [ l ] S h o u ' t h a t ; : / , c o s 0 ' t f u s i n d a n d^ ; * : - / , s i n 0 -^ f r c o s 1.
->L =^ 2w.?*^ +^ >y^ e1^ j^ {^ c"s?+f-u'n^ v a,tt-, >/U (^) ?I TrL. fa-^ ' I
}.w (^) s
av
(^r) y
( b ) f 2 lS o l v e t h e e q u a t i o n s i n p a r t ( a )t o e x p r e s s / , a n d / r i n t e r m s o f0 u l 0 r u t 4 Q u : f 0 0.
r{r
-:-\ t^ n.^ ' l
- 1 (^) a - 4 d w, 4 -.rr, VV
/?uo )e
l;"Jr
(^nn e
D n u i n * =
2 9 *t Lo> g + {l 2,1,.
{*-r" t{lc,n&
= L*.#
(") (^) Lvt Loru
-t-st,az t fl
^r1^
o f
ph- t "
o,, R'^Le
U (^73) iay (^) U=4-t-44|uv)-+h"ovrY
.e-L xl 21 ) y
= ()"+*-,
rhh (eeerevo r Ulfq;a'
@)'r;,'*
-z{L{r*irtLo>v+ ff-,f
Un,,
t
= [k)- t (h)'
) I
t
L _
- &'^pn, 1:h* /. witt unLios^ -{*^ i-/^ 1f
e' 'zs,- U (^) 4e ]-P ft
= (^) k a4 +^ v*
4uI ( =zt,tu?P+ry2q
4 / t g -
that (f )2 + (/r)' -^
( A'' t (0''' t;l
-il*) (^) rrru
4 ttu2f -;r',^r,-+^
\a /
T L{o/
'* {Usin
v) I t
\IATH 251- Summer2006 (^) \Iidterm 2
Question 4. At the point (1,2), the function ,f (t, y) has^ a derir,ativeof 2 in the directionto.,vard(2.2) and a derivativeof -2^ in the direction toward (1.1).
(u) i+lFind/"(i.2) andfo3,2)
= 1 4 t o ) = a -
a ,nL V { {-t,r)' (-/ J.
Ca,u)< 1
Latz) = (,
lL. vector fr"* (+,D^ fo^ Lz,z-)ts^ 12-at^2
+l"t '"r{o'^ f'""' ('tru)^ l-" (+,+) is^ 1a-ot
-t (^) ^A i
a.re "
,)f (^) v..(n.s r s^ o^ Y{^ Ca,z).
(^) )
) )7,-L",2)
L+)+ l-U,z)
(o)= u :\ )
i n U,z),Lo)t !-rt+,r)En)= ^z
'
7
-a)
<./^ n
-J lx-
4 TY
f h') l2l Find the riprivative\ (^) " / l ' l v v (^) v r (^) of. (^) fJ at (1,2) in the directiontoward the point (4,6)r r r r r r L u r r l v l r v l l L v v v a r u r r r q (^) P v r r r r \ *. u /. (^) Fr- (.) (^) V/ I
W
vecle".fr* Lap) f-^
(216),s (^) /.V-It6,-Z>=(
il- ,--"il u.rlor is^ 42t = (^) : ?L,r>^9 q Rt,u;l^ (^) @= T't "h Y{?r,z)-r?,?, = 1*? t "-:2?rrLr:
-7/
(1,t1=/e,zy
3r
L 5
(.) (^) [tL \ihat is the largest value that the derivative of f can have at the point (I.2)?
7l- &rt^ vqL-.-^ {r^ lV{[+,2)l= 2'+ *^ szE
J l t
NIATH 251- Summer2006 Nlidterm 2
Question 6.^ [7] Find the nraxirnurn:and^ minimum values of f (r,A,z)^ :^ 12 + y2 *.22 subject to lhe constraintsr i 29 * 3z :^ 6 and r * 3y * 9z :^ 9.
U,^g t^dtt^ Hr'(FiyLlr- ,)L- .!^-
2'r=*f
"Y
= 2)+]}.{.-
L5- =))*$f-
7+2y+37:L L=:
"-1]y
+?7:
z = t+f^)
1= At?v 3 = )?r-
tP'
q-,4,(tn* r')I 3(j)"3 r,)=(
a(^rr)
n U"?r)i 3(L^.U^ :^3
(^4) - l | e-L+o^ TS , - l v\ (^) {g r6J
2jf o^ L, (^) LL f3z{Z
b zr*9 3
C?s {
Y= I
1 d
T.tQ*r)
Y
3 = 3) * 7r'
7)+n"l*=
Vq)+44f=
X=
)=ry 13
g
53
= l^ r
4 a I -iq {
_
ft
rr=
+(L,r
fq
ui^-
i
4 L 9
ib/
r J
tu) =
-t (^) aL \ n t *^ 4 2 3 2^ 74 v7.l
r
t. z5 rt
)+ 8t
T
