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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 32 "Test 1 solutions (green \+
version)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
41 "1 (10 points).  Calculate the following:\n" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 4 "a)  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "ma
p(int,[t/(t^2+1), exp(-3*t)], t=1..2);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#7$,&-%#lnG6#\"\"&#\"\"\"\"\"#*&#F*F+F*-F&6#F+F*!\"\",&-%$expG6#!
\"'#F0\"\"$*&#F*F7F*-F36#!\"$F*F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
84 "This is the best frm of the answer, but in case you prefer numeric
al approximations:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(
%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$$\"+dOX\"e%!#5$\"+tQ%pd\"!#6
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 2 "b)" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 70 "with(linalg): vec := t -> crossprod([2*t, 4*t ,exp(
t) ], [t ,2*t,0 ]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$vecGf*6#%\"t
G6\"6$%)operatorG%&arrowGF(-%*crossprodG6$7%,$9$\"\"#,$F1\"\"%-%$expG6
#F17%F1F0\"\"!F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "map
(diff,vec(t),t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'vectorG6#7%,&*&
-%$expG6#%\"tG\"\"\"F,F-!\"#*&\"\"#F-F)F-!\"\",&F(F-F)F-\"\"!" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 252 "2 (10 points).  On the curve show
n below, clearly  indicate\n\na)  the unit tangent vector at the point
 P.  Label it T\n\nb) the principal unit normal vector at the point P.
  Label it N\n\nc) at least one point where the curvature is maximized
.  Label it M.\n" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 97 "For part a), the unit tangent vector can point in either \+
direction.  For part b), the unit normal" }}{PARA 0 "" 0 "" {TEXT -1 
99 "vector should point to the inside of the curve and have the same l
ength as T.  For part c, M should" }}{PARA 0 "" 0 "" {TEXT -1 31 "be w
here the curve is sharpest." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 118 "In problems 3 and 4, an object of mass 2 moves
 along the helix\n\n\011r(t) = 2 t^2 i - 3 cos(2 \271 t^2) j + 3 sin(2
 \271 t^2) k\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "r := t ->
 vector([2*t^2,-3*cos(2*Pi*t^2),3*sin(2*Pi*t^2)]);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%\"rGf*6#%\"tG6\"6$%)operatorG%&arrowGF(-%'vectorG6#
7%,$*$)9$\"\"#\"\"\"F4,$-%$cosG6#,$*&%#PiGF5F2F5F4!\"$,$-%$sinGF9\"\"$
F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 24 "First, let's look at it
:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "with(plots): spacecurv
e([2*cos(3*t^2),3*t^2,2*sin(3*t^2)], t=0..2, axes=NORMAL);" }}{PARA 
13 "" 1 "" {GLPLOT3D 349 262 262 {PLOTDATA 3 "6$-%'CURVESG6#7T7%$\"\"#
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/$\"+9q;**>!#6$\"+*pn!)*RF:7%$\"+Gq(z*>F/$\"+#yD\")\\%F:$\"+k\"=K**)F:
7%$\"+P(3O*>F/$\"+`!om*zF:$\"+?'Hwf\"!#57%$\"+8$3W)>F/$\"+$Qz%\\7FK$\"
+a:Y#\\#FK7%$\"+BUrn>F/$\"+6.D*z\"FK$\"+0j6zNFK7%$\"+TTKS>F/$\"+!fz*[C
FK$\"+&fY\"\\[FK7%$\"+pYb)*=F/$\"+>sm)>$FK$\"+64!))G'FK7%$\"+%pP$Q=F/$
\"+*>8$[SFK$\"+5WFxyFK7%$\"+kZOb<F/$\"+Hv\"z*\\FK$\"+%[b[e*FK7%$\"+v2H
X;F/$\"+5-[ZgFK$\"+5'4r8\"F/7%$\"+eo+/:F/$\"+U7+(>(FK$\"+)Q=$=8F/7%$\"
+h*))zK\"F/$\"+C1[Y%)FK$\"+[\\Z&\\\"F/7%$\"+gHs96F/$\"+c$=fz*FK$\"+*))
R0m\"F/7%$\"+\\Y)>j)FK$\"+W9`C6F/$\"+H48/=F/7%$\"+')z^WdFK$\"+()oYz7F/
$\"+Xds:>F/7%$\"+lsB@DFK$\"+m\")RW9F/$\"+kZ/%)>F/7%$!++8)>q*F:$\"+z_K>
;F/$\"+.ak(*>F/7%$!+qJuEYFK$\"+G#[U!=F/$\"+!GZd%>F/7%$!+a^y2$)FK$\"+6q
;**>F/$\"+9vG>=F/7%$!+yBf$=\"F/$\"+I;3/AF/$\"+k_<7;F/7%$!+Eo<+:F/$\"+%
3#**=CF/$\"+F^nA8F/7%$!+(Hrtv\"F/$\"+s$)*Qk#F/$\"+]h,[&*FK7%$!+0oLJ>F/
$\"+'\\+)yGF/$\"+'=db>&FK7%$!+1)z'**>F/$\"+d%)pBJF/$\"+f#['yNF:7%$!+4O
4W>F/$\"+bAfyLF/$!+'=Vdp%FK7%$!+*\\]Lv\"F/$\"+()=[VOF/$!+2Gl@'*FK7%$!+
XiNE9F/$\"+btO=RF/$!+8d'>S\"F/7%$!+cd`[(*FK$\"+e'[K?%F/$!+?xKY<F/7%$!+
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%oF/$\"+_ghg5F/7%$\"+4Zb*=\"F/$\"+uH*p@(F/$\"+a4y2;F/7%$\"+g;r*)\\FK$
\"+IE$=g(F/$\"+DqvO>F/7%$!+8i0WGFK$\"+@\"om*zF/$\"+&3v'z>F/7%$!+$**Q6/
\"F/$\"+[%*\\,%)F/$\"+.'Qwq\"F/7%$!+6%o5k\"F/$\"+5mK;))F/$\"+`[>V6F/7%
$!+r_Pm>F/$\"+2'\\6C*F/$\"+)Ri>l$FK7%$!+7WBP>F/$\"+S%ofn*F/$!+r]9r\\FK
7%$!+l#>[`\"F/$\"+6$y?,\"!\")$!+XUJ#G\"F/7%$!+%=c[:)FK$\"+h$fv0\"Fc_l$
!+RO>E=F/7%$\"+)HI@'*)F:$\"+&**RS5\"Fc_l$!+'*4*z*>F/7%$\"+PE9J**FK$\"+
7-_^6Fc_l$!+#33gt\"F/7%$\"+f!3xo\"F/$\"+8+++7Fc_l$!+8c9t5F/-%*AXESSTYL
EG6#%'NORMALG" 1 2 0 1 10 0 2 1 1 4 2 1.000000 45.000000 45.000000 0 
0 "Curve 1" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 57 "3 (10 points).  \n
\na)   The speed of the object at t= 3 is" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 43 "with(linalg): map(diff,r(t),t): norm(%, 2);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*$-%%sqrtG6#,(*$)-%$absG6#%\"tG\"\"#
\"\"\"F0*(\"\"*F0)%#PiGF/F0)-F,6#*&-%$sinG6#,$*&F4F0)F.F/F0F/F0F.F0F/F
0F0*(F2F0F3F0)-F,6#*&-%$cosGF;F0F.F0F/F0F0F0\"\"%" }}}{EXCHG {PARA 11 
"" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 19 "which sim
plifies to" }{TEXT 256 1 " " }{TEXT -1 37 "4 t sqrt(1+9*Pi^2).  At t=1
/2, it is " }{TEXT 257 16 "2 sqrt(1+9*Pi^2)" }{TEXT -1 1 "." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 112 "b)  Tthe lengt
h of the curve between the points (0,-3,0) and (4,-3 ,0) is the same a
s the curve from t=0 to t=2:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 30 "int(2*sqrt(1+9*Pi^2), t=0..2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#,$*$-%%sqrtG6#,&\"\"\"F)*&\"\"*F))%#PiG\"\"#F)F)F)\"\"%" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 62 "c)  As for the tangent line at t=1/sqrt(3
), it passes through " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "r(
1/sqrt(3));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'vectorG6#7%#\"\"#\"
\"$#F)F(,$*$-%%sqrtG6#F)\"\"\"F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
39 "and has direction given by the velocity" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 17 "map(diff,r(t),t);" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#-%'vectorG6#7%,$%\"tG\"\"%,$*(-%$sinG6#,$*&%#PiG\"\"\")F(\"\"#F2F4F
2F1F2F(F2\"#7,$*(-%$cosGF.F2F1F2F(F2F5" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 15 "at t= 1/sqrt(3)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 21 "subs(t=1/sqrt(3), %);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'vect
orG6#7%,$*$-%%sqrtG6#\"\"$\"\"\"#\"\"%F,,$*(-%$sinG6#,$%#PiG#\"\"#F,F-
F6F-F)F-F/,$*(-%$cosGF4F-F6F-F)F-F/" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 12 "simplify(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'ve
ctorG6#7%,$*$-%%sqrtG6#\"\"$\"\"\"#\"\"%F,,$%#PiG\"\"',$*&F1F-F)F-!\"#
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 17 "Hence the line is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 257 "" 0 "" {TEXT -1 84 "R(u) = (2/3 + 4 u/ sqrt(3)) i + (3/2 +
 6 \271 u) j + (3 sqrt(3)/2 - 2 sqrt(3) pi u) k. " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 258 1 "4" }{TEXT 
-1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 65 "a)  Find the force on the obj
ect in the moving coordinate system." }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 12 "There is no " }{TEXT 259 1 "B" }{TEXT 
-1 63 " component, because the force is just 2* acceleration, which is
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 "     \+
s''  " }{TEXT 260 1 "T" }{TEXT -1 12 " + (s')^2 k " }{TEXT 261 1 "N" }
{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 72 "hence we need to calculate the speed, its derivative, and
 the curvature." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 27 "We already know the speed, " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 33 "speed := t -> 4*t*sqrt(1+9*Pi^2);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%&speedGf*6#%\"tG6\"6$%)operatorG%&arrowGF(,$*&9$\"
\"\"-%%sqrtG6#,&F/F/*&\"\"*F/)%#PiG\"\"#F/F/F/\"\"%F(F(F(" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 135 "so the acceleration is  4*sqrt(1+9*Pi^2)
.   The only tricky part is the curvature, but since the curve is a he
lix, it won't be too bad." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
48 "crossprod(map(diff,r(t),t), map(diff,r(t),t$2));" }}{PARA 12 "" 1 
"" {XPPMATH 20 "6#-%'vectorG6#7%,&**-%$sinG6#,$*&%#PiG\"\"\")%\"tG\"\"
#F/F2F/F.F/F1F/,&*(F)F/)F.F2F/F0F/!#[*(\"#7F/-%$cosGF+F/F.F/F/F/F8*,F8
F/F9F/F.F/F1F/,&*(F9F/F5F/F0F/\"#[*(F8F/F)F/F.F/F/F/!\"\",&*(F9F/F.F/F
1F/F>*(\"\"%F/F1F/F3F/F@,&*&F1F/F<F/FD**F>F/F)F/F.F/F1F/F@" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "simplify(norm(%,2)/(4*t*sqrt(1+9*Pi
^2))^3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$**%#PiG\"\"#,(*&)F%F&\"
\"\")-%$absG6#%\"tG\"\"'F*\"\"*-F-6#*&,&!\"\"F**$)-%$cosG6#,$*&F%F*)F/
F&F*F&F&F*F*F*)F/F0F*F*-F-6#*&F8F*F?F*F*#F*F&F/!\"$,&F*F**&F1F*F)F*F*#
FDF&\"\"$" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 146 "Come on, Maple, can
't you see the common factor of t^3 in the numerator and denominator? \+
 And what about the cancellation of the cosines?  Sheesh!" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "curvature := 3*Pi^2/(1 + 9*Pi^2);" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%*curvatureG,$*&%#PiG\"\"#,&\"\"\"F
**&\"\"*F*)F'F(F*F*!\"\"\"\"$" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "T
he " }{TEXT 262 1 "T" }{TEXT -1 19 " component is      " }{TEXT 264 
16 "8*sqrt(1+9*Pi^2)" }{TEXT -1 12 "    and the " }{TEXT 263 1 "N" }
{TEXT -1 13 " component is" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
25 "2*speed(t)^2 * curvature;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&)
%\"tG\"\"#\"\"\")%#PiGF'F(\"#'*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 0 "" }}}}{MARK "0 0 0" 23 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }
{PAGENUMBERS 0 1 2 33 1 1 }
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