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{SECT 0 {PARA 3 "" 0 "" {TEXT -1 42 "Derivatives of trigonometric func
tions III" }}{PARA 0 "" 0 "" {TEXT -1 58 "by Peter Stone, School of Li
fe and Physical Sciences, RMIT" }}{PARA 0 "" 0 "" {TEXT -1 23 "peter.s
tone@rmit.edu.au" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 18 "Version: 15.8.2004" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 25 "lo
ad calculus procedures " }}{PARA 0 "" 0 "" {TEXT -1 35 "RMIT file path
 to read Maple m-file" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "rea
d \"J:\\\\Class_Notes/Peter Stone/MapleMath/procdrs/calculus.m\";" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT 
-1 18 "Another file path " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 
"read \"E:\\\\MapleMath/procdrs/calculus.m\";" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 1 ";" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";
" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 34 "Derivatives of tan, sec, csc
, cot " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 
{PARA 4 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "Diff([tan*x],x) = sec^2
*x" "6#/-%%DiffG6$7#*&%$tanG\"\"\"%\"xGF*F+*&%$secG\"\"#F+F*" }{TEXT 
-1 1 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 
0 "" {TEXT -1 4 "Let " }{XPPEDIT 18 0 "f(x) = tan*x;" "6#/-%\"fG6#%\"x
G*&%$tanG\"\"\"F'F*" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 5 "The
n " }{XPPEDIT 18 0 "f(x) = sin*x/(cos*x);" "6#/-%\"fG6#%\"xG*(%$sinG\"
\"\"F'F**&%$cosGF*F'F*!\"\"" }{TEXT -1 29 ",  so, by the quotient rule
: " }}{PARA 257 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "`f '`(x) = (Dif
f([sin*x],x)*`.`*cos*x-sin*x*`.`*Diff([cos*x],x))/(cos^2*x);" "6#/-%$f
~'G6#%\"xG*&,&**-%%DiffG6$7#*&%$sinG\"\"\"F'F1F'F1%\".GF1%$cosGF1F'F1F
1**F0F1F'F1F2F1-F,6$7#*&F3F1F'F1F'F1!\"\"F1*&)F3\"\"#F1F'F1F9" }{TEXT 
-1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT -1 
1 " " }{XPPEDIT 18 0 "`` = (cos*x*`.`*cos*x-sin*x*`.`*(-sin*x))/(cos^2
*x);" "6#/%!G*&,&*,%$cosG\"\"\"%\"xGF)%\".GF)F(F)F*F)F)**%$sinGF)F*F)F
+F),$*&F-F)F*F)!\"\"F)F0F)*&F(\"\"#F*F)F0" }{TEXT -1 1 " " }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 
0 "`` = (cos^2*x+sin^2*x)/(cos^2*x);" "6#/%!G*&,&*&%$cosG\"\"#%\"xG\"
\"\"F+*&%$sinGF)F*F+F+F+*&F(F)F*F+!\"\"" }{TEXT -1 1 " " }}{PARA 257 "
" 0 "" {TEXT -1 2 "  " }{XPPEDIT 18 0 "`` = 1/(cos^2*x);" "6#/%!G*&\"
\"\"F&*&%$cosG\"\"#%\"xGF&!\"\"" }{TEXT -1 1 " " }}{PARA 257 "" 0 "" 
{TEXT -1 2 "  " }{XPPEDIT 18 0 "`` = sec^2*x;" "6#/%!G*&%$secG\"\"#%\"
xG\"\"\"" }{TEXT -1 2 ", " }}{PARA 258 "" 0 "" {TEXT -1 7 "since  " }
{XPPEDIT 18 0 "sec*x=1/(cos*x)" "6#/*&%$secG\"\"\"%\"xGF&*&F&F&*&%$cos
GF&F'F&!\"\"" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 42 "The following picture shows the graph of \+
 " }{XPPEDIT 18 0 "f(x) = tan*x;" "6#/-%\"fG6#%\"xG*&%$tanG\"\"\"F'F*
" }{TEXT -1 4 " in " }{TEXT 260 3 "red" }{TEXT -1 20 " and its derivat
ive " }{XPPEDIT 18 0 "`f '`(x) = sec^2*x;" "6#/-%$f~'G6#%\"xG*&)%$secG
\"\"#\"\"\"F'F," }{TEXT -1 4 " in " }{TEXT 256 4 "blue" }{TEXT -1 1 ".
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT -1 1 " " }
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2 0 1 10 0 2 9 1 4 2 1.000000 46.000000 45.000000 0 0 "Curve 1" "Curve
 2" "Curve 3" "Curve 4" "Curve 5" "Curve 6" "Curve 7" }}{TEXT -1 1 " \+
" }}{PARA 0 "" 0 "" {TEXT -1 92 "The following animation shows the con
struction of one branch of the graph of the derivative." }}{PARA 0 "" 
0 "" {TEXT -1 26 "It requires the procedure " }{TEXT 0 9 "derivplot" }
{TEXT -1 24 " from another worksheet." }}{PARA 0 "" 0 "" {TEXT -1 109 
"To run the animation click on the graphic and use the controls in the
 context bar, or use the animation menu." }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 84 "derivplot(tan(x),x=-1
.5..1.5,-5..5,frames=50,\ntickmarks=[3,11],scaling=constrained);" }}
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" {TEXT -1 1 " " }{XPPEDIT 18 0 "Diff([sec*x],x)=sec*x*tan*x" "6#/-%%D
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{TEXT -1 4 "Let " }{XPPEDIT 18 0 "f(x) = sec*x;" "6#/-%\"fG6#%\"xG*&%$
secG\"\"\"F'F*" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 5 "Then " }
{XPPEDIT 18 0 "f(x) = 1/(cos*x);" "6#/-%\"fG6#%\"xG*&\"\"\"F)*&%$cosGF
)F'F)!\"\"" }{XPPEDIT 18 0 "`` = ``(cos*x)^(-1);" "6#/%!G)-F$6#*&%$cos
G\"\"\"%\"xGF*,$F*!\"\"" }{TEXT -1 27 "  , so, by the chain rule: " }}
{PARA 257 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "`f '`(x) = (-1)*(cos*
x)^(-2);" "6#/-%$f~'G6#%\"xG*&,$\"\"\"!\"\"F*)*&%$cosGF*F'F*,$\"\"#F+F
*" }{TEXT -1 1 " " }{XPPEDIT 18 0 "Diff([cos*x],x)" "6#-%%DiffG6$7#*&%
$cosG\"\"\"%\"xGF)F*" }{TEXT -1 1 " " }}{PARA 257 "" 0 "" {TEXT -1 1 "
 " }{XPPEDIT 18 0 "`` = (-1)*(cos*x)^(-2)*`.`*(-sin*x);" "6#/%!G**,$\"
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{TEXT -1 1 " " }}{PARA 257 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "`` =
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" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" 
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{PARA 257 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "`` = sec*x*tan*x;" "6
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0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 41 "The following pictur
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G*&%$secG\"\"\"F'F*" }{TEXT -1 4 " in " }{TEXT 260 3 "red" }{TEXT -1 
20 " and its derivative " }{XPPEDIT 18 0 "`f '`(x) = sec*x*tan*x;" "6#
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{TEXT 256 4 "blue" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
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{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 92 "The following animation \+
shows the construction of one branch of the graph of the derivative." 
}}{PARA 0 "" 0 "" {TEXT -1 26 "It requires the procedure " }{TEXT 0 9 
"derivplot" }{TEXT -1 24 " from another worksheet." }}{PARA 0 "" 0 "" 
{TEXT -1 109 "To run the animation click on the graphic and use the co
ntrols in the context bar, or use the animation menu." }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 84 "derivplot
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{TEXT -1 1 " " }{XPPEDIT 18 0 "`` = ``(-1/(sin*x))*`.`*``(cos*x/(sin*x
));" "6#/%!G*(-F$6#,$*&\"\"\"F**&%$sinGF*%\"xGF*!\"\"F.F*%\".GF*-F$6#*
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-1 0 "" }}{PARA 257 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "`` = -csc*x
*cot*x;" "6#/%!G,$**%$cscG\"\"\"%\"xGF(%$cotGF(F)F(!\"\"" }{TEXT -1 2 
". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 41 "Th
e following picture shows the graph of " }{XPPEDIT 18 0 "f(x) = csc*x;
" "6#/-%\"fG6#%\"xG*&%$cscG\"\"\"F'F*" }{TEXT -1 4 " in " }{TEXT 260 
3 "red" }{TEXT -1 20 " and its derivative " }{XPPEDIT 18 0 "`f '`(x) =
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45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" "Curve 4" "Curve
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{XPPEDIT 18 0 "Diff([cot*x],x)=-csc^2*x" "6#/-%%DiffG6$7#*&%$cotG\"\"
\"%\"xGF*F+,$*&%$cscG\"\"#F+F*!\"\"" }{TEXT -1 1 " " }}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 4 "Let " }
{XPPEDIT 18 0 "f(x) = cot*x;" "6#/-%\"fG6#%\"xG*&%$cotG\"\"\"F'F*" }
{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 5 "Then " }{XPPEDIT 18 0 "f(
x) = cos*x/(sin*x);" "6#/-%\"fG6#%\"xG*(%$cosG\"\"\"F'F**&%$sinGF*F'F*
!\"\"" }{TEXT -1 28 ", so, by the quotient rule: " }}{PARA 257 "" 0 "
" {TEXT -1 1 " " }{XPPEDIT 18 0 "`f '`(x) = (Diff([cos*x],x)*`.`*sin*x
-cos*x*`.`*Diff([sin*x],x))/(sin^2*x);" "6#/-%$f~'G6#%\"xG*&,&**-%%Dif
fG6$7#*&%$cosG\"\"\"F'F1F'F1%\".GF1%$sinGF1F'F1F1**F0F1F'F1F2F1-F,6$7#
*&F3F1F'F1F'F1!\"\"F1*&)F3\"\"#F1F'F1F9" }{TEXT -1 1 " " }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "
`` = ((-sin*x)*`.`*sin*x-cos*x*`.`*cos*x)/(sin^2*x);" "6#/%!G*&,&**,$*
&%$sinG\"\"\"%\"xGF+!\"\"F+%\".GF+F*F+F,F+F+*,%$cosGF+F,F+F.F+F0F+F,F+
F-F+*&F*\"\"#F,F+F-" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 257 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "`` = (-sin^2*x-cos^2
*x)/(sin^2*x);" "6#/%!G*&,&*&%$sinG\"\"#%\"xG\"\"\"!\"\"*&%$cosGF)F*F+
F,F+*&F(F)F*F+F," }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 257 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "`` = -1/(sin^2*x);" "
6#/%!G,$*&\"\"\"F'*&%$sinG\"\"#%\"xGF'!\"\"F," }{TEXT -1 1 " " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT -1 1 " " }
{XPPEDIT 18 0 "`` = -csc^2*x;" "6#/%!G,$*&%$cscG\"\"#%\"xG\"\"\"!\"\"
" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 41 "The following picture shows the graph of " }{XPPEDIT 18 
0 "f(x) = cot*x;" "6#/-%\"fG6#%\"xG*&%$cotG\"\"\"F'F*" }{TEXT -1 4 " i
n " }{TEXT 260 3 "red" }{TEXT -1 20 " and its derivative " }{XPPEDIT 
18 0 "`f '`(x) = -csc^2*x;" "6#/-%$f~'G6#%\"xG,$*&)%$cscG\"\"#\"\"\"F'
F-!\"\"" }{TEXT -1 4 " in " }{TEXT 256 4 "blue" }{TEXT -1 1 "." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT -1 1 " " }
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7$Ff]u$\"\"&Fhaq-Faaq6&FcaqFhaqFhaqFhaq-%*LINESTYLEG6#\"\"$-F$6%7$7$$
\"\"\"FhaqFg]u7$Ff^uFj]uF\\^uF^^u-F$6%7$7$$\"\"#FhaqFg]u7$F]_uFj]uF\\^
uF^^u-%%TEXTG6%7$$\"#GF`]r$FaesF`]rQ\"x6\"-%%FONTG6$%*HELVETICAG\"#5-F
a_u6%7$$Fh]uFaesFj]uQ\"yFh_uFi_u-Fj_u6$%'SYMBOLGF]`u-%*AXESTICKSG6$7*/
$Fd\\lF`]r%&-3p/2G/F`]r%#-pG/$Fh]uF`]r%%-p/2G/$F[^uF`]r%$p/2G/Fg^u%\"p
G/$\"#:F`]r%%3p/2G/F^_u%#2pG/$\"#DF`]r%%5p/2GF^_u-%+AXESLABELSG6%%!GFd
bu-Fj_u6#%(DEFAULTG-%%VIEWG6$;$F`zF`]rFd_u;Fg]uFj]u" 1 2 0 1 10 0 2 9 
1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" "
Curve 4" "Curve 5" "Curve 6" "Curve 7" }}{TEXT -1 1 " " }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 42 "We can check these formulas using Map
le's " }{TEXT 0 4 "diff" }{TEXT -1 3 ".\n " }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 212 "Diff(tan(x),x)= diff(tan(x),x);\n``= simplify(rhs(
%),\{1+tan(x)^2=sec(x)^2\});\nDiff(sec(x),x)=diff(sec(x),x);\nDiff(csc
(x),x)=diff(csc(x),x);\nDiff(cot(x),x)=diff(cot(x),x);\n``= simplify(r
hs(%),\{1+cot(x)^2=csc(x)^2\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%
%DiffG6$-%$tanG6#%\"xGF*,&\"\"\"F,*$)F'\"\"#F,F," }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/%!G*$)-%$secG6#%\"xG\"\"#\"\"\"" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/-%%DiffG6$-%$secG6#%\"xGF**&F'\"\"\"-%$tanGF)F," }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%%DiffG6$-%$cscG6#%\"xGF*,$*&F'\"\"
\"-%$cotGF)F-!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%%DiffG6$-%$co
tG6#%\"xGF*,&!\"\"\"\"\"*$)F'\"\"#F-F," }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#/%!G,$*$)-%$cscG6#%\"xG\"\"#\"\"\"!\"\"" }}}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 
1 {PARA 4 "" 0 "" {TEXT -1 9 "Examples " }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 9 "Example 1" }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT 
294 8 "Question" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 36 "Find \+
the derivative of the function " }{XPPEDIT 18 0 "f(x) = 1/(2*tan*x+cot
*x);" "6#/-%\"fG6#%\"xG*&\"\"\"F),&*(\"\"#F)%$tanGF)F'F)F)*&%$cotGF)F'
F)F)!\"\"" }{TEXT -1 26 ", and plot the graphs of  " }{XPPEDIT 18 0 "f
(x)" "6#-%\"fG6#%\"xG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "`f '`(x)" "
6#-%$f~'G6#%\"xG" }{TEXT -1 21 " in the same picture." }}{PARA 0 "" 0 
"" {TEXT -1 71 "Find the coordinates of the maximum and minimum points
 on the graph of " }{XPPEDIT 18 0 "f(x)" "6#-%\"fG6#%\"xG" }{TEXT -1 
15 " such that the " }{TEXT 299 1 "x" }{TEXT -1 47 " coordinates of th
ese points are between 0 and " }{XPPEDIT 18 0 "2*Pi" "6#*&\"\"#\"\"\"%
#PiGF%" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT 295 8 "Solution" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT 
-1 31 "We can find the derivative of  " }{XPPEDIT 18 0 "f(x) = 1/(2*ta
n*x+cot*x);" "6#/-%\"fG6#%\"xG*&\"\"\"F),&*(\"\"#F)%$tanGF)F'F)F)*&%$c
otGF)F'F)F)!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "``=(2*tan*x+cot*x)^(
-1)" "6#/%!G),&*(\"\"#\"\"\"%$tanGF)%\"xGF)F)*&%$cotGF)F+F)F),$F)!\"\"
" }{TEXT -1 25 " by using the chain rule." }}{PARA 257 "" 0 "" {TEXT 
-1 1 " " }{XPPEDIT 18 0 "`f '`(x) = (-1)*(2*tan*x+cot*x)^(-2);" "6#/-%
$f~'G6#%\"xG*&,$\"\"\"!\"\"F*),&*(\"\"#F*%$tanGF*F'F*F**&%$cotGF*F'F*F
*,$F/F+F*" }{TEXT -1 1 " " }{XPPEDIT 18 0 "Diff([2*tan*x+cot*x],x);" "
6#-%%DiffG6$7#,&*(\"\"#\"\"\"%$tanGF*%\"xGF*F**&%$cotGF*F,F*F*F," }
{TEXT -1 1 " " }}{PARA 257 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "``= \+
(-1)*(2*tan*x+cot*x)^(-2)*`.`*(2*sec^2*x-csc^2*x)" "6#/%!G**,$\"\"\"!
\"\"F'),&*(\"\"#F'%$tanGF'%\"xGF'F'*&%$cotGF'F.F'F',$F,F(F'%\".GF',&*(
F,F'*$%$secGF,F'F.F'F'*&%$cscGF,F.F'F(F'" }{TEXT -1 1 " " }}{PARA 257 
"" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "`` = (csc^2*x-2*sec^2*x)/((2*ta
n*x+cot*x)^2);" "6#/%!G*&,&*&%$cscG\"\"#%\"xG\"\"\"F+*(F)F+*$%$secGF)F
+F*F+!\"\"F+*$,&*(F)F+%$tanGF+F*F+F+*&%$cotGF+F*F+F+F)F/" }{TEXT -1 1 
"." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 13 "The
 graph of " }{XPPEDIT 18 0 "f(x)" "6#-%\"fG6#%\"xG" }{TEXT -1 15 " is \+
plotted in " }{TEXT 260 3 "red" }{TEXT -1 20 " and the derivative " }
{XPPEDIT 18 0 "`f '`(x) = g(x);" "6#/-%$f~'G6#%\"xG-%\"gG6#F'" }{TEXT 
-1 15 " is plotted in " }{TEXT 256 4 "blue" }{TEXT -1 1 "." }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 155 "f :
= x -> 1/(2*tan(x)+cot(x)):\n'f(x)'=f(x);\ng := x -> (csc(x)^2-2*sec(x
)^2)/(2*tan(x)+cot(x))^2:\n'g(x)'=g(x);\nplot([f(x),g(x)],x=-2..6,colo
r=[red,blue]);\n " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"fG6#%\"xG*&
\"\"\"F),&*&\"\"#F)-%$tanGF&F)F)-%$cotGF&F)!\"\"" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/-%\"gG6#%\"xG*&,&*$)-%$cscGF&\"\"#\"\"\"F/*&F.F/)-%$se
cGF&F.F/!\"\"F/,&*&F.F/-%$tanGF&F/F/-%$cotGF&F/!\"#" }}{PARA 13 "" 1 "
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^C@QF-7$Fccm$\"3eg_\\p*y1v%F-7$$\"3/++v.UacfF1$\"3[>,>C)\\8o&F-7$Fhcm$
\"3SH7`*QG))e'F--F]dm6&F_dmFcdmFcdmF`dm-%+AXESLABELSG6$Q\"x6\"Q!F`_o-%
%VIEWG6$;F(Fhcm%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 
45.000000 0 0 "Curve 1" "Curve 2" }}}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 "At the maximum and minimu
m points we have " }{XPPEDIT 18 0 "`f '`(x) = 0;" "6#/-%$f~'G6#%\"xG\"
\"!" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 18 "This occurs when \+
 " }{XPPEDIT 18 0 "csc^2*x-2*sec^2*x=0" "6#/,&*&%$cscG\"\"#%\"xG\"\"\"
F)*(F'F)*$%$secGF'F)F(F)!\"\"\"\"!" }{TEXT -1 16 ", that is when  " }
{XPPEDIT 18 0 "csc^2*x = 2*sec^2*x;" "6#/*&%$cscG\"\"#%\"xG\"\"\"*(F&F
(*$%$secGF&F(F'F(" }{TEXT -1 6 "  or  " }{XPPEDIT 18 0 "1/(sin^2*x)=2/
(cos^2*x)" "6#/*&\"\"\"F%*&%$sinG\"\"#%\"xGF%!\"\"*&F(F%*&%$cosGF(F)F%
F*" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 47 "Multiplying both s
ides of the last equation by " }{XPPEDIT 18 0 "sin^2*x;" "6#*&%$sinG\"
\"#%\"xG\"\"\"" }{TEXT -1 8 " gives: " }}{PARA 257 "" 0 "" {TEXT -1 1 
" " }{XPPEDIT 18 0 "1 = 2*tan^2*x;" "6#/\"\"\"*(\"\"#F$*$%$tanGF&F$%\"
xGF$" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 3 "or " }}{PARA 257 "
" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "tan^2*x=1/2" "6#/*&%$tanG\"\"#%
\"xG\"\"\"*&F(F(F&!\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 
12 "This gives: " }}{PARA 257 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "t
an*x=`` " "6#/*&%$tanG\"\"\"%\"xGF&%!G" }{TEXT 301 1 "+" }{TEXT -1 1 "
 " }{XPPEDIT 18 0 "1/sqrt(2)" "6#*&\"\"\"F$-%%sqrtG6#\"\"#!\"\"" }
{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 8 "so that " }{TEXT 302 1 "
x" }{TEXT -1 1 " " }{TEXT 303 1 "~" }{TEXT -1 21 "  0.6154797087  o r \+
 " }{TEXT 304 1 "x" }{TEXT -1 1 " " }{TEXT 305 1 "~" }{TEXT -1 18 " 2.
526112945  for " }{XPPEDIT 18 0 "0<x" "6#2\"\"!%\"xG" }{XPPEDIT 18 0 "
``<2*Pi" "6#2%!G*&\"\"#\"\"\"%#PiGF'" }{TEXT -1 2 ". " }}{PARA 0 "" 0 
"" {TEXT -1 41 "The corresponding points on the graph of " }{XPPEDIT 
18 0 "y=f(x)" "6#/%\"yG-%\"fG6#%\"xG" }{TEXT -1 85 " are approximately
 ( 0.6154797087, 0.3535533906 ) and ( 2.526112945, -0.3535533906 )." }
}{PARA 0 "" 0 "" {TEXT -1 112 "From the graph we see that the first of
 these two points is a maximum point, and the second is a minimum poin
t. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 121 "[solve((1-cot(x)^2+2*tan(x)^2))];\nmap(_u->`if`(sign
um(_u)>0,_u,_u+Pi),%);\nevalf(evalf(%,13));\nevalf(evalf(map(f,%%),13)
);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$-%'arctanG6#,$*&\"\"#!\"\"F)#
\"\"\"F)F,,$F$F*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$-%'arctanG6#,$*&
\"\"#!\"\"F)#\"\"\"F)F,,&F$F*%#PiGF," }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#7$$\"+(3(zah!#5$\"+XH6ED!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$$
\"+1R`NN!#5$!+1R`NNF&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" 
}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 259 4 "Note" 
}{TEXT -1 56 ": We could have determined the derivative using Maple's \+
" }{TEXT 0 4 "diff" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 60 "Th
e result looks a bit different from the one above because " }{TEXT 0 
4 "diff" }{TEXT -1 23 " gives the derivatives " }{XPPEDIT 18 0 "d/dx" 
"6#*&%\"dG\"\"\"%#dxG!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "[tan*x] = \+
1+tan^2*x;" "6#/7#*&%$tanG\"\"\"%\"xGF',&F'F'*&F&\"\"#F(F'F'" }{TEXT 
-1 5 " and " }{XPPEDIT 18 0 "d/dx" "6#*&%\"dG\"\"\"%#dxG!\"\"" }{TEXT 
-1 1 " " }{XPPEDIT 18 0 "[cot*x] = -1-cot^2*x;" "6#/7#*&%$cotG\"\"\"%
\"xGF',&F'!\"\"*&F&\"\"#F(F'F*" }{TEXT -1 3 " . " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "f := x -> 1/
(2*tan(x)+cot(x)):\n'f(x)'=f(x);\nDiff('f(x)',x)= diff(f(x),x);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"fG6#%\"xG*&\"\"\"F),&*&\"\"#F)-%$
tanGF&F)F)-%$cotGF&F)!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%%Diff
G6$-%\"fG6#%\"xGF*,$*&,&*&\"\"#\"\"\"-%$tanGF)F0F0-%$cotGF)F0!\"#,(F0F
0*&F/F0)F1F/F0F0*$)F3F/F0!\"\"F0F;" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 
"" {TEXT -1 9 "Example 2" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";
" }}}{PARA 0 "" 0 "" {TEXT 266 8 "Question" }{TEXT -1 2 ": " }}{PARA 
0 "" 0 "" {TEXT -1 36 "Find the derivative of the function " }
{XPPEDIT 18 0 "f(x) = x/(1+tan*x);" "6#/-%\"fG6#%\"xG*&F'\"\"\",&F)F)*
&%$tanGF)F'F)F)!\"\"" }{TEXT -1 25 ", and plot the graphs of " }
{XPPEDIT 18 0 "f(x)" "6#-%\"fG6#%\"xG" }{TEXT -1 5 " and " }{XPPEDIT 
18 0 "`f '`(x)" "6#-%$f~'G6#%\"xG" }{TEXT -1 5 " for " }{XPPEDIT 18 0 
"Pi/4 <x" "6#2*&%#PiG\"\"\"\"\"%!\"\"%\"xG" }{XPPEDIT 18 0 "``<3*Pi/4
" "6#2%!G*(\"\"$\"\"\"%#PiGF'\"\"%!\"\"" }{TEXT -1 22 "  in the same p
icture." }}{PARA 0 "" 0 "" {TEXT -1 62 "Estimate the coordinates of th
e maximum point on the graph of " }{XPPEDIT 18 0 "f(x)" "6#-%\"fG6#%\"
xG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT 267 8 "Solution" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 30 
"We can find the derivative of " }{XPPEDIT 18 0 "f(x) = x/(1+tan*x);" 
"6#/-%\"fG6#%\"xG*&F'\"\"\",&F)F)*&%$tanGF)F'F)F)!\"\"" }{TEXT -1 30 "
  by using the quotient rule. " }}{PARA 257 "" 0 "" {TEXT -1 1 " " }
{XPPEDIT 18 0 "`f '`(x) = (1*`.`*(1+tan*x)-x*`.`*sec^2*x)/((1+tan*x)^2
);" "6#/-%$f~'G6#%\"xG*&,&*(\"\"\"F+%\".GF+,&F+F+*&%$tanGF+F'F+F+F+F+*
*F'F+F,F+)%$secG\"\"#F+F'F+!\"\"F+*$),&F+F+*&F/F+F'F+F+F3F+F4" }{TEXT 
-1 2 "  " }}{PARA 257 "" 0 "" {TEXT -1 3 " = " }{XPPEDIT 18 0 "(1+tan*
x-x*sec^2*x)/((1+tan*x)^2);" "6#*&,(\"\"\"F%*&%$tanGF%%\"xGF%F%*(F(F%*
$%$secG\"\"#F%F(F%!\"\"F%*$,&F%F%*&F'F%F(F%F%F,F-" }{TEXT -1 2 ". " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 13 "The graph
 of " }{XPPEDIT 18 0 "f(x)" "6#-%\"fG6#%\"xG" }{TEXT -1 15 " is plotte
d in " }{TEXT 260 3 "red" }{TEXT -1 20 " and the derivative " }
{XPPEDIT 18 0 "`f '`(x) = g(x);" "6#/-%$f~'G6#%\"xG-%\"gG6#F'" }{TEXT 
-1 15 " is plotted in " }{TEXT 256 4 "blue" }{TEXT -1 1 "." }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 304 "f :
= x -> x/(1+tan(x)):\n'f(x)'=f(x);\ng := x -> (1+tan(x)-x*sec(x)^2)/((
1+tan(x))^2):\n'g(x)'=g(x);\np1 := plot([f(x),g(x)],x=-Pi/4..3*Pi/4,y=
-8..8,color=[red,blue]):\np2 := plot([[[-Pi/4,-8],[-Pi/4,8]],[[3*Pi/4,
-8],[3*Pi/4,8]]],\n   color=black,linestyle=3):\nplots[display]([p1,p2
],view=[-.79..2.37,-8..8]); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"f
G6#%\"xG*&F'\"\"\",&F)F)-%$tanGF&F)!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#/-%\"gG6#%\"xG*&,(\"\"\"F*-%$tanGF&F**&F'F*)-%$secGF&\"\"#F*!\"
\"F*,&F*F*F+F*!\"#" }}{PARA 13 "" 1 "" {GLPLOT2D 301 401 401 
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FVFV7$FenFen7$FjnFjn7$F_oF_o7$FdoFdo7$FioFio7$F^pF^p7$FcpFcp7$FhpFhp7$
F]qF]q7$FbqFbq7$FgqFgq7$F\\rF\\r7$FarFar7$FfrFfr7$F[sF[s7$F`sF`s7$FesF
es7$FjsFjs7$F_tF_t7$FdtFdt7$FitFit7$F^uF^u7$FcuFcu7$FhuFhu7$F]vF]v7$Fb
vFbv7$FgvFgv7$F\\wF\\w7$FawFaw7$FfwFfw7$F[xF[x7$F`xF`x7$FexFex7$FjxFjx
7$F_yF_y7$FdyFdy7$FiyFiy7$F^zF^z7$FczFcz-Fhz6&Fjz$\"*++++\"F][l$\"\"!F
i^lFf^l-%+AXESLABELSG6$Q\"x6\"Q!F^_l-%%VIEWG6$;$\"%y))!\"%$\"&&y))!\"&
%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "C
urve 1" "Curve 2" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 76 "By zooming in towards the point of intersection we see th
at it occurs where " }{TEXT 291 1 "x" }{TEXT -1 1 " " }{TEXT 290 1 "~
" }{TEXT -1 9 " 0.88783." }}{PARA 0 "" 0 "" {TEXT -1 19 "The correspon
ding  " }{TEXT 292 1 "y" }{TEXT -1 40 " coordinate is approximately f(
0.88783) " }{TEXT 293 1 "~" }{TEXT -1 13 " 0.39828.    " }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "f(0.887
83);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+%pUG)R!#5" }}}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 19 "We can use Maple's \+
" }{TEXT 0 6 "fsolve" }{TEXT -1 18 " to calculate the " }{TEXT 270 1 "
x" }{TEXT -1 46 " coordinate of the minimum point on the graph " }
{XPPEDIT 18 0 "y = f(x)" "6#/%\"yG-%\"fG6#%\"xG" }{TEXT -1 21 ", which
 occurs where " }{XPPEDIT 18 0 "`f '`(x) = 0;" "6#/-%$f~'G6#%\"xG\"\"!
" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 117 "f := x -> x/(1+tan(x)):\ng := x -> (1+tan(x)-
x*sec(x)^2)/((1+tan(x))^2):\nxmax := fsolve(g(x)=0,x=0.9);\n[xmax,f(xm
ax)];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%xmaxG$\"+S')Gy))!#5" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#7$$\"+S')Gy))!#5$\"+%pUG)RF&" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 87 "The coord
inates of the maximum point are approximately ( 0 .8878288640, 0.39828
42694 )." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 259 
4 "Note" }{TEXT -1 56 ": We could have determined the derivative using
 Maple's " }{TEXT 0 4 "diff" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT 
-1 60 "The result looks a bit different from the one above because " }
{TEXT 0 4 "diff" }{TEXT -1 22 " gives the derivative " }{XPPEDIT 18 0 
"d/dx" "6#*&%\"dG\"\"\"%#dxG!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "[ta
n*x] = 1+tan^2*x;" "6#/7#*&%$tanG\"\"\"%\"xGF',&F'F'*&F&\"\"#F(F'F'" }
{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 87 "f := x -> x/(1+tan(x)):\n'f(x)'=f(x);\nDiff('f(x
)',x)= diff(f(x),x);\n``=simplify(rhs(%));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/-%\"fG6#%\"xG*&F'\"\"\",&F)F)-%$tanGF&F)!\"\"" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%%DiffG6$-%\"fG6#%\"xGF*,&*&\"\"\"F-
,&F-F--%$tanGF)F-!\"\"F-*(F*F-F.!\"#,&F-F-*$)F/\"\"#F-F-F-F1" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#/%!G,$*&,*\"\"\"!\"\"-%$tanG6#%\"xGF)F-F(*&F
-F()F*\"\"#F(F(F(,&F(F(F*F(!\"#F)" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "
" {TEXT -1 9 "Example 3" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";
" }}}{PARA 0 "" 0 "" {TEXT 262 8 "Question" }{TEXT -1 2 ": " }}{PARA 
0 "" 0 "" {TEXT -1 207 "Find the length of the longest beam that can b
e carried horizontally around a right-angled corner from a passage of \+
width 2 metres to a passage of width 1 metre. (Assume that the beam ha
s negligible width.) " }}{PARA 257 "" 0 "" {TEXT -1 1 " " }{GLPLOT2D 
369 309 309 {PLOTDATA 2 "62-%'CURVESG6&7%7$$!\"%\"\"!$F*F*7$F+F+7$F+$!
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-%*LINESTYLEG6#\"\"\"-F$6&7%7$F($F<F*7$$F@F*FE7$FGF.F0F9F=-F$6&7$7$$!3
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F$6&7%7$$!3#*************\\OFO$\"33++++++]=FO7$$!33+++++++OFOFE7$$!3#)
************\\NFOFhn-F16&F3F*F*F*-F:F?F=-F$6&7%7$Ffn$\"3%*************
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O$\"3_V(3dylaf\"FO7$$!3aZ!y3CaX5#FO$\"3r*oK6D#e/;FO7$$!3-8ZKEyU*4#FO$
\"3t7*odhPPh\"FO7$$!3oQ&[DV!G%4#FO$\"3`4jwZVAB;FO7$$!3ka;)e(H&)*3#FO$
\"3m6$3)HrjJ;FO7$$!3i&4)4L())\\3#FO$\"3+kTW7,<T;FO7$$!3M9)*yqKB!3#FO$
\"3/XUp]o!3l\"FO7$$!3g@-QxLxv?FO$\"3QJ8=rK:g;FO7$$!3'fbt**[G=2#FO$\"3y
D'en4!po;FO7$$!3%yG,_kos1#FO$\"3'[#z2)p+*y;FO7$$!3SRh:Ny^j?FO$\"3/J'zC
e*f(o\"FO7$$!3)[?'*er%Gf?FO$\"39@aZ#[txp\"FO7$$!3M%[6\">\\lb?FO$\"3XjE
d7q#oq\"FO7$$!3?Xiuy(*z^?FO$\"3!zMo)f-\"or\"FO7$$!3OQMK8KD[?FO$\"3NPcC
-QOE<FO7$$!3!oZl&)y#oW?FO$\"3Z*)=]I#zjt\"FO7$$!3%=mhI`@:/#FO$\"37/T\"3
e<cu\"FO7$$!3&z<*fs$Q#Q?FO$\"3O]wTJ[ib<FO7$$!3.`n)Q*y'\\.#FO$\"3!eHg]P
kgw\"FO7$$!3%e8*R$QPA.#FO$\"3g(>NEr(=v<FO7$$!3Y]xNs0TH?FO$\"3u_rT6p2&y
\"FO7$$!36Q#G6(RiE?FO$\"3'H9R2rI`z\"FO7$$!34lxP\\)HS-#FO$\"37B8#e+(R0=
FO7$$!34<8E3Yk@?FO$\"3uhS\"p&z;:=FO7$$!3a'HU'f.9>?FO$\"3+-[`$4^g#=FO7$
$!3:#3Xl&*>q,#FO$\"3*[!Gu\\!fe$=FO7$$!3lZS`j>*[,#FO$\"3'G*=n/'fj%=FO7$
$!3,4Rp)G&38?FO$\"3%QPYz^)*e&=FO7$$!36n$eKtU7,#FO$\"3%H\"y<.8Nm=FO7$$!
3h%o/g!ej4?FO$\"35JUPu!3i(=FO7$$!367yWsx33?FO$\"3101rP=`')=FO7$$!390=(
zk/n+#FO$\"3Y8J&\\*Qk'*=FO7$$!3CwhN#*\\R0?FO$\"3g&ou&y#[s!>FO7$$!3\"\\
%QIQwE/?FO$\"33ki0CsZ<>FO7$$!3xK\"H^$4D.?FO$\"3'o4Qv!>&z#>FO7$$!3#e*[y
?)yB+#FO$\"3luA'Q<_$Q>FO7$$!3e![%H>rp,?FO$\"3?Uq/J#=z%>FO7$$!3M'H<*4p0
,?FO$\"3em`mM6*)e>FO7$$!3(>pu$o?h+?FO$\"3Y0Snm?ro>FO7$$!3N`tdZ2F+?FO$
\"3-_z*pR)=z>FO7$$!3*p$ymZE2+?FO$\"3ukvB_*=#*)>FO7$$!3y**************>
FOFE-F16&F3$\")!\\DP\"F6F`al$\")viobF6-%%TEXTG6%7$$!#J!\"\"FGQ)2~metre
s6\"F`o-Feal6%7$$\"\"&Fjal$!$D#FepQ(1~metreF\\blF`o-Feal6%7$$!$H#Fep$
\"$(=FepQ\"tF\\blF`o-%+AXESLABELSG6%Q!6\"F`cl%(DEFAULTG-%*AXESSTYLEG6#
%%NONEG-%%VIEWG6$FbclFbcl" 1 2 0 1 10 0 2 9 1 1 2 1.000000 45.000000 
45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" "Curve 4" "Curve 5" "Curve
 6" "Curve 7" "Curve 8" "Curve 9" "Curve 10" "Curve 11" "Curve 12" "Cu
rve 13" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 263 
8 "Solution" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 192 "The prob
lem is the same as that of determining the shortest length of a line s
egment which pivots around the inside corner, and has its end points t
ouching the outer walls of the two passages." }}{PARA 0 "" 0 "" {TEXT 
-1 82 "Let the angle which the line segment (or beam) makes with the n
arrower passage be " }{TEXT 274 1 "t" }{TEXT -1 26 ", as shown in the \+
picture." }}{PARA 0 "" 0 "" {TEXT -1 38 "Then the length of the line s
egment is" }}{PARA 257 "" 0 "" {TEXT -1 2 "  " }{XPPEDIT 18 0 "x = 2/(
sin*t)+1/(cos*t);" "6#/%\"xG,&*&\"\"#\"\"\"*&%$sinGF(%\"tGF(!\"\"F(*&F
(F(*&%$cosGF(F+F(F,F(" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 8 "
that is," }}{PARA 257 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "x = 2*csc
*t+sec*t;" "6#/%\"xG,&*(\"\"#\"\"\"%$cscGF(%\"tGF(F(*&%$secGF(F*F(F(" 
}{TEXT -1 1 "." }}{PARA 257 "" 0 "" {TEXT -1 1 " " }{GLPLOT2D 341 204 
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*>_^[:Qn!#?7$$\"3++++++++]FAF/Fj\\l-%%TEXTG6%7$$\"#8!\"#F=Q\"26\"F1-Fi
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F`]mF1-Fi\\m6%7$$\"#RF^]m$!#7F^]mFa^mF1-Fi\\m6%7$$!#;Ff]m$\"#vF^]mQ*2~
cosec~tF`]mFE-Fi\\m6%7$$\"#GF^]m$!#ZF^]mQ&sec~tF`]mFE-%+AXESLABELSG6%Q
!F`]mF\\`m-%%FONTG6#%(DEFAULTG-%*AXESSTYLEG6#%%NONEG-%%VIEWG6$F``mF``m
" 1 2 0 1 10 0 2 9 1 1 2 1.000000 46.000000 45.000000 0 0 "Curve 1" "C
urve 2" "Curve 3" "Curve 4" "Curve 5" "Curve 6" "Curve 7" "Curve 8" "C
urve 9" "Curve 10" "Curve 11" }}{TEXT -1 1 " " }}{PARA 0 "" 0 "" 
{TEXT -1 33 "Plotting the graph of the length " }{XPPEDIT 18 0 "x = 2*
csc*t+sec*t;" "6#/%\"xG,&*(\"\"#\"\"\"%$cscGF(%\"tGF(F(*&%$secGF(F*F(F
(" }{TEXT -1 9 " against " }{TEXT 306 1 "t" }{TEXT -1 55 " shows that \+
the length has a minimum for some value of " }{TEXT 273 1 "t" }{TEXT 
-1 15 " between 0 and " }{XPPEDIT 18 0 "Pi/2" "6#*&%#PiG\"\"\"\"\"#!\"
\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 40 "plot(2*csc(t)+sec(t),t=0..Pi/2,x=0..10);" }}
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\\-5?1P(o)*!#<7$$\"31+++vK)[h#F^p$\"3u$H@60y:x)F`q7$$\"35+++.W0ZHF^p$
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i%yF^p$\"3\\w*Qh*4uVUF`q7$$\"35+++W&[D>)F^p$\"3>F%Q;Sn>?%F`q7$$\"3=+++
08-%\\)F^p$\"3-\"\\P9D\"oxTF`q7$$\"35+++Y,i>))F^p$\"3uTEkM(QR;%F`q7$$
\"3a+++4c*f:*F^p$\"3E([3[V$[jTF`q7$$\"3Q+++uL2&[*F^p$\"3q(fz0Aop<%F`q7
$$\"3t*****H.tM!)*F^p$\"3LLD0VW#Q?%F`q7$$\"3%******p:+d,\"F`q$\"3a#eGm
@+3D%F`q7$$\"3)******zEmu/\"F`q$\"33\"RaYHu*4VF`q7$$\"3!******>P$Q\"3
\"F`q$\"3COtJ?1=$R%F`q7$$\"3\"*******4t676F`q$\"3O>zegM4*[%F`q7$$\"3'*
*****4i<d9\"F`q$\"3\"yBgZ+\\-i%F`q7$$\"3-+++'*GLx6F`q$\"3KA!f_>%ytZF`q
7$$\"3%******p'*z.@\"F`q$\"3Q(Gs<1%)G(\\F`q7$$\"36+++?&*oU7F`q$\"3nO!H
?Dre@&F`q7$$\"3%******zi9lF\"F`q$\"3oNx++C\\PbF`q7$$\"30+++FA448F`q$\"
3=pw#fKxb$fF`q7$$\"3*******p_2CM\"F`q$\"3/))[8Le6qkF`q7$$\"3-+++fpWv8F
`q$\"3%e8)>6_^!>(F`q7$$\"3*******fw1eS\"F`q$\"3$Q,__84h6)F`q7$$\"3\"**
****4`-1W\"F`q$\"3'p'y)*y/n>(*F`q7$$\"3%******zFC<Z\"F`q$\"3G)>Q)y]+77
Ffn7$$\"3,+++_l!\\]\"F`q$\"3?X!=J#)>#><Ffn7$$\"3-+++(Q'y?:F`q$\"39&H'y
*e'o+AFfn7$$\"3.+++AimO:F`q$\"3<OVD*3]18$Ffn7$$\"3-+++[()>X:F`q$\"3(Hw
]>o6r5%Ffn7$$\"3#******HFJPb\"F`q$\"3?u)z-MU-1'Ffn7$$\"3-+++Ov*zb\"F`q
$\"35*>K#4yY8!)Ffn7$$\"3/+++)zjAc\"F`q$\"3MEK-Y/+#>\"F77$$\"3-+++IpRk:
F`q$\"3%)RV*fNeEe\"F77$$\"3#******41Ilc\"F`q$\"39]4g=v(RO#F77$$\"3/+++
Emfn:F`q$\"37_\"H]l(HXJF77$$\"3/+++#>j'o:F`q$\"3_cT,$))Rzq%F77$$\"3/++
+e(H(p:F`q$\"3U+'H!p?'eR*F77$$\"3+++lBjzq:F`q$\"3q<;@uZvzJ!\"*-%'COLOU
RG6&%$RGBG$\"#5!\"\"F(F(-%+AXESLABELSG6$Q\"t6\"Q\"xF\\bl-%%VIEWG6$;F($
\"+Fjzq:F`al;F($FfalF)" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 
45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";
" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 41 "This
 minimum occurs where the derivative " }{XPPEDIT 18 0 "dx/dt" "6#*&%#d
xG\"\"\"%#dtG!\"\"" }{TEXT -1 9 " is zero." }}{PARA 0 "" 0 "" {TEXT 
-1 3 "Now" }}{PARA 257 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "dx/dt = \+
-2*csc*t*cot*t+sec*t*tan*t;" "6#/*&%#dxG\"\"\"%#dtG!\"\",&*,\"\"#F&%$c
scGF&%\"tGF&%$cotGF&F-F&F(**%$secGF&F-F&%$tanGF&F-F&F&" }{TEXT -1 1 " \+
" }}{PARA 257 "" 0 "" {TEXT -1 2 "  " }{XPPEDIT 18 0 "`` = -2*cos*t/(s
in^2*t)+sin*t/(cos^2*t);" "6#/%!G,&**\"\"#\"\"\"%$cosGF(%\"tGF(*&%$sin
GF'F*F(!\"\"F-*(F,F(F*F(*&F)F'F*F(F-F(" }{TEXT -1 1 " " }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "
`` = (-2*cos^3*t+sin^3*t)/(sin^2*t*cos^2*t);" "6#/%!G*&,&*(\"\"#\"\"\"
*$%$cosG\"\"$F)%\"tGF)!\"\"*&%$sinGF,F-F)F)F)**F0F(F-F)F+F(F-F)F." }
{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" 
{TEXT -1 1 " " }{XPPEDIT 18 0 "`` = (sin^3*t-2*cos^3*t)/(sin^2*t*cos^2
*t);" "6#/%!G*&,&*&%$sinG\"\"$%\"tG\"\"\"F+*(\"\"#F+*$%$cosGF)F+F*F+!
\"\"F+**F(F-F*F+F/F-F*F+F0" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT 
-1 1 " " }{XPPEDIT 18 0 "dx/dt=0" "6#/*&%#dxG\"\"\"%#dtG!\"\"\"\"!" }
{TEXT -1 6 " when " }}{PARA 257 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 
"sin^3*t = 2*cos^3*t;" "6#/*&%$sinG\"\"$%\"tG\"\"\"*(\"\"#F(*$%$cosGF&
F(F'F(" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT -1 14 "that is, when \+
" }}{PARA 257 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "tan^3*t = 2;" "6#
/*&%$tanG\"\"$%\"tG\"\"\"\"\"#" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" 
{TEXT -1 2 "or" }}{PARA 257 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "tan
*t = 2^(1/3);" "6#/*&%$tanG\"\"\"%\"tGF&)\"\"#*&F&F&\"\"$!\"\"" }
{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 20 " Hence the value of " }{TEXT 307 1 "t" }{TEXT -1 58 " whi
ch makes the length of the line segment a minimum is: " }}{PARA 257 "
" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "t[min] = arctan(2^(1/3));" "6#/&
%\"tG6#%$minG-%'arctanG6#)\"\"#*&\"\"\"F.\"\"$!\"\"" }{TEXT -1 2 ". " 
}}{PARA 0 "" 0 "" {TEXT -1 4 "Now " }}{PARA 257 "" 0 "" {TEXT -1 1 " \+
" }{XPPEDIT 18 0 "sec(t[min]) = sqrt(1+tan^2*``(t[min]));" "6#/-%$secG
6#&%\"tG6#%$minG-%%sqrtG6#,&\"\"\"F/*&%$tanG\"\"#-%!G6#&F(6#F*F/F/" }
{XPPEDIT 18 0 "`` = sqrt(1+2^(2/3));" "6#/%!G-%%sqrtG6#,&\"\"\"F))\"\"
#*&F+F)\"\"$!\"\"F)" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 3 "an
d" }}{PARA 257 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "csc(t[min]) = se
c(t[min])/tan(t[min]);" "6#/-%$cscG6#&%\"tG6#%$minG*&-%$secG6#&F(6#F*
\"\"\"-%$tanG6#&F(6#F*!\"\"" }{XPPEDIT 18 0 "`` = sqrt(1+2^(2/3))/(2^(
1/3));" "6#/%!G*&-%%sqrtG6#,&\"\"\"F*)\"\"#*&F,F*\"\"$!\"\"F*F*)F,*&F*
F*F.F/F/" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 42 "The minimum \+
length of the line segment is " }}{PARA 257 "" 0 "" {TEXT -1 2 "  " }
{XPPEDIT 18 0 "2*csc(t[min])+sec(t[min]);" "6#,&*&\"\"#\"\"\"-%$cscG6#
&%\"tG6#%$minGF&F&-%$secG6#&F+6#F-F&" }{TEXT -1 1 " " }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "`` \+
= 2*sqrt(1+2^(2/3))/(2^(1/3))+sqrt(1+2^(2/3));" "6#/%!G,&*(\"\"#\"\"\"
-%%sqrtG6#,&F(F()F'*&F'F(\"\"$!\"\"F(F()F'*&F(F(F/F0F0F(-F*6#,&F(F()F'
*&F'F(F/F0F(F(" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 257 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "`` = sqrt(1+2^(2/3))*
``(2^(2/3)+1);" "6#/%!G*&-%%sqrtG6#,&\"\"\"F*)\"\"#*&F,F*\"\"$!\"\"F*F
*-F$6#,&)F,*&F,F*F.F/F*F*F*F*" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "`` = \+
``(2^(2/3)+1)^(3/2);" "6#/%!G)-F$6#,&)\"\"#*&F*\"\"\"\"\"$!\"\"F,F,F,*
&F-F,F*F." }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT -1 77 "The length of the longest beam which can be negot
iated around the corner is: " }}{PARA 257 "" 0 "" {TEXT -1 1 " " }
{XPPEDIT 18 0 "[2^(2/3)+1]^(3/2)" "6#)7#,&)\"\"#*&F'\"\"\"\"\"$!\"\"F)
F)F)*&F*F)F'F+" }{TEXT -1 8 " metres " }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 257 "" 0 "" {TEXT -1 1 " " }{TEXT 264 1 "~" }{TEXT -1 21 "  \+
4.161938184 metres." }}{PARA 257 "" 0 "" {TEXT -1 2 "  " }{TEXT 265 
12 "____________" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}
{PARA 0 "" 0 "" {TEXT -1 37 "Maple can perform these calculations." }}
{PARA 0 "" 0 "" {TEXT 259 4 "Note" }{TEXT -1 50 ": We need to remove t
he solutions of the equation " }{XPPEDIT 18 0 "dx/dt = 0" "6#/*&%#dxG
\"\"\"%#dtG!\"\"\"\"!" }{TEXT -1 35 ", which involve the imaginary uni
t " }{TEXT 0 1 "I" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 98 "Sinc
e the symbol for the imaginary unit can be changed, it is perhaps bett
er to refer to it using " }{TEXT 0 10 "Complex(1)" }{TEXT -1 1 "." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
158 "f := t -> 2*csc(t)+sec(t):\n'f(t)'=f(t);\ndf := D(f):\nDiff('f(t)
',t)=df(t);\n[solve(df(t)=0)]:\ntmin := op(remove(has,%,Complex(1)));
\nsimplify(f(tmin));\nevalf(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%
\"fG6#%\"tG,&*&\"\"#\"\"\"-%$cscGF&F+F+-%$secGF&F+" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#/-%%DiffG6$-%\"fG6#%\"tGF*,&*(\"\"#\"\"\"-%$cscGF)F.-
%$cotGF)F.!\"\"*&-%$secGF)F.-%$tanGF)F.F." }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%%tminG-%'arctanG6#*$)\"\"##\"\"\"\"\"$F," }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#,&*&,&\"\"\"F&*$)\"\"##F)\"\"$F&F&#F&F)F(F&F&*$
F%F,F&" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+%=Q>;%!\"*" }}}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 28 "The graph of the d
erivative " }{XPPEDIT 18 0 "`f '`(t) = -2*csc*t*cot*t+sec*t*tan*t;" "6
#/-%$f~'G6#%\"tG,&*,\"\"#\"\"\"%$cscGF+F'F+%$cotGF+F'F+!\"\"**%$secGF+
F'F+%$tanGF+F'F+F+" }{TEXT -1 28 "  can be plotted as follows." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
102 "f := t -> 2*csc(t)+sec(t):\ndf := D(f):\nDiff('f(t)',t)=df(t);\np
lot(df(t),t=0..Pi/2,-10..10,color=blue);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#/-%%DiffG6$-%\"fG6#%\"tGF*,&*(\"\"#\"\"\"-%$cscGF)F.-%$cotGF)F.!
\"\"*&-%$secGF)F.-%$tanGF)F.F." }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 
300 {PLOTDATA 2 "6&-%'CURVESG6#7`o7$$\"35+++,`'*p5!#?$!3q)fMlU!*pu\"!#
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$\"3-+++fpWv8Fbs$\"35J;fqz@jDFdp7$$\"3*******fw1eS\"Fbs$\"3>p]!*4T(Hi$
Fdp7$$\"3\"******4`-1W\"Fbs$\"37->1M5QceFdp7$$\"3%******zFC<Z\"Fbs$\"3
O+=>Q/::5F^o7$$\"3,+++_l!\\]\"Fbs$\"38P\"zf`z.I#F^o7$$\"3-+++(Q'y?:Fbs
$\"3_a2:%pRd*RF^o7$$\"3.+++AimO:Fbs$\"3e`ENW@M#e)F^o7$$\"3-+++[()>X:Fb
s$\"3k(fE$\\'\\f_\"FT7$$\"3#******HFJPb\"Fbs$\"3&*3V*yrvOV$FT7$$\"3-++
+Ov*zb\"Fbs$\"3K.L)*)pwW5'FT7$$\"3/+++)zjAc\"Fbs$\"3K<0QJ;`t8FC7$$\"3-
+++IpRk:Fbs$\"3W&)Q@x&[=W#FC7$$\"3#******41Ilc\"Fbs$\"3Y,3&4!*yT\\&FC7
$$\"3/+++Emfn:Fbs$\"3#)HJ^_TVn(*FC7$$\"3/+++#>j'o:Fbs$\"3[BTM\\Hn(>#F3
7$$\"3/+++e(H(p:Fbs$\"3U%f#36Rn!z)F37$$\"3+++lBjzq:Fbs$\"3cQJpHS365\"
\"!-%'COLOURG6&%$RGBG$F_alF_alFdal$\"*++++\"!\")-%+AXESLABELSG6$Q\"t6
\"Q!F\\bl-%%VIEWG6$;Fdal$\"+Fjzq:!\"*;$!#5F_al$\"#5F_al" 1 2 0 1 10 0 
2 6 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 16 "We can find the " }
{TEXT 272 1 "x" }{TEXT -1 64 " intercept of this graph by a numerical \+
root-finding method via " }{TEXT 0 6 "fsolve" }{TEXT -1 65 ", and use \+
this value to calculate the maximum length of the beam." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "tmin :=
 fsolve(df,0.9);\nf(tmin);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%tminG
$\"+\"[$3***)!#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+&=Q>;%!\"*" }}
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 1 ";" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 5 "Tasks" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "\004;" }}}{SECT 1 {PARA 4 "" 
0 "" {TEXT -1 3 "Q1 " }}{PARA 0 "" 0 "" {TEXT -1 20 "Find the derivati
ve " }{XPPEDIT 18 0 "`f '`(x)" "6#-%$f~'G6#%\"xG" }{TEXT -1 17 " of th
e function " }{XPPEDIT 18 0 "f(x) = x/(1+sec*x);" "6#/-%\"fG6#%\"xG*&F
'\"\"\",&F)F)*&%$secGF)F'F)F)!\"\"" }{TEXT -1 25 ", and plot the graph
s of " }{XPPEDIT 18 0 "f(x)" "6#-%\"fG6#%\"xG" }{TEXT -1 5 " and " }
{XPPEDIT 18 0 "`f '`(x)" "6#-%$f~'G6#%\"xG" }{TEXT -1 5 " for " }
{XPPEDIT 18 0 "-2<=x" "6#1,$\"\"#!\"\"%\"xG" }{XPPEDIT 18 0 "``<=2" "6
#1%!G\"\"#" }{TEXT -1 22 " in the same picture. " }}{PARA 0 "" 0 "" 
{TEXT -1 10 "The graph " }{XPPEDIT 18 0 "y = f(x)" "6#/%\"yG-%\"fG6#%
\"xG" }{TEXT -1 50 " appears to have minimum and maximum points where \+
" }{XPPEDIT 18 0 "x = -1" "6#/%\"xG,$\"\"\"!\"\"" }{TEXT -1 5 " and " 
}{XPPEDIT 18 0 "x = 1" "6#/%\"xG\"\"\"" }{TEXT -1 14 " respectively." 
}}{PARA 0 "" 0 "" {TEXT -1 111 "Determine the coordinates of the maxim
um and minimum points to about 10 digit accuracy to show that the valu
es " }{XPPEDIT 18 0 "x = -1" "6#/%\"xG,$\"\"\"!\"\"" }{TEXT -1 5 " and
 " }{XPPEDIT 18 0 "x = 1" "6#/%\"xG\"\"\"" }{TEXT -1 33 " are only app
roximations for the " }{TEXT 277 1 "x" }{TEXT -1 36 " coordinates of t
he turning points. " }}{PARA 0 "" 0 "" {TEXT -1 42 "__________________
________________________" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
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{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "
" {TEXT -1 2 "Q2" }}{PARA 0 "" 0 "" {TEXT -1 208 "Find the length of t
he longest beam that can be carried horizontally around a right-angled
 corner from a passage of width 3 metres to a passage of width 2 metre
s. (Assume that the beam has negligible width.) " }}{PARA 0 "" 0 "" 
{TEXT -1 42 "__________________________________________" }}{PARA 0 "" 
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 " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 42 "____________
______________________________" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 1 ";" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}
{SECT 1 {PARA 0 "" 0 "" {TEXT -1 18 "Code for pictures " }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 258 "" 0 "" 
{TEXT -1 49 "Code for graph of tangent function and derivative" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
410 "p1 := plot([tan(t*Pi),sec(t*Pi)^2],t=-1.15..2.15,-5..5,discont=tr
ue,\n   color=[red,blue]):\np2 := plot([[[-1/2,-5],[-1/2,5]],[[1/2,-5]
,[1/2,5]],\n   [[3/2,-5],[3/2,5]]],linestyle=3,color=black):\nt1 := pl
ots[textplot]([[2.15,-.3,`x`],[-.05,5,`y`]],font=[HELVETICA,10]):\nplo
ts[display]([p1,p2,t1],xtickmarks=[-1=`-p`,-.5=`-p/2`,.5=`p/2`,1=`p`,1
.5=`3p/2`,\n  2=`2p`],font=[SYMBOL,10],\n  ytickmarks=2,labels=[``,``]
);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 258 "" 0 "" {TEXT -1 48 "Code for
 graph of secant function and derivative" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 448 "p1 := plot([sec(t*Pi
),sec(t*Pi)*tan(t*Pi)],t=-1.15..3.2,-5..5,discont=true,\ncolor=[red,bl
ue]):\np2 := plot([[[-1/2,-5],[-1/2,5]],[[1/2,-5],[1/2,5]],\n   [[3/2,
-5],[3/2,5]],[[5/2,-5],[5/2,5]]],linestyle=3,color=black):\nt1 := plot
s[textplot]([[3.2,-.2,`x`],[-.05,5,`y`]],font=[HELVETICA,10]):\nplots[
display]([p1,p2,t1],xtickmarks=[-1=`-p`,-.5=`-p/2`,.5=`p/2`,1=`p`,1.5=
`3p/2`,\n   2=`2p`,2.5=`5p/2`,3=`3p`],font=[SYMBOL,10],ytickmarks=2,la
bels=[``,``]);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 258 "" 0 "" {TEXT -1 51 "C
ode for graph of cosecant function and derivative " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 418 "p1 := plot(
[csc(t*Pi),-csc(t*Pi)*cot(t*Pi)],t=-1.6..2.8,-5..5,\n   discont=true,c
olor=[red,blue]):\nt1 := plots[textplot]([[2.8,-.2,`x`],[-.05,5,`y`]],
font=[HELVETICA,10]):\np2 := plot([[[-1,-5],[-1,5]],[[1,-5],[1,5]],[[2
,-5],[2,5]]],linestyle=3,color=black):\nplots[display]([p1,p2,t1],xtic
kmarks=[-1.5=`-3p/2`,-1=`-p`,-.5=`-p/2`,.5=`p/2`,1=`p`,1.5=`3p/2`,2=`2
p`,2.5=`5p/2`],font=[SYMBOL,10],ytickmarks=2,labels=[``,``]);" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
1 ";" }}}}{SECT 1 {PARA 258 "" 0 "" {TEXT -1 52 "Code for graph of cot
angent function and derivative " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 410 "p1 := plot([cot(t*Pi),-csc(
t*Pi)^2],t=-1.6..2.8,-5..5,discont=true,\n   color=[red,blue]):\np2 :=
 plot([[[-1,-5],[-1,5]],[[1,-5],[1,5]],[[2,-5],[2,5]]],linestyle=3,col
or=black):\nt1 := plots[textplot]([[2.8,-.2,`x`],[-.05,5,`y`]],font=[H
ELVETICA,10]):\nplots[display]([p1,p2,t1],xtickmarks=[-1.5=`-3p/2`,-1=
`-p`,-.5=`-p/2`,.5=`p/2`,1=`p`,1.5=`3p/2`,2=`2p`,2.5=`5p/2`],font=[SYM
BOL,10],ytickmarks=2,labels=[``,``]);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "n;" }}}}{SECT 1 {PARA 
258 "" 0 "" {TEXT -1 22 "Code for beam pictures" }}{PARA 258 "" 0 "" 
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 588 "with(plots)
:\np1 := plot([[[-4,0],[0,0],[0,-3]],[[-4,2],[1,2],[1,-3]],\n     [[-1
4/5,2],[1,-5/7]],[[-3.65,1.85],[-3.6,2],[-3.55,1.85]],\n     [[-3.65,.
15],[-3.6,0],[-3.55,.15]],\n     [[.12,-1.95],[0,-2],[.12,-2.05]],\n  \+
   [[.88,-1.95],[1,-2],[.88,-2.05]],\n     [[-3.6,0],[-3.6,2]],[[0,-2]
,[1,-2]]],\n     color=[brown$2,red,black$6],thickness=[2$3,1$6],\n   \+
  linestyle=[1$7,2$2]):\np2 := plot([-2.8+0.8*cos(t),2+0.8*sin(t),t=0.
.arctan(-5/7)],color=navy):\nt1 := plots[textplot]([[-3.1,1,`2 metres`
],\n    [.5,-2.25,`1 metre`],[-2.29,1.87,`t`]],color=black):\nplots[di
splay]([p1,p2,t1],axes=none);" }}}{PARA 258 "" 0 "" {TEXT -1 1 " " }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 490 "p1 := plot([[[-14/5,2],[0,2
],[0,0]],[[0,0],[1,0],[1,-5/7]],\n     [[-14/5,2],[1,-5/7]]],\n     co
lor=[black$2,red,black$6],\n     linestyle=[1$7,2$2]):\np2 := plot([[-
2.8+0.8*cos(t),2+0.8*sin(t),t=0..arctan(-5/7)],\n[0.5*cos(t),0.5*sin(t
),t=0..arctan(-5/7)]],color=navy):\nt1 := plots[textplot]([[.13,1,`2`]
,[.6,.15,`1`],\n    [-2.29,1.87,`t`],[.39,-.12,`t`]],color=black):\nt2
 := plots[textplot]([[-1.6,.75,`2 cosec t`],\n    [.28,-.47,`sec t`]],
color=red):\nplots[display]([p1,p2,t1,t2],axes=none);" }}}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}
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1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }