CHAPTER 2 ADDITIONAL AND ADVANCED EXERCISES
3.2 DIFFERENTIATION RULES
x0 := 1;
plot( f(x), x=x0-5..x0+2, color=black, title="Section 3_1, #62(a)" );
q := unapply( (f(x+h)-f(x))/h, (x,h) ); # (b) L := limit( q(x,h), h=0 ); # (c) m := eval( L, x=x0 );
tan_line := f(x0) + m*(x-x0);
plot( [f(x),tan_line], x=x0-2..x0+3, color=black, linestyle=[1,7], title="Section 3.1 #62(d)", legend=["y=f(x)","Tangent line at x=1"] );
Xvals := sort( [ x0+2^(-k) $ k=0..5, x0-2^(-k) $ k=0..5 ] ): # (e) Yvals := map( f, Xvals ):
evalf[4](< convert(Xvals,Matrix) , convert(Yvals,Matrix) >);
plot( L, x=x0-5..x0+3, color=black, title="Section 3.1 #62(f)" );
: (functions and x0 may vary) (see section 2.5 re. RealOnly ):
Mathematica
<<Miscellaneous`RealOnly`
Clear[f, m, x, y, h]
x0= 1/4;
f[x_]:=x 2Cos[x]
Plot[f[x], {x, x0 3, x03}]
q[x_, h_]:=(f[x h] f[x])/h m[x_]:=Limit[q[x, h], h Ä 0]
ytan:=f[x0]m[x0] (xx0)
Plot[{f[x], ytan},{x, x0 3, x03}]
m[x01]//N
m[x01]//N
Plot[{f[x], m[x]},{x, x03, x03}]
10. yœ 4 2xx$ Ê dydxœ # 3x% œ # x3% Ê d ydx œ 0 12x& œx12
#
# &
11. rœ "3s##5s" Ê dsdr œ 32s$#5s# œ3s2$2s5# Ê dsd r# œ2s%5s$œ s2 s5$
#
%
12. rœ12)"4)$)% Ê ddr) œ 12)#12)%4)& œ)12# 12)% )4& Ê dd r)# œ24)$48)&20)'
#
œ 24)$ 48)& 20)'
13. (a) yœa3x#b ax$ x 1 b Ê ywœa3x#b†dxd ax$ x 1bax$ x 1b†dxd a3x#b 3œa x#b a3x#1bax$ x 1 ( 2x)b œ 5x%12x#2x3
(b) yœ x& 4x$x#3x Ê3 ywœ 5x%12x#2x3
14. (a) yœ(x1) xa # x 1 b Ê ywœ(x1)(2x 1) ax# x 1 ( )b " œ3x# (b) yœ(x1) xa # x 1bœx$ Ê1 ywœ3x#
15. (a) yœax#1bˆx 5 "x‰ Ê ywœax#1b†dxd ˆx 5 "x‰ˆx 5 "x‰†dxd ax#1b
xœa #1b a1x#bax 5 x"b(2x)œax# 1 1 x#ba2x#10x2bœ3x#10x 2 x"#
(b) yœx$5x#2x 5 "x Ê ywœ3x#10x 2 x"#
16. yœˆx"x‰ ˆx "x 1‰
(a) ywœaxx"b a† 1x#baxx"1b a1x#bœ2x 1 x"#x2$
(b) yœx# x "x x"# Ê ywœ2x 1 x"# x2$
17. yœ2x3x52; use the quotient rule: uœ2x5 and vœ3x Ê2 uwœ2 and vwœ3 Ê ywœ vuwv#uvw
œ (3x2)(2)(3x(2x2)#5)(3) œ 6x(3x4 #6x)# 15 œ(3x192)#
18. zœ2xx#11 Ê dxdz œ x 1 (2)x(2x1 1)(2x) œ2x x24x1 2xœ 2xx2x12 œ2 xx 1x 1
# #
# # # # # # # #
# # #
a b a b
a b a b a b a b
19. g(x)œ xx 0.5#4; use the quotient rule: uœx 4 and vœ x 0.5 Ê u œ2x and v œ1 Ê g (x)œ vuwv#uvw
# w w w
œ (x0.5)(2x)(x0.5)ax##4 ( )b" œ2x(x x0.5)x# 4 œ x(xx0.5)4#
# # #
20. f(t)œ t#t# "t 2 œ tt tt œ tt "2, tÁ " Ê f (t)œ t t 2t œt # "t 2t œ t "2 # "
" " w # " " "
a ba b a ba b a ba b
a ba b a b2 a b2 a b2
21. vœ(1t) 1a t# "b œ1 t1 t# Ê dvdt œ 1t (" 1)t (1 t)(2t) œ" 1tt2t 2t œt1 "2tt
#
## ## ##
# # #
a b
a b a b a b
22. wœ 2xx57 Ê wwœ (2x7)(1)(2x7)(x# 5)(2) œ2x (2x72x7)# 10 œ(2x177)#
23. f(s)œÈÈss 1 Ê f (s)œ ˆÈs ‰Š ˆÈ‹ ˆÈ‰s 1‰Š ‹ œ ˆÈÈ Èˆ‰ ˆÈ‰ ‰ œ È Èˆ ‰
s 1 2 s s 1 s s 1
s s 1
"
w "
" "
" "
#Ès #Ès
# # #
NOTE: dsd ˆÈs‰œ #È"s from Example 2 in Section 2.1
24. uœ5x#È "x Ê dudx œˆ2Èx (5)‰ 4x(5x1)ŠÈx" ‹ œ 5x4x$Î#1
25. vœ1 xx4Èx Ê vwœ x 1Š È2x‹xˆ#1 x 4Èx‰ œ2Èxx# "
26. rœ2Š " È ‹ Ê rwœ2 œ " "
È
È Š ‹
)
)
) ) )
) (0) 1 #È)" " $Î# "Î#
#È)
27. yœax#1b ax"# x 1b; use the quotient rule: uœ1 and vœax#1b ax# x 1 b Ê uwœ0 and vwœax#1 (2xb 1) ax# x 1 (2x)b œ2x$x#2x 1 2x$2x#2xœ4x$3x#1
Ê dydxœ vuwv#uvw œ 0x 1 4x1 x 3xx 11 œ x 4x1$ x3x#x11
$ #
# # # # # # # #
a b
a b a b a b a b
28. yœ(x(x1)(x1)(x #2)) œxx##3x3x22 Ê ywœ x 3x2 (2x(x3)1) (x# x2)#3x2 (2x3) œ(x6x1) (x##122)#
# #
a b a b
œ (x6 x1) (xa##2b2)#
29. yœ"#x%#3 x# Êx ywœ2x$3x Ê1 ywwœ6x# Ê3 ywwwœ12x Ê yÐ%Ñ œ12 Ê yÐ Ñn œ0 for all n 5 30. yœ1 0"# x & Ê ywœ #"4x % Ê ywwœ"6x $ Ê ywwwœ "#x # Ê yÐ%Ñ œx Ê yÐ&Ñ œ1 Ê yÐ Ñn œ0 for all n 6
31. yœx$x7 œx 7x Ê dxdy œ2x7x œ # x x# Ê dxd y# œ 2 14x œ # x$
# " # ( # $ "%
32. sœ t# 5tt# 1 œ œ 1 5t t"# 1 5t"t# Ê dsdt œ 0 5t#2t$ œ 5t#2t$ œ &t# t#$
Ê d sdt## œ10t$6t% œ"!t$ t'%
33. rœ () ")a))#$ ) 1b œ )) "$ œ " )"$ œ 1 $ Ê ddr) œ $0 %œ $ % œ )$ Ê d rd)# œ 12 &œ "#)&
$ #
) ) ) % )
34. uœax# xb axx%# x 1b œ x(x 1) xax%# x b œ x xax$% 1b œ xx%x œ 1 xx% œ 1 x
" % $
Ê dudxœ 0 3x%œ 3x% œ$x% Ê d udx œ12x& œ"#x
#
# &
35. wœˆ13z3z‰(3 œz) ˆ3"z"1 (3‰ œz) z" œ3" 3 z z" Ê83 z dwdz œ z# œ 0 1 z#1 œ "z# " Ê d wdz# œ2z$ œ0 2z$œ z#$
#
36. wœ (z 1)(z1) za #1bœaz#1b az#1bœz% Ê1 4zdwdz œ $ œ0 4z $ Ê d wdz## œ12z#
37. pœŠq12q#3‹ Šq%q$1‹œ q' q#12q3q% %3 œ 1" q 1" q 4" 4"q Ê dpdqœ 6"q6"q q œ 6"q6q"$ q"&
# # # # % $ &
q 5q
Ê d pdq## œ 6" " œ 6" q"% q&
# % '
# 6
38. pœ(q 1)q$#(q3 1)$ œ q$ 3q# 3q 1q#3q$ 3q# 3q 1 œ 2qq#$36q œ2q qq##33 œ q œ q
#" #" "
a b a b a b
q q
Ê dpdqœ "# #œ #"q# Ê d pdq# œ $œ q"$
#
39. u(0)œ5, u (0)w œ 3, v(0)œ 1, v (0)w œ2
(a) (uv)dxd œuvwvu (uv)w Ê dxd ¸ œu(0)v (0)w v(0)u (0)w œ5 2 œ( 1)( 3) 13
x=0 †
(b) dxd ˆ ‰uv œ vuwv#uvw Ê dxd ˆ ‰uv ¸ œ v(0)u (0)(v(0))u(0)v (0)# œ( )( 3)( 1)#(5)(2)œ 7
w w
"
x=0
(c) dxd ˆ ‰vu œ uvwu#vuw Ê dxd ˆ ‰vu ¸ œu(0)v (0)(u(0))v(0)u (0)# œ (5)(2)(5)( 1)( 3)# œ 257
w w
x=0
(d) (7vdxd 2u)œ7vw2u (7vw Ê dxd 2u)¸ œ7v (0)w 2u (0)w œ7 2 œ2( 3) 20
x=0 †
40. u(1)œ2, u (1)w œ0, v(1)œ5, v (1)w œ 1
(a) (uv)dxd ¸ u(1)v (1) v(1)u (1) 2 ( 1) 5 0 2
x=1 œ w w œ † † œ
(b) dxd ˆ ‰uv ¸ v(1)u ( ) u(1)v (1)(v(1)) 5 0 2 ( 1)(5) 252 x=1œ w" # w œ † †# œ (c) dxd ˆ ‰vu ¸ u(1)v ( ) v(1)u (1)(u(1)) 2 ( 1) 5 0(2) 12
x=1œ w" # w œ † # † œ
(d) (7vdxd 2u)¸x=1œ7v (1)w 2u (1)w œ7 ( 1)† 2 0† œ 7 41. yœx$4x1. Note that (#ß ") is on the curve: 1œ2$4(2)1
(a) Slope of the tangent at (x y) is yß wœ3x# Ê4 slope of the tangent at (#ß ") is y (2)w œ3(2)# œ4 8. Thus the slope of the line perpendicular to the tangent at (#ß ") is Ê"8 the equation of the line perpendicular to to the tangent line at (#ß ") is y œ 1 "8(x2) or yœ 8x 45.
(b) The slope of the curve at x is mœ3x#4 and the smallest value for m is 4 when x œ0 and yœ1.
(c) We want the slope of the curve to be 8 Ê ywœ8 Ê 3x# œ4 8 Ê 3x#œ12 Ê x#œ4 Ê xœ „2. When xœ2, yœ1 and the tangent line has equation y œ1 8(x2) or yœ8x15; when xœ 2,
yœ ( 2)$ œ4( 2) 1 1, and the tangent line has equation y œ1 8(x2) or yœ8x17.
42. (a) yœx$3x Ê2 ywœ3x#3. For the tangent to be horizontal, we need mœywœ0 Ê 0œ3x#3 3x 3 x 1. When x 1, y 0 the tangent line has equation y 0. The line
Ê #œ Ê œ „ œ œ Ê œ
perpendicular to this line at ("ß !) is xœ 1. When xœ1, yœ Ê4 the tangent line has equation yœ 4. The line perpendicular to this line at ("ß %) is xœ1.
(b) The smallest value of y is 3, and this occurs when xw œ0 and yœ 2. The tangent to the curve at (!ß 2) has slope 3 Ê the line perpendicular to the tangent at (!ß 2) has slope "3 Ê œ y 2 "3(x0) or yœ"3x2 is an equation of the perpendicular line.
43. yœx#4x1 Ê dxdy œ x 1 (4)x 1(4x)(2x) œ4xx 418x œ4xx 1 . When xœ0, yœ0 and y œ 4(01 1)
# #
# # # # # #
# #
"
w
a b a b
a b a b a b
, so the tangent to the curve at ( ) is the line y 4x. When x 1, y 2 y 0, so the tangent to the
œ % !ß ! œ œ œ Ê wœ
curve at ("ß2) is the line yœ2.
44. yœx#84 Ê y œ x 4 (0)x 4 8(2x) œ x 16x4 . When xœ2, yœ1 and y œ 216(2)4 œ , so the tangent
#
# # # # # #
w w #
"
a b
a b a b a b
line to the curve at (2ß ") has the equation y œ 1 "#(x2), or yœ #x 2.
45. yœax#bxc passes through (!ß ! Ê) 0œa(0)b(0) Ê œc c 0; yœax#bx passes through ("ß #) 2 a b; y 2ax b and since the curve is tangent to y x at the origin, its slope is 1 at x 0
Ê œ wœ œ œ
y 1 when x 0 1 2a(0) b b 1. Then a b 2 a 1. In summary a b 1 and c 0 so
Ê wœ œ Ê œ Ê œ œ Ê œ œ œ œ
the curve is yœx#x.
46. yœcxx passes through (# "ß ! Ê) 0œc(1) Ê œ1 c 1 Ê the curve is yœ x x . For this curve,# ywœ 1 2x and xœ1 Ê ywœ 1. Since yœ x x and y# œx#axb have common tangents at xœ0, yœx#axb must also have slope 1 at x œ1. Thus ywœ2x Ê œa 1 2 1† Ê œ a a 3
y x 3x b. Since this last curve passes through ( ), we have 0 1 3 b b 2. In summary,
Ê œ # "ß ! œ Ê œ
aœ 3, bœ2 and cœ1 so the curves are yœx#3x2 and yœ x x .#
47. (a) yœx$ Êx ywœ3x#1. When xœ 1, yœ0 and ywœ2 Ê the tangent line to the curve at ("ß !) is yœ2(x1) or yœ2x2.
(b)
(c) y x x x x 2x 2 x 3x 2 (x 2)(x 1) 0 x 2 or x 1. Since
yœ2x2
œ $ Ê $ œ Ê $ œ #œ Ê œ œ yœ2 2a b œ2 6; the other intersection point is (2 6)ß
48. (a) yœx$6x#5x Ê ywœ3x#12x5. When xœ0, yœ0 and ywœ5 Ê the tangent line to the curve at (0 0) is yß œ5x.
(b)
(c) y x 6x 5x x 6x 5x 5x x 6x 0 x (x 6) 0 x 0 or x 6.
yœ 5x
œ$ # Ê $ # œ Ê $ #œ Ê # œ Ê œ œ Since yœ5 6a bœ $!, the other intersection point is (6 30).ß
49. P xa bœa xn nan xn â a x a xa ÊP xa bœna xn n an "ban xn â #a xa
" " # # " ! w " " # # "
50. RœM#ˆC# M3‰œ C#M#"3M , where C is a constant $ Ê dMdR œCMM#
51. Let c be a constant Ê dcdxœ0 Ê dxd (u c)† œu†dcdxc†dudx œu 0† c dudxœc dudx. Thus when one of the functions is a constant, the Product Rule is just the Constant Multiple Rule Ê the Constant Multiple Rule is a special case of the Product Rule.
52. (a) We use the Quotient rule to derive the Reciprocal Rule (with uœ1): dxd ˆ ‰v" œv 0† v#1†dvdx œ"v†#dvdx œ v"# †dxdv.
(b) Now, using the Reciprocal Rule and the Product Rule, we'll derive the Quotient Rule: dxd ˆ ‰uv œ dxd ˆu†v"‰
u (Product Rule) u (Reciprocal Rule)
œ †dxd ˆ ‰v" v" †dudx œ †ˆv#1‰dvdxv dx" du Ê dxd ˆ ‰uv œ u v dvdxv# dudx , the Quotient Rule.
œ v u dudxv# dvdx
53. (a) dxd (uvw)œdxd ((uv) w)† œ(uv)dwdxw†dxd (uv)œuv w u v uv wu wv dwdx ˆ dvdx dudx‰œ dwdx dvdx dudx uvwœ wuv ww u vww
(b) u u u udxd a " # $ %bœ dxd aau u u" # $bu%bœau u u" # $bdudx% u u u u u u u u% dxd a " # $b Ê dxd a " # $ %b u u u u u u u u u u (using (a) above)
œ " # $ du % " # du $ " du $ # du
dx% ˆ dx$ dx# dx"‰
u u u u u u u u u u u u u u u u
Ê dxd a " # $ %bœ " # $ dudx% " # % dudx$ " $ % dudx# # $ % dudx"
u u u uœ " # $ w u u u u" # w %u u u u" w $ %u u u uw # $ %
% $ # "
(c) Generalizing (a) and (b) above, dxd au"âunbœu u" #âun"uwnu u" #âun#uwn"un á u uw" #âun
54. In this problem we don't know the Power Rule works with fractional powers so we can't use it. Remember x (from Example 2 in Section 2.1)
d
dxˆÈ ‰œ #È"x
(a) xdxd dxd x x x dxd x x (x)dxd x x 1 x 3x
x
x 3 x
ˆ $Î#‰ ˆ "Î#‰ ˆÈ ‰ È " È È "Î#
# # # #
œ † œ † œ † È † œ È œ È œ
(b) xdxd dxd x x x dxd x x xdxd x x 2x x 2x 5x
ˆ &Î#‰œ ˆ #† "Î#‰œ # ˆÈ ‰È a b# œ #†Š#È"x‹È † œ"# $Î# $Î#œ# $Î#
(c) xdxd dxd x x x dxd x x xdxd x x 3x x 3x 7x
ˆ (Î#‰œ ˆ $† "Î#‰œ $ ˆÈ ‰È a b$ œ $†Š#È"x‹È † #œ"# &Î# &Î#œ # &Î#
(d) We have dxd ˆx$Î#‰œ 3#x"Î#, dxd ˆx&Î#‰œ 5#x$Î#, dxd ˆx(Î#‰œ7#x&Î# so it appears that dxd axnÎ#bœn#xÐ Î#Ñ"n whenever n is an odd positive integer 3.
55. pœV nbnRT anV##. We are holding T constant, and a, b, n, R are also constant so their derivatives are zero Ê dVdP œ(Vnb) 0(V nb)†(nRT)(1)# V (0)Van (2V) œ(VnRTnb)# 2anV$
# #
# #
a b #
a b
56. A qa bœ kmq cmhq# œakm qb "cmˆ ‰h# qÊ dAdq œ akm qb #ˆ ‰h# œ kmq# Êh# d Adt# œ #akm qb $œ #kmq$
#
3.3 THE DERIVATIVE AS A RATE OF CHANGE 1. sœ $ # Ÿ Ÿ #t# t , 0 t
(a) displacementœ?sœ # s( ) s(0)œ ! # œ #m m m, vavœ ? œ œ " m/sec
?s
t #
#
(b) vœdsdt œ # $ Êt v(0)k kœ l$l œ $ m/sec and v( )k # œk 1 m/sec;
aœd sdt## œ # Ê a(0)œ # m/sec and a( )# # œ # m/sec#
(c) vœ Ê # $ œ0 t 0 Ê œt $#. v is negative in the interval ! t $# and v is positive when $# # Êt the body changes direction at tœ$#.
2. sœ ' t t , # ! Ÿ Ÿ 't
(a) displacementœ?sœ ' s( ) s(0)œ ! m, vavœ ? œ œ ! m/sec
?s
t !
'
(b) vœdsdt œ ' #> Ê v(0)k kœ l 'l œ ' m/sec and v( )k ' œ l'l œ 'k m/sec;
aœd sdt## œ # Ê a(0)œ # m/sec and a( )# ' œ # m/sec#
(c) vœ Ê ' # œ0 t 0 Ê œ $t . v is positive in the interval ! $t and v is negative when $ ' Êt the body changes direction at tœ $.
3. sœ t$ 3t#3t, 0Ÿ Ÿt 3
(a) displacementœ?sœs(3)s(0)œ 9 m, vavœ ? œ œ 3 m/sec
?s 9
t 3
(b) vœdsdt œ 3t# Ê6t 3 v(0)k kœ œk 3k 3 m/sec and v(3)k kœ k 12kœ12 m/sec; aœ d sdt## œ 6t 6 a(0) 6 m/sec and a(3) 12 m/sec
Ê œ # œ #
(c) vœ0 Ê 3t# œ6t 3 0 Ê t# œ2t 1 0 Ê (t1)#œ0 Ê œ t 1. For all other values of t in the interval the velocity v is negative (the graph of vœ 3t# 6t 3 is a parabola with vertex at tœ1 which opens downward Ê the body never changes direction).
4. sœ t4% t$ t , 0# Ÿ Ÿ $t
(a) ?sœ $ s( ) s(0)œ *% m, vavœ ? œ $ œ$% m/sec
?s t
*%
(b) vœ t$ 3t#2t Ê v(0)k kœ0 m/sec and v( )k $ œ 'k m/sec; aœ3t# Ê6t 2 a(0)œ2 m/sec and# a( )$ œ ""m/sec#
(c) vœ0 Ê t$3t# œ2t 0 Ê t(t2)(t1)œ0 Ê œ t 0, 1, 2 Ê vœt(t2)(t1) is positive in the
interval for 0 t 1 and v is negative for 1 t 2 and v is positive for # $ Êt the body changes direction at
tœ1 and at tœ #. 5. sœ 25t# 5t, 1Ÿ Ÿt 5
(a) ?sœs(5)s(1)œ 20 m, vavœ20œ 5 m/sec
4
(b) vœt50$ t5# Ê v(1)k kœ45 m/sec and v(5)k kœ"5 m/sec; aœ150t% 10t$ Ê a(1)œ140 m/sec and# a(5)œ 254 m/sec#
(c) vœ0 Ê 50 5tt$ œ0 Ê œ 50 5t 0 Ê œ t 10 Ê the body does not change direction in the interval 6. sœ t 525 , % Ÿ Ÿt 0
(a) ?sœs(0) œ s( 4) 20 m, vavœ 20 œ 5 m/sec
4
(b) vœ(t 5)25# Ê v( 4)k kœ25 m/sec and v(0)k kœ " m/sec; aœ(t 5)50$ Ê a( 4) œ50 m/sec and# a(0)œ 25 m/sec#
(c) vœ0 Ê (t 5)25# œ0 Ê v is never 0 Ê the body never changes direction
7. sœ t$ 6t#9t and let the positive direction be to the right on the s-axis.
(a) vœ3t#12t9 so that vœ0 Ê t# œ 4t 3 (t 3)(t1)œ0 Ê œ t 1 or 3; aœ 6t 12 Ê a(1) 6 m/sec and a(3) 6 m/sec . Thus the body is motionless but being accelerated left when t 1, and
œ # œ # œ
motionless but being accelerated right when tœ3.
(b) aœ0 Ê 6t12œ0 Ê œ t 2 with speed v(2)k kœk1224 œ9k 3 m/sec
(c) The body moves to the right or forward on 0Ÿ t 1, and to the left or backward on 1 t 2. The positions are s(0)œ0, s(1)œ4 and s(2)œ2 Ê total distanceœks(1)s(0)kks(2)s(1)kœk k4 k 2k
6 œ m.
8. vœ Ê œ t# 4t 3 a 2t 4
(a) vœ0 Ê t# œ4t 3 0 Ê œ t 1 or 3 Ê a(1)œ 2 m/sec and a(3)# œ2 m/sec#
(b) v0 Ê (t3)(t1)0 Ê 0Ÿ t 1 or t3 and the body is moving forward; v0 Ê (t3)(t1)0 t 3 and the body is moving backward
Ê "
(c) velocity increasing Ê a 0 Ê 2t 4 0 Ê t 2; velocity decreasing Ê a 0 Ê 2t 4 0 Ê ! Ÿ t 2 9. smœ1.86t # Ê vmœ3.72t and solving 3.72tœ27.8 Ê ¸ t 7.5 sec on Mars; sjœ11.44t # Ê vjœ22.88t and
solving 22.88tœ27.8 Ê ¸ t 1.2 sec on Jupiter.
10. (a) v(t)œs (t)w œ241.6t m/sec, and a(t)œv (t)w œs (t)ww œ 1.6 m/sec# (b) Solve v(t)œ0 Ê 241.6tœ0 Ê œ t 15 sec
(c) s(15)œ24(15).8(15)#œ180 m
(d) Solve s(t)œ90 Ê 24t.8t# œ90 Ê œ t 30 15„#È2 ¸4.39 sec going up and 25.6 sec going down (e) Twice the time it took to reach its highest point or 30 sec
11. sœ15t"#g t s# Ê vœ15g t so that vs œ0 Ê 15g ts œ0 Ê gsœ 15. Therefore gsœ 15œ œ3 0.75 m/sec#
t 20 4
12. Solving smœ832t2.6t#œ0 Ê t(8322.6t)œ0 Ê œ t 0 or 320 Ê 320 sec on the moon; solving seœ832t16t#œ0 Ê t(83216t)œ0 Ê œ t 0 or 52 Ê 52 sec on the earth. Also, vmœ8325.2tœ0
t 160 and s (160) 66,560 ft, the height it reaches above the moon's surface; v 832 32t 0
Ê œ m œ eœ œ
t 26 and s (26) 10,816 ft, the height it reaches above the earth's surface.
Ê œ e œ
13. (a) sœ17916t # Ê vœ 32t Ê speedœk kv œ32t ft/sec and aœ 32 ft/sec#
(b) sœ0 Ê 17916t#œ0 Ê œ t É17916 ¸3.3 sec
(c) When tœÉ179, vœ 32É179 œ 8È179¸ 107.0 ft/sec
16 16
14. (a) lim v lim 9.8(sin )t 9.8t so we expect v 9.8t m/sec in free fall
)Ä1# œ)Ä1# ) œ œ
(b) aœdvdt œ9.8 m/sec#
15. (a) at 2 and 7 seconds (b) between 3 and 6 seconds: $ Ÿ Ÿt 6
(c) (d)
16. (a) P is moving to the left when 2 t 3 or 5 t 6; P is moving to the right when 0 t 1; P is standing still when 1 t 2 or 3 t 5
(b)
17. (a) 190 ft/sec (b) 2 sec
(c) at 8 sec, 0 ft/sec (d) 10.8 sec, 90 ft/sec (e) From tœ8 until tœ10.8 sec, a total of 2.8 sec
(f) Greatest acceleration happens 2 sec after launch
(g) From tœ2 to tœ10.8 sec; during this period, aœv(10.8) v(2) ¸ 32 ft/sec
10.8 2
#
18. (a) Forward: 0Ÿ t 1 and 5 t 7; Backward: 1 t 5; Speeds up: 1 t 2 and 5 t 6;
Slows down: 0Ÿ t 1, 3 t 5, and 6 t 7
(b) Positive: 3 t 6; negative: 0Ÿ t 2 and 6 t 7; zero: 2 t 3 and 7 t 9 (c) tœ0 and 2Ÿ Ÿt 3
(d) 7Ÿ Ÿt 9
19. sœ490t # Ê vœ980t Ê œa 980
(a) Solving 160œ490t # Ê œ t 47 sec. The average velocity was s(4/7) s(0)4/7 œ280 cm/sec.
(b) At the 160 cm mark the balls are falling at v(4/7)œ560 cm/sec. The acceleration at the 160 cm mark was 980 cm/sec .#
(c) The light was flashing at a rate of 4/717 œ29.75 flashes per second.
20. (a)
(b)
21. Cœposition, Aœvelocity, and Bœacceleration. Neither A nor C can be the derivative of B because B's derivative is constant. Graph C cannot be the derivative of A either, because A has some negative slopes while C has only positive values. So, C (being the derivative of neither A nor B) must be the graph of position.
Curve C has both positive and negative slopes, so its derivative, the velocity, must be A and not B. That leaves B for acceleration.
22. Cœposition, Bœvelocity, and Aœacceleration. Curve C cannot be the derivative of either A or B because C has only negative values while both A and B have some positive slopes. So, C represents position. Curve C has no positive slopes, so its derivative, the velocity, must be B. That leaves A for acceleration. Indeed, A is negative where B has negative slopes and positive where B has positive slopes.
23. (a) c(100)œ11,000 Ê cavœ11,000 œ$110
100
(b) c(x)œ2000100x.1x # Ê c (x)w œ100.2x. Marginal costœc (x)w Êthe marginal cost of producing 100 machines is c (100)w œ$80
(c) The cost of producing the 101 machine is c(101)st c(100)œ10020110 œ$79.90
24. (a) r(x)œ20000 1ˆ "x‰ Ê r (x)w œ 20000x# , which is marginal revenue.
(b) rwa"!! œb 20000100# œ #Þ$
(c) lim r (x)xÄ _ w œx lim Ä _ 20000x# œ0. The increase in revenue as the number of items increases without bound will approach zero.
25. b(t)œ10'10 t% 10 t $ # Ê b (t)w œ10%(2) 10 ta $bœ10 (10$ 2t)
(a) b (0)w œ10 bacteria/hr % (b) b (5)w œ0 bacteria/hr (c) b (10)w œ 10 %bacteria/hr
26. Q(t)œ200(30t)#œ200 900a 60tt #b Ê Q (t)w œ200( 60 2t) Ê Q (10)w œ 8,000 gallons/min is the rate the water is running at the end of 10 min. Then Q(10) Q(0)10 œ 10,000 gallons/min is the average rate the
water flows during the first 10 min. The negative signs indicate water is leaving the tank.
27. (a) yœ6 1ˆ 1t#‰#œ6 1Š 6t 144t# ‹ Ê dydt œ 12t 1
(b) The largest value of dydt is 0 m/h when tœ12 and the fluid level is falling the slowest at that time. The smallest value of dydt is 1 m/h, when t œ0, and the fluid level is falling the fastest at that time.
(c) In this situation, dydt Ÿ0 Ê the graph of y is always decreasing. As dydt increases in value, the slope of the graph of y increases from 1 to 0 over the interval 0Ÿ Ÿt 12.
28. (a) Vœ 431r 4 r $ Ê dVdr œ 1# Ê dVdr¸ œ4 (2)1 #œ16 1ft /ft$
r=2
(b) When rœ2, dVdr œ16 so that when r changes by 1 unit, we expect V to change by approximately 16 .1 1 Therefore when r changes by 0.2 units V changes by approximately (16 )(0.2)1 œ3.21¸10.05 ft . Note$ that V(2.2)V(2)¸11.09 ft .$
29. 200 km/hrœ55 m/sec59 œ 5009 m/sec, and Dœ 109 t # Ê Vœ209 t. Thus Vœ 5009 Ê 209 tœ 5009 Ê œ t 25 sec. When tœ25, Dœ109 (25)#œ 62509 m
30. sœv t! 16t # Ê vœv!32t; vœ0 Ê œ t ; 1900œv t! 16t so that t# œ Ê 1900œ
v v #
32 32 3 64
v v
! ! !# !#
v (64)(1900) 80 19 ft/sec and, finally, 238 mph.
Ê ! œÈ œ È 80 19 ft ¸
sec 1 min 1 hr 5280 ft 60 sec 60 min 1 mi
È † † †
31.
(a) vœ0 when tœ6.25 sec
(b) v0 when 0Ÿ t 6.25 Ê body moves up; v0 when 6.25 Ÿt 12.5 Ê body moves down (c) body changes direction at tœ6.25 sec
(d) body speeds up on (6.25 12.5] and slows down on [0 6.25)ß ß
(e) The body is moving fastest at the endpoints tœ0 and tœ12.5 when it is traveling 200 ft/sec. It's moving slowest at tœ6.25 when the speed is 0.
(f) When tœ6.25 the body is sœ625 m from the origin and farthest away.
32.
(a) vœ0 when tœ 3# sec
(b) v0 when 0Ÿ t 1.5 Ê body moves down; v0 when 1.5 Ÿt 5 Ê body moves up (c) body changes direction at tœ 3# sec
(d) body speeds up on ˆ3#ß &‘ and slows down on <!ß3#‰
(e) body is moving fastest at tœ5 when the speedœkv(5)kœ7 units/sec; it is moving slowest at tœ 3# when the speed is 0
(f) When tœ5 the body is sœ12 units from the origin and farthest away.
33.
(a) vœ0 when tœ 6„3È15 sec
(b) v0 when 63È15 t 63È15 Ê body moves left; v0 when 0Ÿ t 63È15 or 63È15 Ÿt 4 body moves right
Ê
(c) body changes direction at tœ 6„3È15 sec
(d) body speeds up on Š63È15ß # ‹ Š63È15ß %“ and slows down on 0’ ß63È15‹ #ߊ 63È15‹.
(e) The body is moving fastest at tœ0 and tœ4 when it is moving 7 units/sec and slowest at tœ 6„È315 sec (f) When tœ6È315 the body is at position s¸ 6.303 units and farthest from the origin.
34.
(a) vœ0 when tœ 6„3È15
(b) v0 when 0Ÿ t 63È15 or 63È15 Ÿt 4 Ê body is moving left; v0 when t body is moving right
6 15 6 15
3 3
È È
Ê
(c) body changes direction at tœ 6„3È15 sec
(d) body speeds up on Š63È15ß # ‹ Š63È15ß %“ and slows down on ’!ß63È15‹ #ߊ 63È15‹ (e) The body is moving fastest at 7 units/sec when tœ0 and tœ4; it is moving slowest and stationary at
tœ 6„3È15
(f) When tœ63È15 the position is s¸10.303 units and the body is farthest from the origin.
35. (a) It takes 135 seconds.
(b) Average speed œ ??Ft œ ($ !& ! œ ($& ¸ !Þ!') furlongs/sec.
(c) Using a symmetric difference quotient, the horse's speed is approximately ??Ft œ&* $$% # œ #'# ¸ !Þ!(( furlongs/sec.
(d) The horse is running the tastest during the last furlong (between the 9th and 10th furlong markers). This furlong takes only 11 seconds to run, which is the least amount of time for a furlong.
(e) The horse accelerates the fastest during the first furlong (between markers 0 and 1).