CHAPTER 2 ADDITIONAL AND ADVANCED EXERCISES
3.4 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
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).
6. yœ(sin xcos x) sec x Ê dydxœ(sin xcos x) dxd (sec x)sec x dxd (sin xcos x) (sin x cos x)(sec x tan x) (sec x)(cos x sin x)
œ œ (sin x cos x) sin x
cos x cos x
cos x sin x
#
sec xœ sin x cos x sin x cos x cos x sin xœ œ
cos x cos x
# #
# #
" #
Note also that y sin x sec x cos x sec x tan x 1 sec x.
Š œ œ Ê dydx œ # ‹
7. yœ1cot xcot x Ê dxdy œ(1cot x) dxd(cot x)(1cot x)(cot x) # dxd(1cot x) œ (1cot x) csc x(1cot x)(cot x)# csc x
# #
a b a b
œ csc xcsc x cot x csc x cot x œ csc x
(1 cot x) (1 cot x)
# # # #
# #
8. yœ1cos xsin x Ê dxdyœ (1 sin x) (cos x)(1 sin x)(cos x) (1 sin x) œ(1 sin x) (1sin xsin x)(cos x) cos x
dxd dxd
# a b # a b
œ sin x(1sin xsin x)# #cos x# œ (1sin xsin x)1# œ (1(1sin x)sin x)# œ 1"sin x
9. yœcos x4 tan x" œ4 sec xcot x Ê dxdy œ4 sec x tan xcsc x#
10. yœcos xx cos xx Ê dxdy œx( sin x) x#(cos x)(1)(cos x)(1)cos x#x( sin x)œ x sin xx#cos xcos xcos x#x sin x
11. yœx sin x# 2x cos x2 sin x Ê dydx œax cos x# (sin x)(2x)ba(2x)( sin x) (cos x)(2)b2 cos x x cos x 2x sin x 2x sin x 2 cos x 2 cos x x cos x
œ # œ #
12. yœx cos x# 2x sin x2 cos x Ê dydxœax ( sin x)# (cos x)(2x)ba2x cos x(sin x)(2)b 2( sin x) x sin x 2x cos x 2x cos x 2 sin x 2 sin x x sin x
œ # œ #
13. sœtan t Êt dsdt œ dtd (tan t) œ1 sec t# œ1 tan t# 14. sœ t# sec t Ê1 dsdt œ 2t dtd (sec t)œ 2t sec t tan t 15. sœ 11csc tcsc t Ê dsdt œ (1 csc t)( csc t cot t)(1 csc t)( csc t)(csc t cot t)
"
#
œ csc t cot tcsc t cot tcsc t cot tcsc t cot t œ2 csc t cot t
(1 csc t) (1 csc t)
# #
# #
16. sœ 1sin tcos t Ê dsdt œ(1 cos t)(cos t)(1 cos t)(sin t)(sin t) œcos t(1cos tcos t)sin t œ (1cos tcos t) œ 1 cos t
" "
# # #
# #
œ cos t"1
17. rœ 4 )# sin ) Ê ddr) œ ˆ)# dd)(sin )) (sin )(2 )) )‰œ a)# cos )2 sin ) )bœ ) )( cos ) # sin )) 18. rœ) sin )cos ) Ê ddr) œ( cos ) )(sin )(1))) sin )œ) cos )
19. rœsec )csc ) Ê d)dr œ(sec )( csc ) )cot )) (csc )(sec ) )tan ))
secœˆcos sin sin " ‰ ˆ " ‰ ˆ ‰ˆsin cos cos " ‰ ˆ " ‰ ˆ ‰œ sin" cos" œ # csc#
cos sin
) ) )) ) ) )) ) )
# # ) )
20. rœ(1sec ) sin ) ) Ê d)dr œ " ( sec ) cos ) )(sin )(sec ) )tan )) œ(cos ) " ) tan#)œcos )sec#) 21. pœ & cot q" œ 5 tan q Ê dqdp œsec q#
22. pœ(1csc q) cos q Ê dpdqœ(1csc q)( sin q) (cos q)( csc q cot q) œ ( sin q 1) cot q# œ sin qcsc q#
23. pœsin q cos q Ê dp œ (cos q)(cos q sin q) (sin q cos q)( sin q)
cos q dq cos q
#
sec qœ cos q cos q sin q sin q cos q sin qœ œ
cos q cos q
# #
# #
" #
24. pœ tan q Ê dp œ œ sec q tan q sec q tan q sec q œ sec q
1 tan q dq (1 tan q) (1 tan q) (1 tan q)
(1 tan q) sec q (tan q) sec q
a # b a # b
# # #
# # # #
25. (a) yœcsc x Ê ywœ csc x cot x Ê ywwœ a(csc x)acsc x# b(cot x)( csc x cot x) bœcsc x$ csc x cot x# (csc x) csc x cot x (csc x) csc x csc x 1 2 csc x csc x
œ a # # bœ a # # bœ $
(b) yœsec x Ê ywœsec x tan x Ê ywwœ(sec x) sec xa # b(tan x)(sec x tan x)œsec x$ sec x tan x# (sec x) sec x tan x (sec x) sec x sec x 1 2 sec x sec x
œ a # # bœ a # # bœ $
26. (a) yœ 2 sin x Ê ywœ 2 cos x Ê ywwœ 2( sin x)œ2 sin x Ê ywwwœ2 cos x Ê yÐ%Ñ œ 2 sin x (b) yœ9 cos x Ê ywœ 9 sin x Ê ywwœ 9 cos x Ê ywwwœ 9( sin x)œ9 sin x Ê yÐ%Ñ œ9 cos x 27. yœsin x Ê ywœcos x Ê slope of tangent at
xœ 1 is y (w œ1) cos ( œ "1) ; slope of tangent at xœ0 is y (0)w œcos (0)œ1; and slope of tangent at xœ 3#1 is ywˆ ‰3#1 œcos 3#1
0. The tangent at ( ) is y 0 1(x ), œ ß !1 œ 1 or yœ x 1; the tangent at (0 0) isß
y œ0 1(x0), or yœx; and the tangent at 1 is y 1.
ˆ3#1ß ‰ œ
28. yœtan x Ê ywœsec x # Ê slope of tangent at xœ 13 is sec#ˆ13‰œ4; slope of tangent at xœ0 is sec (0)# œ1;
and slope of tangent at xœ13 is sec#ˆ ‰13 œ4. The tangent at ˆ ß13 tanˆ13‰‰œ ß Š 13 È3 is y‹ È3œ4 xˆ 13‰; the tangent at (0 0) is yß œx; and the tangent at ˆ13ßtanˆ ‰13 ‰
3 is y 3 4 x . œŠ13ßÈ ‹ È œ ˆ 13‰
29. yœsec x Ê ywœsec x tan x Ê slope of tangent at xœ 13 is secˆ13‰ tanˆ13‰œ 2È3 ; slope of tangent at xœ 14 is secˆ ‰14 tanˆ ‰14 œÈ2 . The tangent at the point
sec is y 2 3 x ;
ˆ ß13 ˆ13‰‰œ ß #ˆ 13 ‰ œ #È ˆ 13‰ the tangent at the point ˆ1 secˆ ‰1 ‰ Š1 È2 is y‹ È2
4ß 4 œ 4ß
2 xœÈ ˆ 14‰.
30. yœ 1 cos x Ê ywœ sin x Ê slope of tangent at xœ 13 is sin ˆ13‰œ È#3; slope of tangent at xœ 3#1 is sin ˆ ‰31# œ1. The tangent at the point
cosˆ ß " 13 ˆ13‰‰œ ߈ 13 3#‰
is y œ3# È#3ˆx31‰; the tangent at the point
cos 1 is y 1 x
ˆ3#1ß " ˆ ‰3#1 ‰œˆ3#1ß ‰ œ 3#1
31. Yes, yœ x sin x Ê ywœ " cos x; horizontal tangent occurs where 1cos xœ0 Ê cos xœ 1 Ê xœ1
32. No, yœ2xsin x Ê ywœ 2 cos x; horizontal tangent occurs where 2cos xœ0 Ê cos xœ #. But there are no x-values for which cos xœ #.
33. No, yœ x cot x Ê ywœ 1 csc x; horizontal tangent occurs where 1# csc x# œ0 Ê csc x# œ 1. But there are no x-values for which csc x# œ 1.
34. Yes, yœ x 2 cos x Ê ywœ 1 2 sin x; horizontal tangent occurs where 12 sin xœ0 Ê 1œ2 sin x sin x x or x
Ê "# œ Ê œ 16 œ 561
35. We want all points on the curve where the tangent line has slope 2. Thus, yœtan x Ê ywœsec x so# that ywœ2 Ê sec x# œ2 Ê sec xœ „È2
x . Then the tangent line at has Ê œ „14 ˆ14ß "‰ equation y œ1 2 xˆ 14‰; the tangent line at
has equation y 1 2 x . ˆ ß "14 ‰ œ ˆ 14‰
36. We want all points on the curve yœcot x where the tangent line has slope 1. Thus y œcot x
y csc x so that y 1 csc x 1 Ê wœ # wœ Ê # œ
csc x 1 csc x 1 x . The Ê # œ Ê œ „ Ê œ1# tangent line at ˆ1#ß !‰ is yœ x 12.
37. yœ 4 cot x2 csc x Ê ywœ csc x# 2 csc x cot xœ ˆsin x" ‰ ˆ1sin x2 cos x‰ (a) When xœ 1#, then ywœ 1; the tangent line is yœ x 1# 2.
(b) To find the location of the horizontal tangent set ywœ0 Ê 1 2 cos xœ0 Ê xœ 13 radians. When xœ 13, then yœ % È3 is the horizontal tangent.
38. yœ 1 È2 csc xcot x Ê ywœ È2 csc x cot xcsc x# œ ˆsin x" ‰ ŠÈ2 cos xsin x1‹ (a) If xœ 14, then ywœ 4; the tangent line is yœ 4x 1 4.
(b) To find the location of the horizontal tangent set ywœ0 Ê È2 cos x œ1 0 Ê xœ3 radians. When
41
xœ341, then yœ2 is the horizontal tangent.
39. lim sin sin sin 0 0
xÄ2 ˆ"x"#‰œ ˆ"#"#‰œ œ
40. lim 1 cos ( csc x) 1 cos csc 1 cos ( 2) 2
xÄ16 È 1 œÉ ˆ1 16‰œÈ 1† œÈ
41. lim sec cos x tan 1 sec cos 0 tan 1 sec 1 tan 1 sec 1
xÄ ! < 1 ˆ4 sec x1 ‰ ‘œ < 1 ˆ4 sec 01 ‰ ‘œ < 1 ˆ ‰14 ‘œ 1œ
42. lim sin sin sin 1
xÄ ! ˆtan x 2 sec x1tan x ‰œ ˆtan 0 2 sec 01tan 0 ‰œ ˆ1#‰œ 43. lim tan 1 tan 1 lim tan (1 1) 0
tÄ ! ˆ sin tt ‰œ Š tÄ ! sin tt ‹œ œ
44. lim cos cos lim cos cos 1
)Ä ! ˆsin sin 1))‰œ Š1)Ä ! ))‹œ Œ1† lim " 9œ ˆ1†1"‰œ
)Ä!
sin ) )
45. sœ # # sin t Ê vœ dsdt œ 2 cos t Ê œ a dvdt œ2 sin t Ê œ j dadt œ2 cos t. Therefore, velocityœvˆ ‰41 2 m/sec; speed v 2 m/sec; acceleration a 2 m/sec ; jerk j 2 m/sec . œ È œ¸ ˆ ‰14 ¸œÈ œ ˆ ‰14 œÈ # œ ˆ ‰14 œÈ $ 46. sœsin tcos t Ê vœdsdt œcos tsin t Ê œ a dvdt œ sin tcos t Ê œj dadt œ cos tsin t. Therefore
velocityœvˆ ‰1 œ0 m/sec; speedœ¸vˆ ‰1 ¸œ0 m/sec; accelerationœaˆ ‰1 œ È2 m/sec ;
4 4 4 #
jerkœjˆ ‰14 œ0 m/sec .$
47. lim f(x) lim lim 9 9 so that f is continuous at x 0 lim f(x) f(0)
xÄ ! œxÄ ! sin 3xx## œxÄ ! ˆsin 3x3x ‰ ˆsin 3x3x ‰œ œ Ê xÄ ! œ
9 c.
Ê œ
48. lim g(x) lim (x b) b and lim g(x) lim cos x 1 so that g is continuous at x 0 lim g(x)
xÄ !c œxÄ !c œ xÄ !b œxÄ !b œ œ Ê xÄ !c
lim g(x) b 1. Now g is not differentiable at x 0: At x 0, the left-hand derivative is
œ Ê œ œ œ
xÄ !b
(x b)¸ 1, but the right-hand derivative is (cos x)¸ sin 0 0. The left- and right-hand
d d
dx x=0œ dx x=0œ œ
derivatives can never agree at xœ0, so g is not differentiable at xœ0 for any value of b (including bœ1).
49. dxd******(cos x)œsin x because dxd%%(cos x)œcos x Ê the derivative of cos x any number of times that is a multiple of 4 is cos x. Thus, dividing 999 by 4 gives 999œ249 4† Ê3 dxd******(cos x)
(cos x) (cos x) sin x.
œ dxd$$’dxd#%* %#%* %†† “œdxd$$ œ
50. (a) yœsec xœ cos x" Ê dxdy œ (cos x)(0)(cos x)(1)( sin x) œ cos xsin x œ cos x" cos xsin x œsec x tan x
# # ˆ ‰ ˆ ‰
(sec x) sec x tan x Ê dxd œ
(b) yœcsc xœ sin x" Ê dxdy œ(sin x)(0)(sin x)(1)(cos x)# œ sin xcos x# œˆsin x" ‰ ˆcos xsin x‰œ csc x cot x (csc x) csc x cot x
Ê dxd œ
(c) yœcot xœ cos x Ê œ œ sin x cos x œ œ csc x
sin x dx (sin x) sin x sin x
dy (sin x)( sin x) (cos x)(cos x) " #
# # #
# #
Ê dxd (cot x)œ csc x#
51.
As h takes on the values of 1, 0.5, 0.3 and 0.1 the corresponding dashed curves of yœ sin (x h)h sin x get closer and closer to the black curve yœcos x because dxd (sin x)œ lim sin (x hh) sin x œcos x. The same
hÄ !
is true as h takes on the values of 1, 0.5, 0.3 and 0.1.
52.
As h takes on the values of 1, 0.5, 0.3, and 0.1 the corresponding dashed curves of yœ cos (x h)h cos x get closer and closer to the black curve yœ sin x because dxd (cos x)œ lim cos (x h)h cos x œ sin x. The
hÄ !
same is true as h takes on the values of 1, 0.5, 0.3, and 0.1.
53. (a)
The dashed curves of yœsin xa hb#hsin xa hb are closer to the black curve yœcos x than the corresponding dashed curves in Exercise 51 illustrating that the centered difference quotient is a better approximation of the derivative of this function.
(b)
The dashed curves of yœcos xa hb#hcos xa hb are closer to the black curve yœ sin x than the corresponding dashed curves in Exercise 52 illustrating that the centered difference quotient is a better approximation of the derivative of this function.
54. lim lim lim 0 0 the limits of the centered difference quotient exists even
hÄ ! xÄ ! hÄ !
k0 hk k0 hk k kh k kh
2h 2h
œ œ œ Ê
though the derivative of f(x)œk kx does not exist at xœ0.
55. yœtan x Ê ywœsec x, so the smallest value# ywœsec x takes on is y# wœ1 when xœ0;
y has no maximum value since sec x has now # largest value on ˆ ß1 1# #‰; y is never negativew since sec x# 1.
56. yœcot x Ê ywœ csc x so y has no smallest# w value since csc x has no minimum value on # (!ß1); the largest value of y is 1, when xw œ1#; the slope is never positive since the largest value ywœ csc x takes on is 1.2
57. yœsin xx appears to cross the y-axis at yœ1, since lim 1; y appears to cross the y-axis
xÄ !
sin x sin 2x
x œ œ x
at y 2, since lim 2; y appears to
œ x œ œ
Ä ! sin 2xx sin 4xx cross the y-axis at yœ4, since lim œ4.
xÄ ! sin 4x
x
However, none of these graphs actually cross the y-axis since xœ0 is not in the domain of the functions. Also,
lim 5, lim 3, and lim
xÄ ! xÄ ! xÄ !
sin 5x sin kx
x x x
sin ( 3x)
œ œ
k the graphs of y , y , and
œ Ê œsin 5xx œsin ( 3x)x
yœsin kxx approach 5, 3, and k, respectively, as x Ä 0. However, the graphs do not actually cross the y-axis.
58. (a) h
1 .017452406 .99994923
0.01 .017453292 1 0.001 .017453292 1 0.0001 .017453292 1
sin h sin h 180
h ˆ h ‰ ˆ 1 ‰
lim lim lim lim ( h )
hÄ ! xÄ ! hÄ ! Ä !
sin h°
h h h 180 180
sin h sin h sin
œ ˆ†1801‰ œ 1801 1ˆ††1801‰ œ 1801 œ œ
180 )
)
) 1 ) † 1
(converting to radians) (b) h
1 0.0001523
0.01 0.0000015
0.001 0.0000001
0.0001 0
cos h 1 h
lim 0, whether h is measured in degrees or radians.
hÄ ! cos h 1
h œ
(c) In degrees, dxd (sin x)œ lim sin (x h)h sin x œ lim (sin x cos h cos x sin h)h sin x
hÄ ! hÄ !
lim sin x lim cos x (sin x) lim (cos x) lim
œ œ
hÄ ! ˆ †cos hh1‰ hÄ ! ˆ †sin hh ‰ †hÄ ! ˆcos hh1‰ †hÄ ! ˆsin hh ‰ (sin x)(0) (cos x) cos x
œ ˆ 1 ‰œ 1
180 180
(d) In degrees, dxd (cos x)œ lim cos (x h)h cos x œ lim (cos x cos h sin x sin h)h cos x
hÄ ! hÄ !
lim lim cos x lim sin x
œ œ
hÄ ! hÄ ! hÄ !
(cos x)(cos h 1) sin x sin h
h h h
cos h1 sin h
ˆ † ‰ ˆ † ‰
(cos x) lim (sin x) lim (cos x)(0) (sin x) sin x
œ œ œ
hÄ ! ˆcos hh1‰ hÄ ! ˆsin hh ‰ ˆ1801 ‰ 1801
(e) dxd##(sin x)œdxd ˆ1801 cos x‰œ ˆ1801 ‰# sin x; dxd$$(sin x)œ dxd Šˆ1801 ‰# sin x‹œ ˆ1801 ‰$ cos x;
(cos x) sin x cos x; (cos x) cos x sin x
d d d d
dx dx 180 180 dx dx 180 180
# $
# œ ˆ 1 ‰œ ˆ 1 ‰# $ œ Šˆ 1 ‰# ‹œˆ 1 ‰$