RELATED RATES

CHAPTER 2 ADDITIONAL AND ADVANCED EXERCISES

3.7 RELATED RATES

eval( q1, pt );

q2 := implicitdiff( q1, y, x );

m := eval( q2, pt );

tan_line := y = 1 + m*(x-2);

p2 := implicitplot( tan_line, x=-5..5, y=-5..5, color=green ):

p3 := pointplot( eval([x,y],pt), color=blue ):

display( [p1,p2,p3], ="Section 3.6 #77(c)" );

: (functions and x0 may vary):

Mathematica

Note use of double equal sign (logic statement) in definition of eqn and tanline.

<<Graphics`ImplicitPlot`

Clear[x, y]

{x0, y0}={1, 1/4};

eqn=x + Tan[y/x]==2;

ImplicitPlot[eqn,{ x, x03, x03},{y, y03, y03}]

eqn/.{xÄx0, yÄy0}

eqn/.{ yÄy[x]}

D[%, x]

Solve[%, y'[x]]

slope=y'[x]/.First[%]

m=slope/.{xÄx0, y[x]Äy0}

tanline=y==y0m (xx0)

ImplicitPlot[{eqn, tanline}, {x, x03, x03},{y, y03, y0 + 3}]

Ê ^ds_dt œ _Èx# ^x_y# _z# ^dx_{dt Èx}# ^y_y# _z# ^dy_{dt Èx}# ^z_y# _z# ^dz_dt

(b) From part (a) with ^dx_dt 0 ^ds_dt ^{y dy}_dt ^{z dz}_dt

x y z x y z

œ Ê œ _È # # # _È # # #

(c) From part (a) with ^ds_dt œ0 Ê 0œ2x ^dx_dt 2y ^dy_dt 2z ^dz_dt Ê ^dx_{dt x dt}^{y dyz dz}_{x dt} œ0

9. (a) Aœ ^"_#ab sin ) Ê ^dA_dt œ^"_#ab cos ) ^d_dt⁾ (b) Aœ ^"_#ab sin ) Ê ^dA_dt œ^"_#ab cos ) ^d_dt^{) "}_#b sin ) ^da_dt (c) Aœ ^"_#ab sin ) Ê ^dA_dt œ^"_#ab cos ) ^d_dt^)"_#b sin ) ^da_dt ^"_#a sin ) ^db_dt

10. Given Aœ1r , ^{# dr}_dt œ0.01 cm/sec, and rœ50 cm. Since ^dA_dt œ2 r 1 ^dr_dt, then ^dA_dt¸ œ2 (50)1 ˆ₁₀₀^" ‰

r=50

œ1cm /min.^#

11. Given ^d_dt^j œ 2 cm/sec, ^dw_dt œ2 cm/sec, j œ12 cm and wœ5 cm.

(a) Aœ j Êw ^dA_dt œ j w ^dw_dt ^d_dt^j 12(2)Ê ^dA_dt œ œ5( 2) 14 cm /sec, ^# increasing (b) Pœ j 2 2w Ê ^dP_dt œ2 ^d_dt^j 2 ^dw_dt œ 2( 2) 2(2)œ0 cm/sec, constant

(c) DœÈw^# j œ^# aw^# j^#b^"Î# Ê œ ^"_#aw^# j^#b^"Î#ˆ2w 2 j ^j‰ Ê œ ^j

j

dD dw d dD

dt dt dt dt

w w

dw d

dt dtj

# #

È

cm/sec, decreasing œ ⁽⁵⁾⁽²⁾_È25^{(12)( 2)}₁₄₄ œ ¹⁴₁₃

12. (a) Vœxyz Ê yz xz xy ^dV_dt œ ^dx_dt ^dy_dt ^dz_dt Ê ^dV_dt¸ œ(3)(2)(1)(4)(2)( 2) (4)(3)(1)œ2 m /sec

(4 3 2)ß ß $

(b) Sœ2xy2xz2yz Ê (2y^dS_dt œ 2z) (2x^dx_dt 2z) (2x^dy_dt 2y) ^dz_dt Ê ^dS¸ œ(10)(1)(12)( 2) (14)(1)œ0 m /sec

dt (4 3 2)ß ß #

(c) j œÈx^#y^#z^#œax^#y^#z^{# "Î#}b Ê ^d_dt^j œ _Èx# ^x_{y# z#} ^dx_{dt Èx#} ^y_{y# z#} ^dy_{dt Èx#} ^z_{y# z#} ^dz_dt Ê ^d_dt^j¸_{(4 3 2)ß ß} œŠ_È⁴₂₉‹(1)Š_ȳ₂₉‹( 2) Š_Ȳ₂₉‹(1)œ0 m/sec

13. Given: ^dx_dt œ5 ft/sec, the ladder is 13 ft long, and xœ12, yœ5 at the instant of time

(a) Since x^#y^#œ169 Ê ^dy_dt œ _{y dt}^{x dx} œ ˆ ‰¹²₅ (5)œ 12 ft/sec, the ladder is sliding down the wall (b) The area of the triangle formed by the ladder and walls is Aœ ^"_#xy Ê ^dA_dt œˆ ‰ Š^"_# x ^dy_dt y ^dx_dt‹. The area

is changing at [12( 12)^"_# 5(5)]œ ¹¹⁹_# œ 59.5 ft /sec.^#

(c) cos )œ ₁₃^x Ê sin ) ^d_dt⁾ œ₁₃^" †^dx_dt Ê ^d_dt⁾ œ _{13 sin} ^" ₎†^dx_dt œ ˆ ‰₅^" (5)œ 1 rad/sec

14. s^#œy^#x ^# Ê 2s 2x 2y ^ds_dt œ ^dx_dt ^dy_dt Ê ^ds_dt œ^"_sŠx y ^dx_dt ^dy_dt‹ Ê ^ds_dt œ_È^"₁₆₉[5( 442) 12( 481)]

614 œ knots

15. Let s represent the distance between the girl and the kite and x represents the horizontal distance between the girl and kite Ê s^#œ(300)^#x ^# Ê ^ds_dt œ ^{x dx}_{s dt} œ^400(25)₅₀₀ œ20 ft/sec.

16. When the diameter is 3.8 in., the radius is 1.9 in. and ^dr_dt œ₃₀₀₀^" in/min. Also Vœ6 r 1^# Ê ^dV_dt œ12 r 1 ^dr_dt 12 (1.9) 0.0076 . The volume is changing at about 0.0239 in /min.

Ê ^dV_dt œ 1 ˆ₃₀₀₀^" ‰œ 1 ^$

17. Vœ ^"₃1r h, h^# œ ₈³(2r)œ ^3r₄ Ê œ r ^4h₃ Ê Vœ ^"₃1ˆ ‰^4h₃ ^#hœ ^{16 h}₂₇^{1 $} Ê ^dV_dt œ ^{16 h}₉^{1 # dh}_dt (a) ^dh¸ ˆ ⁹ ‰(10) ⁹⁰ 0.1119 m/sec 11.19 cm/sec

dt h=4œ 16 4₁ # œ 256₁ ¸ œ

(b) rœ ^4h₃ Ê ^dr_dt œ ^{4 dh}_{3 dt} œ ⁴₃ˆ₂₅₆⁹⁰₁‰œ₃₂¹⁵₁ ¸0.1492 m/secœ14.92 cm/sec

18. (a) Vœ ^"₃1r h and r^# œ^15h_# Ê Vœ^"₃1ˆ^15h_# ‰^#hœ^{75 h}₄^1$ Ê ^dV_dt œ^{225 h}₄^{1 # dh}_dt Ê ^dh_dt¸ œ _{225 (5)}^{4( 50)}_{1 #} œ ₂₂₅⁸₁

h=5

0.0113 m/min 1.13 cm/min

¸ œ

(b) rœ ^15h_# Ê ^dr_dt œ^{15 dh}_{# dt} Ê ^dr_dt¸ œˆ ‰ ˆ¹⁵_{# 225}⁸ ‰œ ₁₅⁴ ¸ 0.0849 m/secœ 8.49 cm/sec

h=5 1 1

19. (a) Vœ ¹₃y (3R^# y) Ê ^dV_dt œ ¹₃c2y(3R y) y ( 1) ^# d ^dy_dt Ê ^dy_dt œ¹₃a6Ry3y^{# "}b‘ ^dV_dt Ê at Rœ13 and yœ8 we have ^dy_dt œ ₁₄₄^"₁( 6) œ ₂₄^"₁ m/min

(b) The hemisphere is on the circle r^#(13y)^#œ169 Ê œ r È26yy m^#

(c) rœa26yy^#b^"Î# Ê œ ^"_#a26yy^#b^"Î#(262y) Ê œ Ê ¸ œ ˆ_#^"‰

dr dr dr 13 8

dt dt dt dt dt 4

dy 13 y dy

26y y 26 8 64

È # y=8 È † 1

œ ₂₈₈⁵₁m/min

20. If Vœ⁴₃1r , S^$ œ4 r , and 1^{# dV}_dt œkSœ4k r , then 1^{# dV}_dt œ4 r 1^{# dr}_dt Ê 4k r1^#œ4 r 1^{# dr}_dt Ê ^dr_dt œk, a constant.

Therefore, the radius is increasing at a constant rate.

21. If Vœ⁴₃1r , r^$ œ5, and ^dV_dt œ100 ft /min, then 1 ^{$ dV}_dt œ4 r 1^{# dr}_dt Ê ^dr_dt œ1 ft/min. Then Sœ4 r 1^# Ê ^dS_dt 8 r 8 (5)(1) 40 ft /min, the rate at which the surface area is increasing.

œ 1 ^dr_dt œ 1 œ 1 ^#

22. Let s represent the length of the rope and x the horizontal distance of the boat from the dock.

(a) We have s^# x^# 36 . Therefore, the boat is approaching the dock at œ Ê ^dx_dt œ ^{s ds}_{x dt} œ ^{s ds}_dt

s 36 È#

( 2)^dx_dt¸ ¹⁰ 2.5 ft/sec.

10 36

s=10œ _{È #} œ

(b) cos )œ ⁶_r Ê sin ) ^d_dt⁾ œ _r^{6 dr}# _dt Ê ^d_dt⁾ œ_{r sin} #⁶ ₎ ^dr_dt. Thus, rœ10, xœ8, and sin )œ ₁₀⁸ ( 2)Ê ^d_dt⁾ œ _10#⁶_{ˆ ‰}8 œ ₂₀³ rad/sec

10 †

23. Let s represent the distance between the bicycle and balloon, h the height of the balloon and x the horizontal distance between the balloon and the bicycle. The relationship between the variables is s^#œh^#x^# h Ê ^ds_dt œ _s^"ˆ ^dh_dt x ^dx_dt‰ Ê [68(1)^ds_dt œ₈₅^" 51(17)]œ11 ft/sec.

24. (a) Let h be the height of the coffee in the pot. Since the radius of the pot is 3, the volume of the coffee is Vœ9 h 1 Ê ^dV_dt œ9 1^dh_dt Ê the rate the coffee is rising is ^dh_dt œ ₉^"₁ ^dV_dt œ ¹⁰₉₁ in/min.

(b) Let h be the height of the coffee in the pot. From the figure, the radius of the filter rœ^h_# Ê Vœ ₃^"1r h^# , the volume of the filter. The rate the coffee is falling is ( 10) in/min.

œ ¹₁^h#^{$ dh}_dt œ₁⁴_h# ^dV_dt œ #₅⁴₁ œ ₅⁸₁

25. yœQD^" Ê ^dy_dt œD^{" dQ}_dt QD^{# dD}_dt œ₄₁^" (0)₍₄₁₎²³³#( 2) œ ₁₆₈₁⁴⁶⁶ L/min Ê increasing about 0.2772 L/min 26. (a) 3x^dc_dt œa ^#12x15 3(2)b ^dx_dt œa ^#12(2)15 (0.1)b œ0.3, 9 9(0.1)^dr_dt œ ^dx_dt œ œ0.9, 0.9^dp_dt œ 0.3œ0.6 (b) ^dc_dt œa3x^#12x45x^#b^dx_dt œa3(1.5)^#12(1.5)45(1.5)^#b(0.05) =1.5625, 70 70(0.05)^dr_dt œ ^dx_dt œ œ3.5, 3.5^dp_dt œ ( 1.5625)œ5.0625

27. Let P(x y) represent a point on the curve yß œx and the angle of inclination of a line containing P and the^# ) origin. Consequently, tan ) œ ^y_x Ê tan )œ ^x_x^# œx Ê sec^#)^d_dt⁾ œ^dx_dt Ê ^d_dt⁾ œcos^#) ^dx_dt . Since ^dx_dt œ10 m/sec and cos^#)kx=3œ _y#^x#_x# œ ₉#^3#₃# œ ₁₀^" , we have ^d_dt)¸x=3œ1 rad/sec.

28. yœ ( x)^"Î# and tan )œ^y_x Ê tan )œ^{( x)}_x^"Î# Ê sec^#)^d_dt) œˆ ‰^{"# ( x) "Î#( 1)x}_x# ^{( x)"Î#(1) dx}_dt

cos . Now, tan cos cos . Then Ê ^d_dt⁾ œŒ _x ^x a b ˆ ‰^dx_dt œ ²₄ œ Ê œ ²₅ Ê œ ⁴₅

ccx

2È x # #

#"

È

# ) ) ) È )

( 8)^d_dt⁾ œŠ⁴⁴₁₆²‹ ˆ ‰⁴₅ œ ²₅ rad/sec.

29. The distance from the origin is sœÈx^#y and we wish to find ^{# ds}_dt¸

(5 12)ß

xœ ^"_#a ^#y^{# "Î#}b Š2x ^dx_dt 2y ^dy_dt‹¹ œ^{(5)( 1)}₂₅^{(12)( 5)}₁₄₄ œ 5 m/sec

(5 12)ß È

30. When s represents the length of the shadow and x the distance of the man from the streetlight, then sœ ³₅x.

(a) If I represents the distance of the tip of the shadow from the streetlight, then Iœ Ês x ^dI_dt œ ^ds_dt ^dx_dt (which is velocity not speed) Ê ¸ ¸^dI_dt œ¸^{3 dx}_{5 dt} ^dx_dt¸œ¸ ¸ ¸ ¸⁸₅ ^dx_dt œ⁸₅k œ5k 8 ft/sec, the speed the tip of the shadow is moving along the ground.

(b) ^ds_dt œ ^{3 dx}_{5 dt} œ³₅( 5) œ 3 ft/sec, so the length of the shadow is decreasing at a rate of 3 ft/sec.

31. Let sœ16t represent the distance the ball has fallen,^# h the distance between the ball and the ground, and I the distance between the shadow and the point directly beneath the ball. Accordingly, s œh 50 and since the triangle LOQ and triangle PRQ are similar we have Iœ ₅₀^30h_h Ê hœ5016t and I^# œ ₅₀^{30 50a}_a50^16t_16t^#b#_b

30 1500 ft/sec.

œ ¹⁵⁰⁰_16t# Ê ^dI_dt œ ¹⁵⁰⁰_8t$ Ê ^dI_dt¸ œ

t=¹₂

32. Let sœdistance of car from foot of perpendicular in the textbook diagram Ê tan )œ₁₃^s_# Ê sec^#) ^d_dt⁾ œ ₁₃^"_# ^ds_dt

; 264 and 0 2 rad/sec. A half second later the car has traveled 132 ft Ê ^d_dt⁾ œ ^cos₁₃₂^{#) ds}_dt ^ds_dt œ )œ Ê ^d_dt⁾ œ

right of the perpendicular Ê k k) œ ¹₄, cos^#)œ ^"_#, and ^ds_dt œ264 (since s increases) Ê ^d_dt⁾ œ ₁₃₂^{ˆ ‰"#} (264)œ1 rad/sec.

33. The volume of the ice is Vœ⁴₃1r^$4₃14 ^$ Ê ^dV_dt œ4 r 1^{# dr}_dt Ê ^dr_dt¸ œ₇₂⁵ in./min when ^dV_dt œ 10 in /min, the^$

r=6 1

thickness of the ice is decreasing at ₇₂⁵₁ in/min. The surface area is Sœ4 r 1^# Ê ^dS_dt œ8 r 1 ^dr_dt Ê ^dS_dt¸ œ481ˆ₇₂⁵₁‰

r=6

in /min, the outer surface area of the ice is decreasing at in /min.

œ ¹⁰₃ ^{# 10}₃ ^#

34. Let s represent the horizontal distance between the car and plane while r is the line-of-sight distance between the car and plane Ê œ 9 s^# r ^# Ê ^ds_dt œ ^{r dr}_dt Ê ^ds_dt œ ⁵ ( 160) œ 200 mph

r 9 16

È# ¸ È

r=5

speed of plane speed of car 200 mph the speed of the car is 80 mph.

Ê œ Ê

35. When x represents the length of the shadow, then tan )œ⁸⁰_x Ê sec^#) ^d_dt⁾ œ ^{80 dx}_x# _dt Ê ^dx_dt œ ^{x sec}₈₀ ^{) )d}_dt.

# #

We are given that ^d_dt⁾ œ0.27°œ _#³₀₀₀¹ rad/min. At xœ60, cos )œ³₅ Ê ft/min 0.589 ft/min 7.1 in./min.

¸ ¸^dx_dt œ¹^{x sec#}₈₀ ^{#) )d}_dt¹¹ œ ³₁₆¹ ¸ ¸

Š^ddt)=2000³1 and sec =) ⁵3‹

36. Let A represent the side opposite and B represent the side adjacent . tan ) ) )œ ^A_B Ê sec^#) ^d_dt⁾ œ _B^{" dA}_{dt B}^{A dB}# _dt

t Ê at Aœ10 m and Bœ20 m we have cos )œ ²⁰ œ ² and ^d œ ( 2) ¹⁰ (1) ⁴

10È5 È5 dt) ˆ ‰_#^"0 ˆ400 ‰‘ ˆ ‰5

œˆ^"_{10 40}^"‰ ˆ ‰₅⁴ œ ₁₀^" rad/secœ ^18°₁ /sec¸ 6°/sec

37. Let x represent distance of the player from second base and s the distance to third base. Then ^dx_dt œ 16 ft/sec (a) s^#œx^#8100 Ê 2s ^ds_dt œ2x ^dx_dt Ê ^ds_dt œ ^{x dx}_{s dt} . When the player is 30 ft from first base, xœ60

s 30 13 and ( 16) 8.875 ft/sec Ê œ È ^ds œ ⁶⁰ œ ³² ¸

dt 30È13 È13

(b) cos )"œ ⁹⁰_s Ê sin )" ^d_dt^)" œ ⁹⁰_s# †^ds_dt Ê ^d_dt^)" œ_{s sin} #⁹⁰₎" †^ds_dt œ⁹⁰_sx †^ds_dt. Therefore, xœ60 and sœ30È13 rad/sec; sin cos

Ê ^d_dt œ ^{90 32} œ ₆₅⁸ œ⁹⁰_s Ê ^d_dt œ ⁹⁰_s ^ds_dt Ê ^d_dt œ_{s cos} ^{90 ds}_dt

30 13 (60) 13

) ) )

" # # )

# # #

Š È ‹ †ŠÈ ‹ )# )# † †

. Therefore, x 60 and s 30 13 rad/sec.

œ _sx⁹⁰†^ds_dt œ œ È Ê ^d_dt^)# œ ₆₅⁸

(c) ^d_dt^)" _{s sin #}⁹⁰_{) dt}^ds _s^{90 x}_s ^dx_dt ⁹⁰_s# ^dx_{dt x#} ⁹⁰₈₁₀₀ ^dx_dt lim ^d_dt^)"

" #

œ † œ _{ˆ †}x_‰† † œ œ Ê

s

ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰

xÄ !

lim ( 15) rad/sec;

œ œ œ œ œ

xÄ !ˆ_x# ⁹⁰₈₁₀₀‰ ₆ ^d_dt# _{s cos}# ^{90 ds}_dt Š_s#⁹⁰‹ˆ ‰ ˆ ‰^x_s ^dx_dt ˆ _s⁹⁰# ‰ ˆ ‰^dx_dt

" ) #

) † _†^x

s

lim rad/sec œˆ_x#⁹⁰₈₁₀₀‰^dx_dt Ê ^d_dt^# œ^"₆

xÄ ! ⁾

38. Let a represent the distance between point O and ship A, b the distance between point O and ship B, and D the distance between the ships. By the Law of Cosines, D^#œa^#b^#2ab cos 120°

2a 2b a b . When a 5, 14, b 3, and 21, then

Ê ^dD_dt œ_#^"_D ^da_dt ^db_dt ^db_dt ^da_dt‘ œ ^da_dt œ œ ^db_dt œ ^dD_dt œ ⁴¹³_2D

where Dœ7. The ships are moving ^dD_dt œ29.5 knots apart.