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Unformatted text preview: . ....o.o.u.occo ooooo g  o... llll FERENTIATING EQUATIONS
NI LATE RATES T0 3.7 In Ilti.\ .\(‘(‘lll)ll we will study related rules problems. In such problems one tries mjiml
Iltt' l'llll‘ (II which smite quantity is changing by relating the quantity to other quantities
u‘ltu.w mm oftImage are known. Figure 3.7.! shows a liquid draining through a conical ﬁlter. As the liquid drains. its volume
V. height h. and radius r are functions of the elapsed lime t, and at each instant these
variables are related by the equation V = Z{Fl/1 3 If we differentiate both sides of this equation with respect to t, then we obtain dV 7: ,dh dr 17 db dr —=— —+h 2— =— 2— 21— m 3 [’ a: (211)] 3 ( dt + "all Thus. ii. at a given instant we have values for r. h, and two of the three rates in this equation.
then we can solve for the value of the third rate at this instant. In this section we present
some speciﬁc examples that use this basic idea. t—Hi l’ h ‘_ __ ‘A Figure 5‘7  Example 3! Assume that oil spilled from a ruptured tanker spreads in a circular pattern
whose radius increases at a constant rate of 2 ft/s. How fast is the area of the spill increasing
when the radius of the spill is 60 ft? STubN THE Exemmes ANb 30L“; THE commas Peoeueus AT THE ENb. 220 The Derivative _ Solution. Let
_Ol_:"' ' r = number of seconds elapsed from the time of the spill
pl r = radius oi the spill In feet alter.t seconds .l = area of the spill in square feet after.r seconds G?! [Ell . 1 . . . . . . 'I
C3— tthure 3.7.3). We know the rate at which the radius ts increasing. and we want to hint rl _
r rate at which the area is increasing at the instant when r = 60: that is. we want to ﬁlhi
o'A , d" .
— given that i = 2 l‘l/s
(I! ;=(,[; (If
mm 313 From the formula for the area of a circle we obtain .l = Jr."2 W  Because A and r are functions off. we can differentiate both sides oft  J with respect In .
to obtain (M (Ir
_ = “Hr——
th (I!
Thus. when r = 60 the area ofthe spill is increasing at the rate of
[A
‘— = memo) = 240ni13/s
d" r=6ll
or approximately 754 flz/s. ‘ With only minor variations. the method used in Example 1 can be used to solve a varictv
of related rates problems. The method consists of ﬁve steps: A Strategy for Solving Related Rates Problems Step 1. identify the rates ofchange that are known and the rate of change that
is to be found. Interpret each rate as a derivative ot'a variable with
respect to time. and provide a description of each variable involved. Step 2. Find an equation relating those quantities whose rates are identiﬁed
in Step 1. In a geometric problem. this is aided by drawing an appro
priately labeled ﬁgure that illustrates a relationship involving these
quantities. Step 3. Obtain an equation involving the rates in Step I by differentiating
both sides of the equation in Step 2 with respect to the time variable. Step 4. Evaluate the equation found in Step 3 using the known values for the
quantities and their rates of change at the moment in question. Step 5. Solve for the value of the remaining rate of change at this moment. w.\l<NtNG. Do not substitute prematurely; that is. always perform the differentiation in
Step 3 before performing the substitution in Step 4. Example 2 A baseball diamond is a square whose sides are 90 ft long (Figure 3.7.3).
Suppose that a player running from second base to third base has a speed of 30 ft/s at the
instant when he is 20 ft from third base. At what rate is the player‘s distance from home
plate changing at that instant? _ Solution. The rate we wish to ﬁnd is the rate of change of the distance from the player
Home to home plate. We are given the speed of the player as he moves along the base path from
Figure 3.7.3 second to third base. which tells us both the speed with which he is moving away from Rocket £1 3/
Elevation
angle
 3000 ll Hm—e
Launching
pad moo ft 3.7 Related Rates 221 second base and the speed with which he is approaching third base. Let
t = number of seconds since the player left second base
at = distance in feet from the player to second base
.r = distance in loot from the player to third base
_1‘ 2 distance in feet from the player to home plate
Thus. we want to lind
rt'r _ dw . (1x .
—' glven that —— = 30 lt/s and — = ~30 it/s
‘1' ..=2u ‘1' .tzzo ‘ ’ .\'=20 [Note that (£l‘_\‘/(l'.\')'\=:” is negative because .r is decreasing with respect to r.]
From the Theorem of Pythagoras, x3 + 902 = y3 (2)
Differentiating both sides of this equation with respect to r yields
(Lt d \' dx (1 t'
_ = 2 ._  __ = ._
‘ d1 '\ (it or i (It '\ d: (3) To evaluate (3) at the instant when .l‘ = 20 we need a value for _\‘ at this instant. Substituting
J: = 20 into (2) yields 400+amo=tittum}2 or _r_,=3.,=\/850 = tot/85 Then, evaluating (3) when .t‘ = 20 yields 1’ ' —6l)0 60
or I '1 —— t —6.5l ft/s .t=2tt a .r=2o = IOV 85 — Jig The negative sign in the answer tells us that _v is decreasing, which makes sense in the
physical situation of the problem (Figure 3.7.3). 4 d v
20 . H30) 2 tot/ss (1—: l‘t In it i' stint it ‘I<. In our solution for Example 2 we chose to relatex and y. An alternative
approach would be to relate it: and _v. Solve the problem using this alternative approach. Example 3 In Figure 3.7.4 we have shown a camera mounted at a point 3000 ft from the
base of a rocket launching pad. if the rocket is rising vertically at 880 11/8 when it is 4000
ft above the launching pad. how fast must the camera elevation angle change at that instant
to keep the camera aimed at the rocket? Solution. Let
t = number of seconds elapsed from the time of launch
if) = camera elevation angle in radians after: seconds
It = height of the rocket in feet afterr seconds (Figure 3.7.5). At each instant the rate at which the camera elevation angle must change is
dip/(ll. and the rate at which the rocket is rising is dlt/dr. We want to ﬁnd I H .
ﬂ given that S = 880 tt/s
d" h=4auu ‘ 3' h=auou
From Figure 3.7.5 we see that
h
lat = (4)
I “b 3000 Because d) and h are functions of r. we can differentiate both sides of (4) with respect to t' to obtain
dd) l (Ht (SEC‘tﬁlz = ME (5)
When It = 4000. it follows that
5000 5
{SEC ¢Jrr=4ooo = m = 3 222 The Derivative 5000 4000 \ab
3000
Figure 3.7.6 I.
i In em
_v
Funnel to 1—
hold filter Figure 3.7.7 A
—u
‘—
4— . The same volume has drained, but
the change in height is greater near
the bottom than near the top. Figure 3. 7. 8 (Figure 3.7.6). 50 that from (5) Y 5 do _ J 22
3 a: _ 3000 ' _ 5
h=4000
I 22 9
(—qbl =—_.—_=—_ktlead/smollﬁdcgfs 4'
dr h_ ﬁlm 7:» 2:) 625 Example 4 Suppose that liquid is to be cleared ot'sediment by allowing it to drain through _
a conical litter that is 16 cm high and has a radius ol'4 cm at the top (Figure 3.7.7). Suppose .
also that the liquid ﬂows out of the cone at a constant rate of 2 cm'Vmin. (a) Do you think that the depth of the liquid will decrease at a constant rate‘.’ Give a verbal .
argument that justifies your conclusion. (b) Find a formula that expresses the rate at which the depth of the liquid is changing in I
terms of the depth. and use that formula to determine whether your conclusion in part .
(a) is correct. (c) At what rate is the depth of the liquid changing at the instant when the liquid in the 
cone is 8 cm deep“? Solution ((1). For the volume of iiquid to decrease by aﬁxed amount, it requires a greater .
decrease in depth when the cone is close to empty than when it is almost full (Figure 3.7.3). '=
This suggests that for the volume to decrease at a constant rate. the depth must decrease at .
an increasing rate. Solution ([2). Let = time elapsed from the initial observation (min)
volume of liquid in the cone at time i (cm'l‘)
depth of the liquid in the cone at timer (cm)
radius of the liquid surface at time t (cm) It I
V
i
1 (Figure 3.7.7). At each instant the rate at which the volume of liquid is changing is dV/dt
and the rate at which the depth is changing is dy/dt. We want to express d‘v/dt in terms of
_v given that dV/dr has a constant value of (IV/d! 2 —2. (We must use a minus sign here
because V decreases as t increases.) From the formula for the volume of a cone. the volume V. the radius r, and the depth r
are related by V = %JTI‘1_V to! If we differentiate both sides 0H6) with respect to r. the right side will involve the quantity
(tr/cit. Since we have no direct information about dr/dr, it is desirable to eliminate r front
(6) before differentiating. This can be done using similar triangles. From Figure 3.7.? we
see that r 4 
 = _, or r 2 —V
_v l6 4'
Substituting this expression in (6) gives
1T 1 .
v = —\l‘ [7]
48' Differentiating both sides of (7) with respect to t we obtain (N If ( ,dv)
— = — 3\"—
(It 48 ' d! or 2 32 ' ii_ _ l_6 dV =_6;(92) : a tit
dt JT—V' dt Jrv rry and
d_\'
d t 32 .
fry a’y 32 Was?) (It )3“ In lixercises I—4. both .r and _\' denote functions of: that are
related by the given equation. Use this equation and the given
derivative information to find the specilied derivative. (i l.)liquation: _r = 11' + 5. la) Given that (Ir/(l: = 2. ﬁnd (Ir/ill when .r .
th) Given that (Ix/(It = — I. find (Ix/(1': when .r = (I.
Gliquation: .r + 4)" = 3.
(a) Given that (hr/(It = I. ﬁnd (Ir/d! When .t'
th) Given that div/rt"! = 4. lind (Ix/(It when .t' ll II II
P” I" ta) Given that d.r/dr = I. Iind ctr/m when I J? 5‘ T th) Given that (Ir/(h = —2. ﬁnd (ﬁr/alt when J5 «a T' T 9 Equation: .1'3 + y: = 2.1. In) Given that (Ix/(h = ~3. find (Ir/(It when
(any) = (I. I). rht Given that (Lt/ill = 3. lind (ix/(It when 2 + \/E J5 __ T 1 (.t'. _\') = (.r. _\'I = a .ct A be the area ol'a square whose sides have length .t'. and assume that .r varies with the time I. no Draw a picture of the square with the labels A and .r
placed appropriately. il'I Write an equation that relates A and .r. w Use the equation in part (ht to Iind an equation that
relates (IA Mr and air/(It. ull At a certain instant the sides are 3 ft long and increasing
at a rate of 2 lit/min. How last is the area increasing at
that instant? 3.7 Related Rates 223 which expresses (Ir/(l! in terms of _\'. The minus sign tells tts that Ir is decreasing with time. tells us how fast _r is decreasing. From this formula we see that ldy/drl increases as y
decreases. which conﬁrms our conjecture in part tail that the depth of the liquid decreases
more quickly as the liquid drains through the ﬁller. Solution (cl. The rate at which the depth is changing when the depth is 8 cm can be
obtained from (8) with _v = 8: —— R: —l). l6 cm/min 4
231 ©Let A he the area ol‘a circle of radius r. and assume that : increases with the time I. (a) Draw a picture of the circle with the labels A and :
placed appropriately. (h) Write an equation that relates A and r. (ct Use the equation in part (b) to ﬁnd an equation that
relates (M Mr and (Ir/(It. It!) At a certain instant the radius is 5 cm and increasing at
the rule of 3 cut/s. How fast is the area increasing at
that instant? C73 Let V be the volume ol'a cylinder having height It and radius
r. and assume that It and r vary with time.
(a) How are (IV/(Ir. (Ht/ch. and (Ir/d! related?
(b) At a certain instant. the height is 6 in and increasing at I
in/s. while the radius is I0 in and decreasing at I in/s.
How fast is the volume changing at that instant? Is the
volume increasing or decreasing at that instant? .Let I be the length of a diagonal ol'a rectangle whose sides
have lengths .r and .\‘. and assume that .t' and _\' vary with
time. (a) How are (ll/(It. (Ix/(h. and (fr/(It related? (h) l‘.r increases at a constant rate ole ft/s and _r decreases
at a constant rate of 41 I't/s. how fast is the size of the
diagonal changing when .r = 3 It and _r = 'I It? Is the
diagonal increasing or decreasing at that instant? 9. Let H tin radians) he an acute angle in a right triangle. and
let .r and _\'. respectively. he the lengths ol'the sides adjacent
to and opposite H. Suppose also that .r and _\' vary with time.
[at How are dH/it'r. rt'.r/rlt. and (Ir/{It related?
lb] At a certain instant. .r = 2 units and is increasing at l unit/s. while ‘t‘ = 2 units and is decreasing at :1; unit/s.
How fast is H changing at that instant? Is H increasing or decreasing at that instant?
@ Suppose that :. = 3533. where both .r and _\ are changing
with time. At a certain instant when .r = I and _r = 2. .r is
decreasing at the rate of 2 units/s. and r is increasing at the 224 The Derivative rate ol‘3 units/s. How last is :. ch
increasing or decreasing '3 angittg at this instant? ls :. H. The mintttc hand of a certain clock is 4 in long
from the moment when the hand is pointing
how fast is the area of tlte sector that is sWept hattd increasing at any instant dttring the
the hand? . Starting
straight up.
ottt by the
next revolution of A stone dropped into a still pond sends out a circular rip ple whose radius increases at a constant rate ol'3 t‘t/s. How
rapidly is the area enclosed by
end of l() s‘.’ @Oil spilled from a ruptured tanker spreads in ac
area increases at a const radius of the spill incre the ripple increasing at the ircle whose
ant rate ot'b mi3/h. How fast is the
asing when th ‘ area is 9 mil? A spherical balloon is inﬂated so th
ing at the rate ol’3 l't“/min. How la
balloon increasing w l . A spheri CFCZISCS at its volume is increas st is the diameter of the
hen the radius is l l't'.’ cal balloon is to be dell
at a constant rate of l5 c
air be removed when the r 16. A l7l'l ladder is leaning against a wall. ll‘tltc bottom ol‘the ladder is pulled along the grottnd away from the wall at a
constant rate ofS lit/s. how fast will the top ol‘the ladder be
moving down the wall when it is 8 ti above the ground? A l3t‘t ladder is leaning ag
ladder slips down the wall at
the foot be moving away
above the ground? @A l0l't plank is leaning against a wall. ll‘ at a certain instant
the bottom of the plank is 2 ft from the wall attd is being
pushed toward the wall at the rate ol‘6 in/s. how fast is the
acute angle that the plank tnakes with the grottnd increasing? A softball diamond is a square whose side
Suppose that a player running from ﬁrst to s
speed of 25 l't/s at the instant whc
base. At what rate is the play changing at that instant? ated so that its radius de nt/min. At what rate tnttst
aditts is 9 cm? ainst a wall. If the top of the
a rate of2 ft/s. how fast will
from the wall when the top is 5 it s are ()0 ft long.
econd base has a
n she is ID ft from second
er’s distance from home plate A rocket. rising vertically
is on the ground 5
rocket rising whe . is tracked by a radar station that
mi from the launchpad. How l nit is l mi high and its dist
radar station is increasing 'ast is the
ancc from the
at a rate ol‘2(}()tl mi/h‘.’ 'a and rocket shown in Figure 3.7.4.
is the cameratorocket distance
is 4000 It up 21. Forthe cantet at what rate
changing when the rocket
and rising \crticall} at 880 It, s'.‘ 12. For the camera and rocket shown in Figure 3.7.4. at what I 'illC
is the rocket rising \\ hen the elevation angle is .7/4 radians
and increasing at a rate ol‘t).2 radian/s?
23. A satellite is in an elliptical orbit around the Earth. lts dis
l ance r tin miles) l'ront the center of the Earth is given by
4095 l+0.l3cosH r= 26. Grain pouring from 28. Wheat is poured through a chute
and falls in a c where H is the angle measured from the
nearest the Earth's surl'ace t see the accompanying tigut'c).
ta) Find the altitude of the satellite at perigee (the poir.
nearest the surface ol'the Earth)
l'arthest from the surl'ace
the radius of the Earth.
(b) At the instant when (J is I20 . the
at the rate of 2.7 /min. Find the
lite and the rate at which the
instant. Express the r point on the orb? ol'the Earth). Use 3960 mi at: angle (J is increasing
altitude of the satel
altitttde is changing at this
ate in units ol‘ mi/ntin. Apogee »— ~ —  — — 3 Perigee Figure lis~23 24. An aircraft is ﬂying horizontally at a constant height ttl'400ll
it above a lixed observation point t see the accompanying ﬁg»
ure). At a certain instant the angle ol‘elevation H is 30 and
decreasing. and the speed of the aircraft is 300 mi/h. (a) How fast is (1 decreasing at this inst
result in units ol‘degrecs/s. t b) How fast is the distance between the aircraft and th
observation point changing at this instant? Express the
resttlt in units of l‘t/s. Use l mi = 5280 ll. ant'.’ Express the ﬁzz: 4000 it
’ tr
rs t Figure E\2l a A conical water tank with \ertex down has a radius of“) It at the top and is 34 It high. It \
at a rate ot~ 3t) I'l"/min. how 1
increasing when the water is l rater lltm’s into the ta:
'ast is the depth ol‘ the was
(1 ti deep'.’ a chute at the rate ol‘ X it ' "min torn;
conical pile “hose aititude is always to ice its radius. HR last is the altitude ol‘ the pile incr=asing at the instant whs
the pile is 6 ft high? Sand pouring from a chute t‘orms a cortical pile \t'hose berg
is always equal to the diameter. it" the height increases a
constant rate ol'S l't/ min. at what rate is sand pouring in
the chute when th pile is it) ft high? at the rate ol' t0 l't"/n‘;
onical pile whose bottom r adios is a\\a_\ s i3; and at apogee t the point l.‘ the altitude. How fast will the circumference of the base be
1; increasing when the pile is 8 ft high? 2". An aircraft is climbing at a 30 angle to the horizontal. How
fast is the aircraft gaining altitude if its speed is 500 mi/h? . .A boat is pulled into a dock by tneans ofa rope attached to a
t! ' ' pulley on the dock tsee the accompanying ﬁgure). The rope
is attached to the bow ofthe boat at a point 10 ft below the
pulley. lt'the rope is pulled through the pulley at a rate of 20
ft/min, at what rate will the boat be approaching the dock
when 125 ft of rope is out? Pulley
'  Boat Dock
l‘igure Eat3t] ,tl lior the boat in Exercise 30. how fast mttst the rope be pulled
if we want the boat to approach the dock at a rate of l 2 ft/min
at the instant when 25 ft of rope is out? .A. .A man 6 ft tail is walking at the rate of 3 ft/s toward a
streetlight 13 ft high (see the accompanying tigttrct. tttl At what rate is his shadow length changing? th) How fast is the tip of his shadow moving? l‘igure Bit32 . all. A beacon that makes one revolution every l0 s is located on
it ship anchored 4 kilometers from a straight shoreline. How
last is the beam moving along the shoreline when it makes
.in angle of 45 with the shore? 39 \n aircraft is ﬂying at a constant altitude with a constant é? , , peed of 600 mi/h. An untiaircraft missile is tired on a ' ' ult'ttlgl'll line perpendicular to the ﬂiglu path ofthe aircraft so that it will hit the aircraft at a point P tsec the accompanying . insure). At the instant the aircraft is 2 mi from the impact point P the missile is 4 mi from P and ﬂying at lltttl mi/h. ‘ \t that instant. how rapidly is the distance between missile
.Illtl aircraft decreasing'.’ P z 1 nature Ex34 3.7 Related Rates 225 35. Solve Exercise 34 under the assumption that the angle be
tween the ilight paths is l20 instead ofthe assumption that
the paths are perpendicular. IHim: Use the law of cosines.] @A police helicopter is ﬂying due north at IOO mi/h and at a
constant altitude of 4 mi. Below. a car is traveling west on a
highway at 75 mm: At the moment the helicopter crosses
over the highway the car is 2 mi east ofthe helicopter. (a) How fast is the distance between the car and helicopter
changing at the moment the helicopter crosses the high
way? (b) ls the distance between the car and helicopter increasing
or decreasing at that moment? 37. A particle is moving along the curve whOse equation is
‘
.rr' 8 l+y2 _§ Assume that the .rcoordinate is increasing at the rate of6 units/s when the particle is at the point (1.2). (a) At what rate is the _vcoordinate ot‘ the point changing
at that instant? (b) Is the particle rising or falling at that instant? 38. A point P is moving along the curve whose equation is
_v = VJ"? + [7. When I" is at (2. 5). _t‘ is increasing at the
rate of2 units/s. How fast is .t' changing? 39. A point P is moving along the line whose equation is
r = 23'. How fast is the distance between P and the point
(3.0) changing at the instant when P is at (1(3) ifx is
decreasing at the rate of2 units/s at that instant? 40. A point P is moving along the curve whose equation is
y = ﬂ. Suppose that .r is increasing at the rate of4 units/s
when .t’ = 3. (a) How fast is the distance between P and the point (2. 0)
changing at this instant? (b) How fast is the angle ofinclination ofthe line segment
from P to (2. 0) changing at this instant? @ A particle is moving along the curve 3‘ = .r/tx2 + l). Find
all values of .r at which the rate ofchange of; with respect
to time is three times that of y. lAssume that dx/dt is never
zero] @A particle is moving along the curve lt‘n‘3 + 9_\'2 = 144.
Find all points U. y] at which the rates of change of.t‘ and
y with respect to time are equal. IAssulnc that (Ix/(ft and
ply/(It are never both zero at the satne point.] @The thin lens equation in physics is
l l l ;+§=7 where s is the object distance from the lens. 5 is the image
distance from the lens. and f is the focal length of the lens.
Suppose that a certain lens has a focal length of 6 cm and
that an object is moving toward the lens at the rate 0f2 crn/s.
How fast is the image distance changing at the instant when
the object is 10 cm from the lens? Is the image moving away
from the lens or toward the lens? ...
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