Related Rates: Part 2
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Transcript of Related Rates: Part 2
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Strategy for Solving Related Rates
Step 1: Make a drawing of thesituation if possible.
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Step 2: Use letters to represent thevariables involved in the situation -
say x y.
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Step !: "dentify all rates of #hangegiven and those to be deter$ined
Use the #al#ulus notation %d&'dt
dydt et#( to represent the$.
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Step ): *eter$ine an e+uation that
involves both
,he variables fro$ step two
,he derivative of step three
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Step : *ifferentiate %by i$pli#itdifferentiation( the e+uation of step
four
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Step /: Substitute all know valuesinto the differentiated e+uation
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Step 0: Use algebrai# $anipulationif ne#essary to solve for the
unknown rate or +uantity
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or$ulas ou May 3eed o 4now
!V a= 2V r h= V lwh=
( )!)!
rV
=
( )2
!
r h
V
=
( )
!
bhV=
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5&a$ple 61
7 ladder 18 feet long is resting against a wall. "f
the botto$ of the ladder is sliding away fro$ the
wall at a rate of 1 foot per se#ond how fast is thetop of the ladder $oving down when the botto$
of the ladder is 9 feet fro$ the wall
irst draw the pi#ture:
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;e have d&'dt is one foot per se#ond. ;e
want to find dy'dt.
Xand y are related by the
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Suppose that the radius of a sphere is changing at an
instantaneous rate of 0.1 cm/sec. How fast is its volume
changing when the radius has reached 10 cm?
!)
!V r=First, we need the volume formula:
Net, what are we loo!ing for?
dt
dV
"#emem$er: How fast is its volumechanging?%
&hen, what do we !now?
r= 18 #$
se#'1.8 cmdt
dr =
(
!
)% !rV
dt
d=Now what do we do?
and'
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!#$)8
se#
dV
dt=
!
)8 #$ ' se#
&he sphere is growing at a rate of .
Suppose that the radius of a sphere is changing at an
instantaneous rate of 0.1 cm/sec. How fast is its volume
changing when the radius has reached 10 cm?
2)dV dr
rdt dt
=
( )2 #$
) 18#$ 8.1se#
dV
dt
=
(!
)
%!
rVdt
d=
r= 18 #$ se#'1.8 cmdt
dr=
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(How fast is the water
level dropping) meansthat we need to solve
for'
*ater is draining from a c+lindrical tan!
of radius 0 cm at - liters/second. How
fast is the water level dropping?
>!
se#
dV
dt=
!#$!888
se#=
Finddh
dt
2V r h=
1 liter 1000 cm-
dt
dh
hV )88=
dt
dh
dt
dV)88=
dt
dhcm)88
se#!888
!
=2
!
)88
se#!888
cm
cm
dt
dh
=
()88% hVdt
d=hV )88=
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*ater is draining from a c+lindrical tan!
of radius 0 cm at - liters/second. How
fast is the water level dropping?
1 liter 1000 cm-
2
!
)88
se#!888
cm
cm
dt
dh
=
se#2
1 cm
dt
dh
=
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Steps for Related Rates Problems:
1. raw a picture "s!etch%.
. *rite down !nown information.
-. *rite down what +ou are loo!ing for.
. *rite an euation to relate the varia$les.
2. ifferentiate $oth sides with respect to t.
3. 4valuate.
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)x=
!y=
5
6
z=
Truck Problem:
&ruc! 6 travels east at 0 mi/hr.
How fast is the distance $etween the
truc!s changing 3 minutes later?
r t d =
2 2 2! ) z+ = z=
3 minutes hr
18
1
1)8 )
18 = miles
1!8 !
18 = miles
Since x, y, and zare alwa+s
changing, the+ are all varia$les.
So the euation we will have to
$egin with will $e'
&ruc! 5 travels north at -0 mi/hr
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) )8 ! !8 dz
dt + =
)x=
!y=
!8dy
dt=
)8dx
dt
=
5
6
z=
Truck Problem:
How fast is the distance $etween the
truc!s changing 3 minutes later?
2 2 2x y z+ =
28 dz
dt= 8
dz
dt= $iles.8
hour
dt
d" %
2 2 2dx dy dz
x y zdt dt dt
+ =
&ruc! 6 travels east at 0 mi/hr.
&ruc! 5 travels north at -0 mi/hr.
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9 #$
hrV 2
!
1=
? #$
But Wait!
*hat do we !now $esides
the dimensions of the cone?
$in'2 !cmdt
dV= We have three variables
Thats one variable too many
6 cone filter of diameter 7 cm and
height 8 cm is draining at a rate of
cm-/min. Find the rate at which the
fluid depth hdecreases when h 2cm.
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9 #$
hrV 2
!
1=
But Wait!
*hat do we !now $esides
the dimensions of the cone?
$in'2 !cmdt
dV=
As eve done in the past
lets see if one variable can
be substituted for the other"
Thats one variable too many
6 cone filter of diameter 7 cm and
height 8 cm is draining at a rate of
cm-/min. Find the rate at which the
fluid depth hdecreases when h 2cm.
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9 #$
hrV 2
!
1=
) #$
? #$
r
h
h
r
=?)
Similar triangles
hr ?
)
=
hrV 2
!
1= hhV
2
?
)
!
1
=
V=1
!
1/
91h
2
h
!
2)!
1/hV =
Now we have two varia$les9
6 cone filter of diameter 7 cm and
height 8 cm is draining at a rate of
cm-/min. Find the rate at which the
fluid depth hdecreases when h 2cm.
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9 #$
hrV 2
!
1=
!
2)!
1/
hV = dtdh
hdt
dV
(!%2)!
1/ 2=
dt
dhhcm
2!
91
1/$in'2 =
dt
dh
h=
29
91
dt
dh=
29
91
dt
dhcm=
$in288
91
6 cone filter of diameter 7 cm and
height 8 cm is draining at a rate of
cm-/min. Find the rate at which the
fluid depth hdecreases when h 2cm.