related rates: part 2

7/23/2019 Related Rates: Part 2

1/21

Strategy for Solving Related Rates

Step 1: Make a drawing of thesituation if possible.


2/21

Step 2: Use letters to represent thevariables involved in the situation -

say x y.


3/21

Step !: "dentify all rates of #hangegiven and those to be deter$ined

Use the #al#ulus notation %d&'dt

dydt et#( to represent the$.


4/21

Step ): *eter$ine an e+uation that

involves both

,he variables fro$ step two

,he derivative of step three


5/21

Step : *ifferentiate %by i$pli#itdifferentiation( the e+uation of step

four


6/21

Step /: Substitute all know valuesinto the differentiated e+uation


7/21

Step 0: Use algebrai# $anipulationif ne#essary to solve for the

unknown rate or +uantity


8/21

or$ulas ou May 3eed o 4now

!V a= 2V r h= V lwh=

( )!)!

rV

=

( )2

!

r h

V

=

( )

!

bhV=


9/21

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:


10/21

;e have d&'dt is one foot per se#ond. ;e

want to find dy'dt.

Xand y are related by the


11/21

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'


12/21

!#$)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=


13/21

(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=


14/21

*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

=


15/21

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.


16/21

)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


17/21

) )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.


18/21

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.


19/21

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






20/21

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






21/21

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





related rates: part 2

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