4.6: related rates
DESCRIPTION
4.6: Related Rates. Remember this problem?. A square with sides x has an area. If a 2 X 2 square has it ’ s sides increase by 0.1, use differentials to approximate how much its area will increase. A square with sides x has an area. - PowerPoint PPT PresentationTRANSCRIPT
4.6: Related Rates
x
x
A square with sides x has an area
2)( xxA
If a 2 X 2 square has it’s sides increase by 0.1, use differentials to approximate how much its area will increase.
2x ?dx 1.0
xdx
dA2 dxxdA 2
4.0)1.0)(2(2 dAx dx
dA
Remember this problem?
cm/sec
A square with sides x has an area
2)( xxA
What if we were to observe the square getting bigger and each side were increasing at a rate of 0.1 cm/sec
2x ?dx 1.0
This would no longer simply be called dx because it is a rate of change. So what should we call it?
Hint: It’s a change in length with respect to time. We did this before when talking about units of velocity…
dt
dx
x
x
cm2/sec
Before we work on that, what would the units for this rate of change be?
A square with sides x has an area
2)( xxA
What if we were to observe the square getting bigger and each side were increasing at a rate of 0.1 cm/sec
2x ?dx 1.0 cm/sec
How we label this will always be decided by…
UNITS!
dt
dx
At what rate would the area of the square be changing?
Try taking the derivative of A with respect to…
2xAdt
d
dt
dA
cm2/sec
Given what we’ve just seen, how would we label this change?
How would we find ?dt
dA
t
sec
2cm
Remember the Chain Rule…
xxdx
d22 But how?
Try taking the derivative of A with respect to…
2xAdt
d
t
2xAdt
d
dt
dxx
dt
dA2
Now let’s find the answer we were trying to find.
sec)/1.0)(2(2 cmcmdt
dA
sec/4.0 2cmdt
dA …this is an exact answer,
not an approximation.
Because this is an instantaneous rate of change...
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?
34
3V rFirst, we need the volume formula:
Next, what are we looking for?
dt
dV
(Remember: How fast is its volume changing?)
Then, what do we know?
r = 10 cm
sec/1.0 cmdt
dr
)3
4( 3rV
dt
d Now what do we do?
and…
3cm40
sec
dV
dt
340 cm / secThe 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?
24dV dr
rdt dt
2 cm4 10cm 0.1
sec
dV
dt
)3
4( 3rV
dt
d
r = 10 cm sec/1.0 cmdt
dr
“How fast is the water level dropping” means that we need to solve for…
Water is draining from a cylindrical tank of radius 20 cm at 3 liters/second. How fast is the water level dropping?
L3
sec
dV
dt
3cm3000
sec
Finddh
dt
2V r h
1 liter = 1000 cm3
dt
dh
hV 400
dt
dh
dt
dV 400dt
dhcm 400sec
30003
2
3
400sec
3000
cm
cm
dt
dh
)400( hVdt
d hV 400
Water is draining from a cylindrical tank of radius 20 cm at 3 liters/second. How fast is the water level dropping?
1 liter = 1000 cm3
2
3
400sec
3000
cm
cm
dt
dh
sec2
15 cm
dt
dh
Steps for Related Rates Problems:
1. Draw a picture (sketch).
2. Write down known information.
3. Write down what you are looking for.
4. Write an equation to relate the variables.
5. Differentiate both sides with respect to t.
6. Evaluate.
4x
3y
B
A
5z
Truck Problem:Truck A travels east at 40 mi/hr.
How fast is the distance between the trucks changing 6 minutes later?
r t d
2 2 23 4 z 5 z
6 minutes = hr10
1
140 4
10 miles
130 3
10 miles
Since x, y, and z are always changing, they are all variables. So the equation we will have to begin with will be…
Truck B travels north at 30 mi/hr
4 40 3 30 5dz
dt 4x
3y
30dy
dt
40dx
dt
B
A
5z
Truck Problem:
How fast is the distance between the trucks changing 6 minutes later?
2 2 2x y z
250 5dz
dt 50
dz
dt miles
50hour
dt
d( )
2 2 2dx dy dz
x y zdt dt dt
Truck A travels east at 40 mi/hr.Truck B travels north at 30 mi/hr.
8 cm
hrV 2
3
19 cm
But Wait!
What do we know besides the dimensions of the cone?
min/2 3cmdt
dV We have three variables
Whatever shall we do?
That’s one variable too many
A cone filter of diameter 8 cm and height 9 cm is draining at a rate of 2 cm3/min. Find the rate at which the fluid depth h decreases when h = 5 cm.
8 cm
hrV 2
3
1
But Wait!
What do we know besides the dimensions of the cone?
min/2 3cmdt
dV
As we’ve done in the past, let’s see if one variable can be substituted for the other.
That’s one variable too many
A cone filter of diameter 8 cm and height 9 cm is draining at a rate of 2 cm3/min. Find the rate at which the fluid depth h decreases when h = 5 cm.
8 cm
hrV 2
3
1
4 cm
9 cm
r
h
h
r
9
4
Similar triangles
hr9
4
hrV 2
3
1 hhV2
9
4
3
1
3
243
16hV
Now we have two variables!
A cone filter of diameter 8 cm and height 9 cm is draining at a rate of 2 cm3/min. Find the rate at which the fluid depth h decreases when h = 5 cm.
8 cm
hrV 2
3
1
3
243
16hV
dt
dhh
dt
dV)3(
243
16 2
dt
dhhcm 23
81
16min/2 dt
dh
h
28
81
Will this rate increase or decrease as h gets lower?
dt
dh
258
81
dt
dhcm
min200
81
A cone filter of diameter 8 cm and height 9 cm is draining at a rate of 2 cm3/min. Find the rate at which the fluid depth h decreases when h = 5 cm.
8 cm
hrV 2
3
1dt
dhhcm 23
81
16min/2
dt
dh
258
81
Show that this is true by comparing
both when h = 5 cm and when h = 3 cm.
dt
dh
dt
dh
238
81
dt
dhcm
min200
81
dt
dhcm
min8
9
As h gets smaller,
dt
dhgets faster
because h is in the
denominator.
8 cm
hrV 2
3
1dt
dhhcm 23
81
16min/2
dt
dh
258
81
Show that this is true by comparing
both when h = 5 cm and when h = 3 cm.
dt
dh
dt
dh
238
81
dt
dhcm
min200
81
dt
dhcm
min8
9
Problems like this surface often so remember your geometric relationships such as similar triangles, etc.
htan500 )tan500( hdt
d
Hot Air Balloon Problem:
Given:4
rad
0.14min
d
dt
How fast is the balloon rising?
Finddh
dt
tan500
h
h
500ft
dt
dh
dt
d 2sec500 dt
dh14.0
4sec500 2
Hot Air Balloon Problem:
Given:4
rad
0.14min
d
dt
How fast is the balloon rising?
Finddh
dt
tan500
h
h
500ft
2
2 0.14 500dh
dt
1
12
4
sec 24
ft140
min
dh
dt
)tan500( hdt
d
dt
dh
dt
d 2sec500 dt
dh14.0
4sec500 2