Transcript 2.6 Related Rates

2.6 Related Rates
Today we will stretch even farther!
2.6 Related Rates
Suppose x and y are differentiable functions of t related
by the equation
y  x 3
2
dy
dx
Find
when x  1, given
 2 when x  1.
dt
dt
dy
dx
y  x 3
 2x  0
dt
dt
dy
 2 1 2   4
dt
2
2.6 Related Rates
A plane is flying over a radar tracking station. If
s is decreasing at a rate of 400 mph when s  10 mi,
what is the speed of the plane?
Geometric
Model
dx/dt?
Solve on
Board
s 2  x 2  36
ds
dx
2s  2 x  0
dt
dt
dx 2 10  400 

dt
2x
x?
10  6  x
2
2
 x8
2
ds
 400
dt
s  10
dx
 500 mph
dt
speed  500 mph
dx
?
dt
Given h  t  =50t 2 , (h in feet and t in seconds),
find the rate of change in the  of elevation of the
camera at 10 seconds after lift-off.
h  t  =50t
x  20002  50002
You cannot
put in values
for variables
until you have
differentiated!
d
?
dt
x
Variable
Constant
Geometric Model
h
tan  
2000
Implicit
Differentiation
d
1 dh


dt 2000 dt
1
d
1 dh



2
cos  dt 2000 dt
d  co s 2  d h


dt
2000 dt
sec 2  
h
2
h 10 =5000
Variable
cos  
2000
20002  50002
dh
 100  10  1000
dt t 10
2

radians per second
29
In an engine, a 7” connecting rod is fastened to a
crankshaft of radius 3”. The crankshaft rotates
counterclockwise at a constant rate of 200 revolutions
per minute. Find the velocity of the piston when

 .
3
Constants  3,7
Variables   ,x
Crankshaft = 3”
Connecting Rod = 7”

Piston = x”
Given Rate: 200 Revolutions per minute
d

 200(2 )  400
d
dt
 400
dt
In an engine, a 7” connecting rod is fastened to a
crankshaft of radius 3”. The crankshaft rotates
counterclockwise at a constant rate of 200 revolutions

per minute. Find the velocity of the piston when

3
.
Connecting Rod = 7”
Crankshaft = 3”

dx

Find:
when  =
dt
3
Piston = x”
Do not substitute
before you
differentiate!!
Equation: Find an equation that relates  and x.
Law of Cosines: b  a  c  2ac cos 
2
2
2
 7 2  32  x 2  (2)(3) x cos 
d
 400
dt Law of Cosines
In an engine, a 7” connecting rod is fastened to a
crankshaft of radius 3”. The crankshaft rotates
counterclockwise at a constant rate of 200 revolutions

per minute. Find the velocity of the piston when

3
.
Connecting Rod = 7”
Crankshaft = 3”

Piston = x”
Implicit Differentiation:
49  9  x  6 x cos 
2
Solution
49  9  x  6 x cos 
2
dx
dx
d
0  0  2 x( )  6( cos   x( sin  ) )
dt
dt
dt
dx
dx
d
0  2 x( )  6 cos  ( )  6 x sin  ( )
dt
dt
dt
You could
divide
every term
by 2.
dx
dx
d
6 cos  ( )  2 x( )  6 x sin  ( )
d
dt
dt
dt
6 x sin  ( )
dx
d
dx
d
t
(6 cos   2 x)  6 x sin  ( )

dt
dt
dt
6cos   2 x
In an engine, a 7” connecting rod is fastened to a
crankshaft of radius 3”. The crankshaft rotates
counterclockwise at a constant rate of 200 revolutions

per minute. Find the velocity of the piston when

Crankshaft = 3”
Connecting Rod = 7”
3
.

Piston = x”
d
3x sin  ( )
dx
dt

dt
3cos   x
Before we substitute,
we need to find a value for x.
49  9  x  6 x cos
2

3
1
40  x  6 x( )
2
2
40  x  3x  0  x  3x  40
2
 0  ( x  8)( x  5)
2
x8
d
3x sin  ( )
dx
dt

dt
3cos   x

3
(3)(8) sin (400 ) (24)( )(400 )
dx
2
3



3
/
2

8
dt
3(cos )  (8)
3
in.
 4018
 Length is decreasing
min .
Diagram
2.6 Related Rates
HW 2.6/1-9odd,31-34,41-44,52
Related Rates
1. Diagram
2. Geometric Model
3. Differentiate
d?
4. Solve for
dt
5. Substitute
6. Answer Question
9.4 Law of Cosines
Objective
To use the Law of Cosines to find unknown parts of a .
The Law of Cosines SAS & SSS
In ABC , c 2  a 2  b2  2ab cos C
 OPP 
2
  ADJ1    ADJ 2   2  ADJ1  ADJ 2  cos   
2
2
Problem