Drill: find the derivative of the
following
• 2xy + y2 = x + y
• xsin(y)= 1 – xy
• 2xy’ +2y + 2yy’ = 1 + y’ • xcos(y)y’ + sin (y) = -(xy’ +y)
• 2xy’ + 2yy’ – y’ = 1 – 2y
• y’(2x + 2y – 1) = 1 – 2y
• y’ = (1-2y)/(2x + 2y -1)
•
•
•
•
xcos(y)y’ + sin (y) = -xy’ –y
xcos(y)y’+xy’= -y – sin (y)
y’ (xcos (y) + x) = -y – sin (y)
y’ = - (y + sin(y))/(xcos(y) + x)
Related Rates
Section 4.6
Objectives
• Students will be able to
– solve related rate problems.
Strategy for Solving Related Rate
Problems
• Understand your problem
• Develop a mathematical model of the problem
• Write an equation relating the variable whose rate of
change you seek with the variables whose rate of
change you know
• Differentiate both sides of the equation implicitly
with respect to time t.
• Substitute values for any quantities that depend on
time
• Interpret the solution
Finding related rate equations
4 3
V  r
3
dV
2 dr
 4r
dt
dt
• Assume the radius r
of a sphere is a
differentiable
function of t and let
V be the volume of
the sphere. Find an
equation that related
dV/dt and dr/dt
Finding related rate equations
1 2
V  r h
• Assume that the
3
radius and height h
1
2
of a cone are
V   ( r h)
3
differentiable
dV 1
dr functions of t and let
2 dh
  (r
 2rh ) V = volume of the
dt 3
dt
dt
cone. Find an
equation that relates
dV/dt, dr/dt and
dh/dt.
Example 1 Finding Related Rate Equations
Assume that the width a and the length b of a
rectangle are functions of t and let A be the area of the
rectangle. Find an equation that relates dA , da , and db .
dt dt
dt
Example 1 Finding Related Rate Equations
Assume that the width a and the length b of a
rectangle are functions of t and let A be the area of the
rectangle. Find an equation that relates dA , da , and db .
dt dt
dt
A  ab
d
d
A  ab 
dt
dt
dA da
db

b   a
dt dt
dt
Example 2 A Rising Balloon
A hot-air balloon rising straight up from a level field
is tracked by a range finder 700 feet from the lift-off
point. At the moment the range finder’s elevation
angle is  , the angle is increasing at the rate of
6
0.16 radians per minute. How fast is the balloon
rising at that moment?
Example 2 A Rising Balloon
A hot-air balloon rising straight up from a level field
is tracked by a range finder 700 feet from the lift-off
point. At the moment the range finder’s elevation
angle is  , the angle is increasing at the rate of
6
0.16 radians per minute. How fast is the balloon
rising at that moment?
dh
We want to find :
dt
d
We know :
 0.16 rad/min
dt
h
tan  
700
h

700 feet
Example 2 A Rising Balloon
dh
We want to find :
dt
d
We know :
 0.16 rad/min
dt
h
dh
2 
tan  
 700 sec
 0.16 
700
dt
6
h  700 tan 
 149 .33 ft/min
dh d
 700 tan  
dt dt
h
dh
d
2
 700 sec  
dt
dt

700 feet
Example 3 A Highway Chase
A police cruiser approaching a right-angled intersection from
the north is chasing a speeding car that has turned the corner
and is now moving straight east. When the cruiser is 0.70 mile
north of the intersection and the car is 0.80 mile to the east, the
police determine with radar that the distance between them and
the car is increasing at 15 mph. If the cruiser is moving at 45
mph at the instant of measurement, what is the speed of the car?
Example 3 A Highway Chase
A police cruiser approaching a right-angled intersection from
the north is chasing a speeding car that has turned the corner
and is now moving straight east. When the cruiser is 0.70 mile
north of the intersection and the car is 0.80 mile to the east, the
police determine with radar that the distance between them and
the car is increasing at 15 mph. If the cruiser is moving at 45
mph at the instant of measurement, what is the speed of the car?
Example 3 A Highway Chase
A police cruiser approaching a right-angled intersection from
the north is chasing a speeding car that has turned the corner
and is now moving straight east. When the cruiser is 0.70 mile
north of the intersection and the car is 0.80 mile to the east, the
police determine with radar that the distance between them and
the car is increasing at 15 mph. If the cruiser is moving at 45
mph at the instant of measurement, what is the speed of the car?
Let x = distance of the speeding car from
the intersection
Let y = distance of the police cruiser from
the intersection
Let z = distance between the car and the
cruiser
Example 3 A Highway Chase
dx
We want to know:
dt
dz
We know:
 15 mph
dt
dy
 45 mph
dt
z
y
x
x y z
d 2
d 2
2

x  y   z 
dt
dt
dx
dy
dz
2x  2 y
 2z
dt
dt
dt
dx
dy
dz
x y
z
dt
dt
dt
2
2
2
Example 3 A Highway Chase
x = .8 and y = .6
dy
dz
 45 mph
 15 mph
dt
dt
We can now substitute, remembering that z2 = x2 + y2 ; so z =(x2 + y2)1/2
y
dx
dy
dz
x y
z
dt
dt
dt
dx
2
2
0.8  0.7  45   0.8  0.7  15 
dt
dx
dx
0.8  31.5  15.9 0.8  47.4
dt
dt
dx
z
 59.25
dt
At that moment, the car’s
speed was about 59.3 mph.
x
Example 4 Filling a Conical Tank
Water runs into a conical tank at the rate of 12 ft3 /min. The
tank stands point down and has a height of 12 ft and a base
radius of 9 ft. How fast is the water level rising when the water
is 8 ft deep?
Example 4 Filling a Conical Tank
Water runs into a conical tank at the rate of 12 ft3/min. The tank
stands point down and has a height of 12 ft and a base radius of
9 ft. How fast is the water level rising when the water is 8 ft
deep?
dh
We want to find :
dt
dV
We know :
 12 ft 3 / min
dt
1 2
V  r h
3
2
1 3 
V    h h
3 4 
r
h
r 9

h 12
3
r h
4
Example 4 Filling a Conical Tank
dh
We want to find :
dt
dV
We know :
 12 ft 3 / min
dt
r
3
V   h3
16
d
d3
3
V   h 
dt
dt  16

2
dV 19  32 dh

V   h h  h
3  4 dt
dt 16

h
r 9

h 12
3
r h
4
Example 4 Filling a Conical Tank
dh
We want to find :
dt
dV
We know :
 12 ft 3 / min
dt
r
h
9
2 dh
12   8
16
dt
dh
12  36
dV
9 dt2 dh
1  dh  h
dt  16  0.106
dt ft/min
3 dt
r 9

h 12
3
r h
4
Homework
• Page 251: 9-21 odd