Slide 1 ___________________________________ 2.7 Related Rates
___________________________________ ___________________________________ ___________________________________ ___________________________________ ___________________________________ ___________________________________ Slide 2 ___________________________________ Strategy For Solving Related Rates Problems:
1. Read the problem carefully.
2. Draw a picture and name variables and constants. Use t
for time.
3. Identify all given quantities and those to be determined.
4. Write an equation involving the variables whose rates of
either are given are to be determined.
5. Use the Chain Rule to implicitly differentiate both sides of
the equation with respect to time, t.
6. Substitute into the equation in step 5 all known values for
the variables and their rates of change.
7. Solve for the required rate of change.
___________________________________ ___________________________________ ___________________________________ ___________________________________ ___________________________________ ___________________________________ Slide 3 ___________________________________ EXAMPLE 1:
A child throws a stone into a still millpond
causing a circular ripple to spread. If the radius
increases at the constant rate of 0.5 meter per
second, how fast is the area of the ripple
increasing when the radius of the ripple is 20
meters?
___________________________________ ___________________________________ How fast is the area changing
when the radius is 20 meters?
r = 20
rate of change of radius:
Given: dr/dt = 0.5
Note: rates of change are
derivatives.
___________________________________ rate of change of area:
Find: dA/dt when r = 20
___________________________________ ___________________________________ ___________________________________ Slide 4 ___________________________________ We need the formula that connects the radius of
the circle with its area.
A = πr2
Next, we differentiate both sides of the equation
with respect to t:
dA
d
=π (r 2 )
dt
dt
dA d (π r 2 )
=
dt dt
note:
r’ =
dr/dt
___________________________________ dA
=π ⋅ 2r r ′
dt
___________________________________ dA
dr
= π ⋅ 2r
dt
dt
π(2)(20)(0.5) = 20π m 2 /s
___________________________________ or 62.8 m2/s
___________________________________ ___________________________________ ___________________________________ Slide 5 ___________________________________ EXAMPLE 2:
Two cars start moving from the same point.
One travels south at 60mi/h and the other
travels west at 25mi/h. At what rate is the
distance between the cars increasing two
hours later?
Given: dy/dt =60 and dx/dt = 25
25 m/h
Find: dz/dt when t = 2h
z
___________________________________ ___________________________________ ___________________________________ 60 mi/h
___________________________________ ___________________________________ ___________________________________ Slide 6 ___________________________________ What equation relates the three sides of a right
triangle? x2 + y2 = z2
dx
dy
dz
d 2
2x +2y
= 2z
( x ) + dtd ( y 2 ) = dtd ( z2 )
dt
dt
dt
dt
___________________________________ After 2 hours: x = 50 and y = 120.
z =
502 +1202 =130
___________________________________ Subst gives: 2(50)(25) + 2(120)(60) = 2(130)(dz/dt)
After 2 hours
dz 2500 +14400
=
= 65m/h
dt
2(130)
___________________________________ ___________________________________ ___________________________________ ___________________________________ Slide 7 EXAMPLE 3
___________________________________ • A ladder 10 ft long rests against a vertical
wall. If the bottom of the ladder slides away
from the wall at a rate of 1 ft/s, how fast is the
top of the ladder sliding down the wall when
the bottom of the ladder is 6 from the wall?
dy/dt
10 ft
6 ft
Given: dx/dt = 1 ft/s
Find: dy/dt when x = 6 ft
x2 + y2 = 102
d 2 d 2 d
x + y = (100)
dt
dt
dt
dx
dy
dx/dt = 1 ft/s
2x
+ 2y
= 0 eqa 1
dt
dt
___________________________________ ___________________________________ ___________________________________ ___________________________________ ___________________________________ ___________________________________ Slide 8 When x = 6: 62 + y2 = 100
___________________________________ y = 100 − 36 = 8
dy
Substitute into eqa 1: (2)(6)(1) + 2(8)( ) = 0
dt
16(dy/dt) = - 12
___________________________________ dy/dt = -3/4 ft/s
___________________________________ ___________________________________ ___________________________________ ___________________________________ ___________________________________ Slide 9 ___________________________________ Example 4:
The altitude of a triangle is increasing at a rate of 1
cm/min while the area of the triangle is increasing at a
rate of 2 cm2/min. At what rate is the base of the
triangle changing when the altitude is 10 cm and the
area is 100 cm2
Given: dh/dt = 1 cm/min
dA/dt = 2 cm2/min
h = altitude
___________________________________ ___________________________________ Find: db/dt when h = 10cm
and A = 100cm2
A = (1/2)bh
100 = (1/2)b(10)
b = 20
___________________________________ ___________________________________ ___________________________________ ___________________________________ Slide 10 ___________________________________ dA d 1
= ( bh) = 1 d (bh) = 1 ⎛ db ⋅ h+b dh ⎞
dt dt 2
2 dt
2 ⎜⎝ dt
dt ⎟⎠
___________________________________ 1 ⎛ db
⎞
2= ⎜
⋅ 10 + 20 ⋅ 1⎟ → 2 = 5 db + 10 → − 8 = 5 db
2 ⎝ dt
⎠
dt
dt
→ − 1.6cm/min =
db
dt
___________________________________ ___________________________________ ___________________________________ ___________________________________ ___________________________________ Slide 11 ___________________________________ Ex 5: A water tank in the shape
of an inverted circular cone has
a base radius of 5 ft and height
dh/dt = ?
r
10 ft. If water is pumped into
when h = 6 ft
the tank at a rate of 9 ft3/min,
h
find the rate at which the water
level is rising when the water is
6 ft deep.
3
To find a relationship
Given: dV/dt = 9 ft /min
between r and h, use
Find: dh/dt when h = 6 ft
similar triangles:
1
V = π r 2 h eqa 1
3
r
h
=
5 10
r=
___________________________________ ___________________________________ ___________________________________ h
2
___________________________________ ___________________________________ ___________________________________ Slide 12 ___________________________________ 2
dV d 1 2
d ⎛ 1 ⎛ h ⎞ ⎞ d ⎛ 1 h2 ⎞
= ( πr h) = ⎜ π ⎜ ⎟ h ⎟ → ⎜ π h ⎟
dt dt 3
dt ⎜⎝ 3 ⎝ 2 ⎠ ⎟⎠ dt ⎝ 3 4 ⎠
dV
π ⎛ 2 dh ⎞
dV π d 3
=
=
h →
⎜ 3h
⎟
dt 12 ⎝
dt ⎠
dt 12 dt
π⎛
dh ⎞
Substituting gives: 9 =
3 ⋅ 62
12 ⎜⎝
dt ⎟⎠
108π dh
9(12) dh
9=
→
=
12 dt
108π dt
___________________________________ ( )
___________________________________ ___________________________________ dh 1
= ≈ 0.32 ft/min
dt π
___________________________________ ___________________________________ ___________________________________ Slide 13 ___________________________________ ___________________________________ ___________________________________ ___________________________________ ___________________________________ ___________________________________ ___________________________________ ___________________________________ Slide 14 ___________________________________ ___________________________________ ___________________________________ ___________________________________ ___________________________________ ___________________________________