(1) The radius of a balloon increases faster when first starting to pump air in compared to just before bursting. When first pumping, the radius is small so a given increase in volume causes a larger increase in radius. Near bursting, the radius is large so the same volume increase causes a smaller radius increase. (2) To solve related rates problems, write an equation relating the variables, take the derivative of the equation with respect to time, and substitute given rate information. Be careful not to substitute values before differentiating, as this prevents variables from changing over time. (3) Geometry formulas are useful for setting up related rates equations involving shapes like spheres or cones. Implicit differentiation allows finding rates of

1. 2.6 Related Rates
Solving Real-Life Problems

2. You pump air at a steady rate into
a deflated balloon until the balloon
bursts. Does the radius of the
balloon change faster when you
first start pumping the air, or just
before the balloon bursts? Why?

3. Ex 1 p. 149 Two rates that are related
Suppose x and y are two differentiable functions of t
and are related by equation y = x3 – 2
Find dy/dt , given that dx/dt = 2 when x = 1
Implicitly derive x and y with respect to t
y = x3 − 2
dy 2 dx
= 3x Now substitute in what you know
dt dt
dy
= 3(1) 2 (2) =6
dt
In this problem you were given the equation that relates y
and x. Most times you have to create the equation that
relates variables. Luckily we know geometry and trig!

4. Ex 2 p. 150 Ripples in a pond
A pebble is dropped in a calm pond, causing ripples in
the form of concentric circles. The radius r of the outer
ripple is increasing at a constant rate of 1 foot per
second. When the radius is 5 feet, at what rate is the
total area A of the disturbed water changing?
Equation: A = π r
2
r
dr
Given rate: = 1 ft/sec
dt
A = π r2
Find dA/dt when r = 5 ft
dA dr Substitute what you know!
= π (2r ) dA
dt dt = 2π (5)(1) = 10π
dt
dA dr
= 2π r When radius is 5 ft, the Area is changing
dt dt at rate of 10π ft2/sec

5. Ex 3 p. 151 An Inflating Balloon
Air is being pumped into a spherical balloon at a rate of 5
cubic feet per minute. Find the rate of change of the radius
when the radius is 1.5 feet.
Volume is changing with time, and so is the radius
Equation: 4 3
V = πr
3
Given dV/dt = 5 ft3/min,
Find dr/dt when r = 1.5 ft
dV  4  2 dr dV 2 dr
=  π ÷3r = 4π r
dt  3  dt dt dt
Now plug in the given info after a cleanup
dr dr 5
5 = 4π (1.5)
2
= ft / min ≈ 0.177 ft / min
dt dt 9π

6. Ex 5 p.152 Changing Angle of Elevation
Find the rate of change in the angle of elevation of the
camera 10 seconds after lift-off 2 2
hyp = s + 2000
2
Solution: Let θ be angle of elevation.
Equations: tan θ = s
2000
s = 50t 2
Given t = 10 so s(10) = 5000 ft
ds
= 100t = velocity
dt
dθ 1 ds dθ (100t )
sec 2 θ = = cos 2 θ
dt 2000 dt dt 2000
2000
cos θ =
2 s 2 + 20002
dθ  2000  (100t ) 2000 ( 100 ) ( 10 ) 2
= ÷ = = rad/sec
dt  s 2 + 20002  2000 50002 + 20002 29

7. Ex 6 Conical tanks and water going into or
out of a tank
When water goes into a
conical tank, the
volume, the radius, and
the height of the water
are all a function of time.
We need to find an equation that relates volume, radius
and height, namely π r 2h
V=
3
(You might recall that a cylinder with same height and
radius is 3 times as much volume as a cone.)

8. A conical tank is 8 inches high and
8inches across the top. If water is
flowing into the tank at rate of 5 in3/min,
find the rate of change of the depth
when the water is 6 inches deep. dV
= 5in3 / min
Given: dt
Implicitly derive with respect
r 4
to time: π r 2h What you can observe: =
V= h 8
3 or r = 1 h dr 1 dh
=
2 dt 2 dt
dV π  2 dh dr  What you are solving for:
dh
= r + h(2r ) ÷ dt
dt 3  dt dt  when h = 6 in.
Replace all r’s with expressions
dV π  2 dh dr 
= r + 2rh ÷ in h, and all dr/dt’s with
dt 3  dt dt  expressions in dh/dt
Yuck!!!!

9. Let’s start again. dV 1
We can figure out: = 5in3 / min r= h
dt 2
π r 2h
V= Substitute in for r
3
π ( 1 2 h) 2 h π h3
V= =
3 12
dV π 2 dh π h 2 dh
= (3h ) =
dt 12 dt 4 dt
π 62 dh dh
5= = 9π
4 dt dt
Solution: 5 dh
=
9π dt
or the height is changing
5
9π inches /minute

10. •Related rate problems often involve a situation in
which you are asked to calculate the rate at which
one quantity changes with respect to time from the
rate at which a second quantity changes with
respect to time.
•Related rate problems can be recognized because
the rate of change of one or more quantities with
respect to time is given and the rate of change with
respect to time of another quantity is required.
http://www.mathdemos.org/mathdemos/relatedrates/relatedrates.html

12. DO NOT MAKE ANY SUBSTITUTIONS before you
differentiate! That creates a problem where things can’t
change with time.
If a rate is decreasing or getting smaller over time, it is a
negative rate!
Geometry formulas are in the back of your book.
2.6 p. 154 #1-8, 13-27 odd, 35, 36,
45, 46