200
3.5. RELATED RATES
we can answer questions such as: how fast is the height of the sandpile changing at the
moment the radius is 4 feet?
The information that the height is always half the radius tells us that for all values of
t, h =
1
2 r. Differentiating with respect to t, it follows that
dh
dt =
1
2
dr
dt . These relationships
enable us to relate
dV
dt exclusively to just one of r or h. Substituting the expressions
involving r and
dr
dt for h and
dh
dt , we now have that
dV
dt
=
1
3
πr
2 ·
1
2
dr
dt
+
2
3
πr ·
1
2
r ·
dr
dt
.
(3.1)
Since sand falls from the conveyor at the constant rate of 10 cubic feet per minute, this tells
us the value of
dV
dt , the rate at which the volume of the sand pile changes. In particular,
dV
dt = 10 ft 3 /min. Furthermore, since we are interested in how fast the height of the pile is
changing at the instant r = 4, we use the value r = 4 along with
dV
dt = 10 in Equation (3.1),
and hence find that
10 =
1
3
π4
2 ·
1
2
dr
dt
r=4
+
2
3
π4 ·
1
2
4 ·
dr
dt
r=4
=
8
3
π
dr
dt
r=4
+
16
3
π
dr
dt
r=4
.
With only the value of
dr
dt
r=4 remaining unknown, we solve for
dr
dt
r=4 and find that
10 = 8π
dr
dt
r=4 , so that
dr
dt
r=4
=
10
8π
≈ 0.39789
feet per second. Because we were interested in how fast the height of the pile was changing
at this instant, we want to know
dh
dt when r = 4. Since
dh
dt =
1
2
dr
dt for all values of t, it
follows
dh
dt
r=4
=
5
8π
≈ 0.19894 ft/min.
Note particularly how we distinguish between the notations
dr
dt and
dr
dt
r=4 . The former
represents the rate of change of r with respect to t at an arbitrary value of t, while the
latter is the rate of change of r with respect to t at a particular moment, in fact the moment
r = 4. While we don’t know the exact value of t, because information is provided about
the value of r, it is important to distinguish that we are using this more specific data.
The relationship between h and r, with h =
1
2 r for all values of t, enables us to transition
easily between questions involving r and h. Indeed, had we known this information at the
problem’s outset, we could have immediately simplified our work. Using h =
1
2 r, it follows
that since V =
1
3 πr 2 h, we can write V solely in terms of r to have
V =
1
3
πr
2
1
2
h
=
1
6
πr
3 .
3.5. RELATED RATES
we can answer questions such as: how fast is the height of the sandpile changing at the
moment the radius is 4 feet?
The information that the height is always half the radius tells us that for all values of
t, h =
1
2 r. Differentiating with respect to t, it follows that
dh
dt =
1
2
dr
dt . These relationships
enable us to relate
dV
dt exclusively to just one of r or h. Substituting the expressions
involving r and
dr
dt for h and
dh
dt , we now have that
dV
dt
=
1
3
πr
2 ·
1
2
dr
dt
+
2
3
πr ·
1
2
r ·
dr
dt
.
(3.1)
Since sand falls from the conveyor at the constant rate of 10 cubic feet per minute, this tells
us the value of
dV
dt , the rate at which the volume of the sand pile changes. In particular,
dV
dt = 10 ft 3 /min. Furthermore, since we are interested in how fast the height of the pile is
changing at the instant r = 4, we use the value r = 4 along with
dV
dt = 10 in Equation (3.1),
and hence find that
10 =
1
3
π4
2 ·
1
2
dr
dt
r=4
+
2
3
π4 ·
1
2
4 ·
dr
dt
r=4
=
8
3
π
dr
dt
r=4
+
16
3
π
dr
dt
r=4
.
With only the value of
dr
dt
r=4 remaining unknown, we solve for
dr
dt
r=4 and find that
10 = 8π
dr
dt
r=4 , so that
dr
dt
r=4
=
10
8π
≈ 0.39789
feet per second. Because we were interested in how fast the height of the pile was changing
at this instant, we want to know
dh
dt when r = 4. Since
dh
dt =
1
2
dr
dt for all values of t, it
follows
dh
dt
r=4
=
5
8π
≈ 0.19894 ft/min.
Note particularly how we distinguish between the notations
dr
dt and
dr
dt
r=4 . The former
represents the rate of change of r with respect to t at an arbitrary value of t, while the
latter is the rate of change of r with respect to t at a particular moment, in fact the moment
r = 4. While we don’t know the exact value of t, because information is provided about
the value of r, it is important to distinguish that we are using this more specific data.
The relationship between h and r, with h =
1
2 r for all values of t, enables us to transition
easily between questions involving r and h. Indeed, had we known this information at the
problem’s outset, we could have immediately simplified our work. Using h =
1
2 r, it follows
that since V =
1
3 πr 2 h, we can write V solely in terms of r to have
V =
1
3
πr
2
1
2
h
=
1
6
πr
3 .