SME
Calculus I · Problem type
The error in a sphere’s volume
Use this for “A ball’s radius is measured as 10 cm with a possible error of 0.1 cm. Estimate the error in its volume.”
Chart
Tangent at x = 10: slope f′(10) = 1,257, y = 1,257x − 8,378
Linear approximation L(10.1) = 4,314; the true f(10.1) = 4,316, off by 1.261
This is an example. Type your own values to start a new problem.
V
=
4
3
π
(
r (cm)
)
3
=
V (cm³)
dV
=
4π
(
r (cm)
)
2
×
dr (cm)
=
dV (cm³)
100 × dV ÷ V
Relative error in the volume, in percent
worked out (not typed) · %
ΔV
Exact change in the volume
worked out (not typed) · cm³
r₁
Radius with the error added, r + dr
worked out (not typed) · cm
Formulas
V
=
4
3
×
π
×
r
3
4,188.7902
=
4
3
×
π
×
10
3
dV
=
4π
×
r
2
×
dr
125.6637
=
4π
×
10
2
×
0.1
100
×
dV
V
=
300
×
dr
r
3
=
300
×
0.1
10
r₁
=
r
+
dr
10.1 = 10 + 0.1
ΔV
=
4
3
×
π
×
dr
×
(3
×
r
2
+
3
×
r
×
dr
+
dr²)
126.9245
=
4
3
×
π
×
0.1
×
(3
×
10
2
+
3
×
10
×
0.1
+
0.1
2
)
Assumptions
•
V = (4 ÷ 3)πr³, so dV/dr = 4πr². The differential dV = 4πr² dr is the change along the tangent line.
•
dV is close to the true change ΔV when dr is small next to r. ΔV also keeps the dr² and dr³ terms.
•
Dividing, dV ÷ V = 3dr ÷ r: V grows as r³, so the relative error in V is three times the one in r.
Step-by-step
WE KNOW
r = 10 cm, dr = 0.1 cm
FIND
volume (V), estimated error in the volume (dV), relative error in the volume, in percent (100 × dV ÷ V), radius with the error added, r + dr (r₁), exact change in the volume (ΔV)
Step 1 · Find volume (V)
Use V = 4 ÷ 3 × π × r³
The volume of the sphere as measured.
V
=
4
3
×
π
×
10
3
V
=
4
3
×
π
×
1,000
V
=
1.3333
×
π
×
1,000
V
=
4.1887
×
1,000
V
=
4,188.79
cm³
Step 2 · Find estimated error in the volume (dV)
Use dV = 4π × r² × dr
Power
rule
on
(
4
3
)πr³
gives
dV/dr
=
4πr²;
times
the
small
step
dr,
that
is
the
rise
along
the
tangent.
dV
=
4π
×
10
2
×
0.1
V
(
r
)
=
4
3
×
π
×
r
3
→
dV/dr
=
4
×
π
×
r
2
dV = 125.6637 cm³
Step 3 · Find relative error in the volume, in percent (100 × dV ÷ V)
Use 100 × dV ÷ V = 300 × dr ÷ r
Divide
4πr²
dr
by
(
4
3
)πr³:
the
π
and
two
powers
of
r
cancel,
leaving
dV
V
=
3d
r
r
.
Times
100
gives
it
in
percent.
100
×
dV
V
=
300
×
0.1
10
100
×
dV
V
=
30
10
100
×
dV
V
=
3%
Step 4 · Find radius with the error added, r + dr (r₁)
Use r₁ = r + dr
The radius if the whole error is added.
r₁
=
10
+
0.1
r₁
=
10.1
cm
Step 5 · Find exact change in the volume (ΔV)
Use ΔV = 4 ÷ 3 × π × dr × (3 × r² + 3 × r × dr + dr²)
The
true
change
is
V
at
r
+
dr
take
away
V
at
r
.
Expanding
(
r
+ dr)
3
−
r
3
cancels
the
r
3
and
leaves
dr(
3
r
²
+
3
r
dr
+
dr²).
ΔV
=
4
3
×
π
×
0.1
×
(3
×
10
2
+
3
×
10
×
0.1
+
0.1
2
)
(
r
+ dr)
3
−
r
3
=
3
r
²
dr
+
3
r
dr²
+
dr³
=
dr(
3
r
²
+
3
r
dr
+
dr²)
3 ×
10
2
+
3
×
10
×
0.1
+
0.1
2
=
303.01
ΔV = 126.9245 cm³
Check
Put the numbers back in. Both sides should match.
4,188.79
=
4
3
×
π
×
10
3
✓
125.6637
=
4π
×
10
2
×
0.1
✓
3
=
300
×
0.1
10
✓
10.1 = 10 + 0.1 ✓
126.9245
=
4
3
×
π
×
0.1
×
(3
×
10
2
+
3
×
10
×
0.1
+
0.1
2
)
✓
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