SME
Calculus II · Problem type
Arc length of y = c·x³ᐟ²
Use this for “Find the length of the curve y = 2x³ᐟ² from x = 0 to x = 1.”
Chart
f
(3)
=
2
√
3
≈
3.464
This is an example. Type your own values to start a new problem.
y
=
c
x
3
2
,
0
≤
x
≤
b
L
=
y(b)
Height at the right end
worked out (not typed)
d
Length of the straight chord
worked out (not typed)
k
Coefficient inside the root, 9c² ÷ 4
worked out (not typed)
Formulas
y(
b
)
=
c
×
b
×
√
b
3.4641
=
2
3
×
3
×
√
3
d
=
√(
b
2
+
y
(
b
)
2
)
4.5826
=
√
3
2
+
3.4641
2
k
=
9
×
c
2
4
1
=
9
×
(
2
3
)
2
÷
4
L
=
(
2
3 ×
k
)
×
((1
+
k
×
b
)
×
√
1 +
k
×
b
−
1)
4.6667
=
(
2
3 × 1
)
×
((1
+
1
×
3)
×
√
1 + 1 × 3
−
1)
Assumptions
•
A short piece of the curve over dx is nearly straight, so its length is √(dx² + dy²) = √(1 + (dy/dx)²) dx.
•
Here dy/dx = (3c ÷ 2)√x, so 1 + (dy/dx)² = 1 + kx with k = 9c² ÷ 4, and its square root integrates by u-substitution.
•
The curve starts at the origin with c positive. The arc is never shorter than the straight chord from (0, 0) to its end.
Step-by-step
WE KNOW
c = 2/3, b = 3
FIND
height at the right end (y(b)), length of the straight chord (d), coefficient inside the root, 9c² ÷ 4 (k), length of the arc (L)
Step 1 · Find height at the right end (y(b))
Use y(b) = c × b × √(b)
Put
x
=
b
into
the
curve,
using
x
3
ᐟ
2
=
x√x,
to
find
where
the
arc
ends.
y(
b
)
=
2
3
×
3
×
√
3
y(
b
)
=
2
3
×
3
×
1.732
y(
b
)
=
2
×
1.732
y(
b
)
=
3.4641
Step 2 · Find length of the straight chord (d)
Use d = √(b² + y(b)²)
The chord runs straight from (0, 0) to the end of the arc; the arc found below must be at least this long.
d
=
√
3
2
+
3.4641
2
d
=
√
9 + 12
d
=
√
21
d
=
4.5826
Step 3 · Find coefficient inside the root, 9c² ÷ 4 (k)
Use k = 9 × c² ÷ 4
The
slope
is
dy/dx
=
(
3
c
2
)√x,
so
its
square
is
(dy/dx)
2
=
(
9
c
²
4
)x
=
kx.
The
arc
length
adds
up
√
1 + kx
dx.
k
=
9
×
(
2
3
)
2
÷
4
k
=
9
×
4
9
÷
4
k
=
4
4
k
=
1
Step 4 · Find length of the arc (L)
Use L = (2 ÷ (3 × k)) × ((1 + k × b) × √(1 + k × b) − 1)
With
u
=
1
+
kx,
du
=
k
dx,
so
√
1 + kx
integrates
to
(
2
3
k
)
(1 + kx)
3
2
.
Take
its
value
at
0
from
its
value
at
b
.
L
=
(
2
3 × 1
)
×
((1
+
1
×
3)
×
√
1 + 1 × 3
−
1)
∫
from
0
to
3
of
√
1 + x
dx
=
[(
2
3
)
(1 + x)
3
2
]
from
0
to
3
=
4.6667
L
=
4.6667
Check
Put the numbers back in. Both sides should match.
3.4641
=
2
3
×
3
×
√
3
✓
4.5826
=
√
3
2
+
3.4641
2
✓
1
=
9
×
(
2
3
)
2
÷
4
✓
4.6667
=
(
2
3 × 1
)
×
((1
+
1
×
3)
×
√
1 + 1 × 3
−
1)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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