Sa
sk
The length of the curve (x(t), y(t)) as t varies from t0 to t1 is given by
L=
t=t1 t=t0
DEO
PAT-
ET
RIE
n
U
Arc Length
of
versity
ni
atc he w
a
2
2
(x (t)) + y (t) dt
If we wish to find the length of a curve which is the graph of a function y = f (x), as x
runs from a to b,
we let x(t) = t, y(t) = f (x(t)) = f (x) and we get x (t) = 1, and y (t) = f (x(t))x (t) =
f (x), so we have a simple formula for the length:
L=
x=b x=a
2
1 + (f (x)) dx =
b
a
2
1 + (f (x)) dx =
b
Similarly, if we have a curve x = g(y) with y running from c to d we get
L=
y=d y=c
Example:
2
1 + g (y) dy =
d
c
1 + g (y) dy =
Consider the curve given by
x(t) = cos t, y(t) = sin t , 0 ≤ t ≤ π .
Its length is
L=
t=π t=π
t=0
t=0
x (t)2
+
y (t)2 dt
t=π =
(− sin t)2 + (cos t)2 dt =
t=0
1dt = t|π
0 = π
1
2
a
2
1 + y dx
d
c
1 + (x )2 dy
2
1
a
b
3 1 2
2
x
1+
dx =
2
a
b
9
1 + xdx =
4
a
0
3 b
9
2
(1
+
x)
4
4
=
3
9
2
a
b
32 b
32 b
3
9
8 4 + 9x 1
2
1+ x =
= 27 (4 + 9x) =
4
27
4
a
a
sk
3
b
2
We have L =
1 + y dx =
8
27
Sa
3
y = x2, 0 ≤ a ≤ x ≤ b
y
a
3
3
1 (4 + 9b) 2 − (4 + 9a) 2
27
2
0
1
2
x
DEO
PAT-
ET
RIE
n
U
Find the length of the curve
of
Example:
versity
ni
atc he w
a
x = y 2, 0 ≤ c ≤ y ≤ b
1
0
x
-1 0
-1
1
2
3
4
-2
y=d y=d y=d 2
2
We have L =
1 + (x ) dy =
1 + 2y dy =
1 + 4y 2 dy
y=c
y=c
y=c
Making the substitution y = 12 tan θ, we have
1
dy = 2 sec2 θdθ, 1 + 4y 2 = 1 + tan2 θ = sec2 θ,
θc = arctan 2c when y = c and θd = arctan 2d when y = d, so
L=
1
2
θ=θd θ=θc
sec2
1
1
θ sec2 θdθ =
2
2
θ=θd
θ=θc
sec3 θdθ =
θ=θd
1
=
(sec θ tan θ + ln | sec θ + tan θ|) 2
θ=θc
1
1
[sec θd tan θd + ln | sec θd + tan θd |] − [sec θc tan θc + ln | sec θc + tan θc |] =
4
4
1
1 1 + 4d2 2d + ln | 1 + 4d2 + 2d| −
1 + 4c 2 2c + ln | 1 + 4c 2 + 2c| =
4
4
√
√
√
d 1 + 4d2 − c 1 + 4c 2 1
1 + 4d2 + 2d
+ ln √
2
4
1 + 4c 2 + 2c
Note that if we let c = 0, we get the formula for the distance along the parabola to the
point (d2 , d):
√
d 1 + 4d2 1
L=
+ ln( 1 + 4d2 + 2d)
2
4
√
and if we let b = d2 , we get the (equivalent) formula for the distance from (0,0) to (b, b):
√ √
b 1 + 4b 1 L=
+ ln 1 + 4b + 2 b
2
4
3
sk
DEO
PAT-
ET
RIE
n
U
2
Find the length of the curve
Sa
Example:
of
y
versity
ni
atc he w
a
U
2
1 2
x +2 , 0≤x ≤3
3
Example:
1
2
Find the length of the curve y =
Sa
12
13 2
x + 2 2x = x x 2 + 2 , so
32
2
2
2
1+ y
= 1 + x 2 (x 2 + 2) = x 2 + 2x 2 + 1 = x 2 + 1 , and
3
3
3
3
x
2
2
L=
+ x =
x + 1dx =
(x 2 + 1) dx =
3
0
0
0
03
33
+3 −
+ 0 = 12
3
3
y =
1
2
sk
6
3
x
0123
3 4 3 2
x 3 − x 3 + 5, 1 ≤ x ≤ 8
y
4
8
16
Solution:
12
1
1
3 4 1 3 2 −1
1
x3 −
x 3 = x 3 − x − 3 , so
43
83
4
2 2
2
1
1
1
1
1
1
8
1 + y = 1 + x 3 − x− 3
= x 3 + x − 3 , and
4
4
4
2 8
8 8
2
3
3
4
1
1
1
1
x
1
1
1x L=
x 3 + x − 3 dx =
x 3 + x − 3 dx = 4 +
2 =
4
4
4 3 1
1
3
y =
8 3 4 3 2
3 4 3 2
3 4 3 2
3 +
3
3 +
3
L = x3 + x3
8
8
1
1
=
−
=
4
8 1
4
8
4
8
3 4 3 2
3 3
3 9
2 + 2 −
+
= 12 + − =
4
8
4 8
2 8
12
4
3
8
1
0
x
012345678
PAT-
ET
RIE
atc he w
9
0
DEO
n
3
Find the length of the curve y =
y
Solution:
of
Example:
versity
ni
a