Practice Problems: Arc Length
Written by Victoria Kala
vtkala@math.ucsb.edu
December 6, 2014
Solutions to the practice problems posted on November 30.
Find the exact length of the curve for the problems below.
1. x =
1√
3 y(y
− 3), 1 ≤ y ≤ 9
Solution: Use the formula L =
x=
R9q
1
2
1 + ( dx
dy ) dy. Need to find
1√
1
dx
1
y(y − 3) = (y 3/2 − 3y 1/2 ) ⇒
=
3
3
dy
3
Then find
s
dx
dy :
3 1/2 3 −1/2
y
− y
2
2
=
1
2
√
1
y− √
y
q
2
1 + ( dx
dy ) :
1
1+
4
√
1
y− √
y
2
2
r
=
1
1
1
+ +
=
4y 2 4y
s 2
1 √
1
1 √
1
y+ √
=
y+ √
4
y
2
y
Now find L:
Z
L=
9
s
1+
1
dx
dy
1
dy =
2
Z
1
9
√
1
y+ √
y
1
dy =
2
2 3/2
y
+ 2y 1/2
3
9
=
1
32
3
2. y = 3 +
1
2
cosh 2x, 0 ≤ x ≤ 1
Solution: Use the formula L =
R1q
0
dy 2
1 + ( dx
) dx. Need to find
dy
dx :
dy
1
= sinh 2x · 2 = sinh 2x
dx
2
Then find
q
dy 2
1 + ( dx
) :
p
1 + sinh2 2x =
p
cosh2 2x = cosh 2x
Now find L:
s
2
Z 1
Z 1
dy
1
L=
1+
dx =
cosh 2xdx = sinh 2x
dx
2
0
0
1
1
=
0
1
1
(sinh 2 − sinh 0) = sinh 2
2
2
3. y = ln(1 − x2 ), 0 ≤ x ≤
1
2
Solution: Use the formula L =
R 1/2 q
0
dy 2
1 + ( dx
) dx. Need to find
dy
dx :
dy
−2x
1
· −2x =
=
dx
1 − x2
1 − x2
Then find
q
dy 2
1 + ( dx
) :
s
4x2
=
1+
(1 − x2 )2
s
(1 − x2 )2 + 4x2
=
(1 − x2 )2
s
1 + 2x2 + x4
=
(1 − x2 )2
s
1 + x2
(1 + x2 )2
=
2
2
(1 − x )
1 − x2
Now find L:
Z 1/2 r
Z 1/2 Z 1/2 Z 1/2
dy 2
2
2
1 + x2
L=
1 + ( ) dx =
dx =
−1 +
dx =
−1 − 2
dx
dx
1 − x2
1 − x2
x −1
0
0
0
0
We need to use partial fractions to integrate
2
x2 −1 .
x2 − 1 factors into (x − 1)(x + 1):
2
A
B
=
+
(x − 1)(x + 1)
x−1 x+1
2 = A(x + 1) + B(x − 1) ⇒ 2 = (A + B)x + A − B
(
A+B =0
We get the system of equations
. Solving this system yields the solution A =
A−B =2
1, B = −1. So then
Z 1/2 Z 1/2 1/2
1
2
1
L=
dx =
+
dx = −x−ln |x−1|+ln |x+1|
−1 − 2
−1 −
x −1
x−1 x+1
0
0
0
1
1
3
1
= − − ln
+ ln
= − + ln 3
2
2
2
2
4. y = 1 − e−x , 0 ≤ x ≤ 2
Solution: Use the formula L =
R2q
0
dy 2
1 + ( dx
) dx. Need to find
dy
dx :
dy
= e−x
dx
Then find
q
dy 2
1 + ( dx
) :
p
Now find L:
Z
L=
0
2
1 + (e−x )2
r
dy
1 + ( )2 dx =
dx
2
Z
0
2
p
1 + (e−x )2 dx
Let u = e−x . Then du = −e−x dx ⇒ −ex du = dx ⇒ − u1 du = dx. When x = 0, u = 1. When
x = 2, u = e−2 :
Z e−2 p
1
1 + u2 du
−
u
1
Need to use trig substitution. Let u = tan θ, then du = sec2 θdθ (we won’t change the bounds
on this one):
Z p
Z p
Z
Z
1
1
(1 + tan2 θ) sec θ
sec3 θ
1 + u2 du =
1 + tan2 θ
sec2 θdθ =
dθ =
dθ
u
tan θ
tan θ
tan θ
Z Z
sec θ
=
dθ = (sec θ tan θ + csc θ) dθ = sec θ + ln | csc θ − cot θ|
sec θ tan θ +
tan θ
√
p
1 + u2 − 1
2
= 1 + u + ln
u
So then
Z
−
1
e−2
p
p
1
1 + u2 du = − 1 + u2 − ln
u
√
√
1 + u2 − 1
u
e−2
1
√
√
1 + e−4 − 1
− − 2 − ln 2 − 1
−2
e
p
p
√
√
= − 1 + e−4 − ln( 1 + e−4 − 1) + ln e−2 + 2 + ln( 2 − 1)
p
p
√
√
= − 1 + e−4 − ln( 1 + e−4 − 1) − 2 + 2 + ln( 2 − 1)
p
= − 1 + e−4 − ln
3