Length of a Plane Curve (Arc Length)
SUGGESTED REFERENCE MATERIAL:
As you work through the problems listed below, you should reference Chapter 6.4 of the recommended textbook (or the equivalent chapter in your alternative textbook/online resource)
and your lecture notes.
EXPECTED SKILLS:
• Be able to find the arc length of a smooth curve in the plane described as a function
of x or as a function of y.
PRACTICE PROBLEMS:
For problems 1-3, compute the exact arc length of the curve over the given
interval.
3
1. y = 4x 2 − 1 from x =
1
2
to x =
12
9
19
54
x2 ln(x)
−
for 2 ≤ x ≤ 4
2
4
1
6 + ln 2; Detailed Solution: Here
4
2. y =
2
3. y = (x2 − 1)3/2 for 1 ≤ x ≤ 3
3
46
3
4. Consider the curve defined by y =
√
4 − x2 for 0 ≤ x ≤ 2.
(a) Compute the arc length on the interval [0, t] for 0 ≤ t < 2. (Your arc length will
depend on t.)
t
−1
2 sin
2
(b) Use your answer from part (a) to compute the arc length on the interval [0, 2].
(Hint: You will need to introduce a limit.)
π
1
(c) Confirm your answer from part (b) by using geometry.
1
On the interval [0, 2], the curve is of a circle with a radius of 2. So, the length
4
1
1
1
should be of the circumference; that is, Length = · 2πr
= · 2π(2) = π.
4
4
4
r=2
Z
5. Consider F (x) =
x
√
t2 − 1 dt. Compute the arc length on [1, 3]
1
4; Detailed Solution: Here
6. Consider the curve defined by f (x) = ln x on 1, e3
(a) Set up but do not evaluate an integral which represents the length of the curve
by integrating with respect to x.
Z e3 r
1
L=
1 + 2 dx
x
1
(b) Set up but do not evaluate an integral which represents the length of the curve
by integrating with respect to y.
Z 3√
L=
1 + e2y dy
0
h π πi
7. Consider the curve defined by f (x) = tan x on − ,
3 4
(a) Set up but do not evaluate an integral which represents the length of the curve
by integrating with respect to x.
Z π √
4
1 + sec4 x dx
L=
− π3
(b) Set up but do not evaluate an integral which represents the length of the curve
by integrating with respect to y.
Z 1 s
1
L= √
1+
dy
(1 + y 2 )2
− 3
8. Consider the curve defined by y = sin x for 0 ≤ x ≤ π.
(a) Set up but do not evaluate an integral which represents the length of the curve.
Z π√
1 + cos2 x dx
0
2
int
2
,x
sqrt 4 Kx2
2 arcsin
1
x
2
(1)
with Student Calculus1
AntiderivativePlot, AntiderivativeTutor, ApproximateInt, ApproximateIntTutor, ArcLength,
ArcLengthTutor, Asymptotes, Clear, CriticalPoints, CurveAnalysisTutor, DerivativePlot,
DerivativeTutor, DiffTutor, ExtremePoints, FunctionAverage, FunctionAverageTutor,
FunctionChart, FunctionPlot, GetMessage, GetNumProblems, GetProblem, Hint,
InflectionPoints, IntTutor, Integrand, InversePlot, InverseTutor, LimitTutor,
MeanValueTheorem, MeanValueTheoremTutor, NewtonQuotient, NewtonsMethod,
2
NewtonsMethodTutor, PointInterpolation, RiemannSum, RollesTheorem, Roots, Rule, Show,
ShowIncomplete, ShowSolution, ShowSteps, Summand, SurfaceOfRevolution,
SurfaceOfRevolutionTutor, Tangent, TangentSecantTutor, TangentTutor,
TaylorApproximation, TaylorApproximationTutor, Understand, Undo, VolumeOfRevolution,
VolumeOfRevolutionTutor, WhatProblem
(2)
(b) Estimate the value of your integral from part (a) by using a Midpoint Approximation with three rectangles of equal width.
√
Below is the graph of y = 1 + cos x on the interval [0, π] along with three
rectangles of equal width whose heights were determined by the function value
at the midpoint of each resulting subinterval.
RiemannSum sqrt 1 C cos x
2
, x = 0 ..Pi, method = midpoint, partition = 3, output = plot
1.4
1.2
1
0.8
0.6
0.4
0.2
0
K0.2
π
8
π
4
3π
8
π
2
x
5π
8
3π
4
7π
8
π
K0.4
Z
Using these rectangles,
π
√ √
π
1+ 7
1 + cos2 x dx ≈
A midpoint Riemann sum approximation of 3
f x dx, where
π
0
0
f x = 1 Ccos x 2 and the partition is uniform. The approximate
value of the integral is 3.817821847. Number of subintervals used: 3.
9. Recall the definitions of Hyperbolic Sine & Hyperbolic Cosine from Math 121:
Hyperbolic Sine
sinh x =
Hyperbolic Cosine
ex − e−x
2
cosh x =
ex + e−x
2
The sketches of y = cosh x and y = sinh x are shown below. The dashed curves are
called “Curvilinear Asymptotes,” which describe the end behavior of the functions.
3
(a) Show that cosh2 x − sinh2 x = 1
cosh2 x − sinh2 x = (cosh x + sinh x)(cosh x − sinh x)
x
x
e + e−x ex − e−x
e + e−x ex − e−x
=
+
−
2
2
2
2
x
−x
= (e )(e )
=1
(b) Verify that f (x) = sinh x is an odd function. (Hint: Recall an odd function
satisfies the identity f (−x) = −f (x).)
To verify that a function is off, we check that f (−x) = −f (x). We compute by
appealing to the definition of sinh x from above.
e−x − e−(−x)
sinh (−x) =
2
e−x − ex
=
2 x
e − e−x
=−
2
= − sinh x
Thus, f (x) = sinh x is odd.
4
d
(c) Show that
(sinh x) = cosh x and deduce that
dx
Z
cosh x dx = sinh x + C.
d
d ex − e−x
(sinh x) =
dx
dx
2
d 1 x 1 −x
e − e
=
dx 2
2
1 x 1 −x
= e + e
2
2
x
−x
e +e
=
2
= cosh x
Z
Thus, as a result,
cosh x dx = sinh x + C. (We could have also verified this
integration formula by integrating the given definition of cosh x.)
Z
d
(cosh x) = sinh x and deduce that sinh x dx = cosh x + C.
(d) Show that
dx
d
d ex + e−x
(cosh x) =
dx
dx
2
d 1 x 1 −x
=
e + e
dx 2
2
1
1
= ex − e−x
2
2
ex − e−x
=
2
= sinh x
Z
Thus, as a result,
sinh x dx = cosh x + C. (We could have also verified this
integration formula by integrating the given definition of sinh x.)
(e) A telephone wire which is supported only by two telephone poles will sag under
its own weight and form the shape of a catenary as shown below.
5
Consider a telephone wire that is supported by two poles (one at x = b and the
other at x = −b), as in the diagram below.
x
The shape of the sagging wire can be modeled by y = a cosh
, where a > 0
a
and −b ≤ x ≤ b. What is the length of the wire?
b
A = 2a sinh
a
6