Length of a Plane Curve (Arc Length) - Drexel University

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

2

1. y = 4x 2 - 1 from x = to x =

12

9

19

54

x2 ln(x)

2. y = -

for 2 x 4

24

1 6 + ln 2; Detailed Solution: Here

4

3. y = 2 (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.)

2 sin-1 t 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.)

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 = ? 2r = ? 2(2) = .

4

4

r=2 4

x

5. Consider F (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.

e3

1

L=

1 + dx

1

x2

(b) Set up but do not evaluate an integral which represents the length of the curve by integrating with respect to y.

3

L=

1 + e2y dy

0

7. Consider the curve defined by f (x) = tan x on - ,

34

(a) Set up but do not evaluate an integral which represents the length of the curve by integrating with respect to x.

4

L=

1 + sec4 x dx

-

3

(b) Set up but do not evaluate an integral which represents the length of the curve by integrating with respect to y.

1

1

L= -3

1+

dy

(1 + y2)2

8. Consider the curve defined by y = sin x for 0 x .

(a) Set up but do not evaluate an integral which represents the length of the curve.

1 + cos2 x dx

0

2

int

2 sqrt 4 Kx2

,x

2 arcsin 1 x

(1)

2

with Student Calculus1

AntiderivativePlot, AntiderivativeTutor, ApproximateInt, ApproximateIntTutor, ArcLength,

(2)

(b) Estimate the value of your integral from part (a) by using a Midpoint ArcLengthTutor, Asymptotes, Clear, CriticalPoints, CurveAnalysisTutor, DerivativePlot, ApproxiDerivativeTutor, DiffTutor, ExtremePoints, FunctionAverage, FunctionAverageTutor, mation with three rectangles of equal width. FunctionChart, FunctionPlot, GetMessage, GetNumProblems, GetProblem, Hint, InflectionPoints, IntTutor, Integrand, InversePlot, InverseTutor, LimitTutor, Below is the graph of y = 1 + cos x on the interval [0, ] along with three MeanValueTheorem, MeanValue2TheoremTutor, NewtonQuotient, NewtonsMethod, NewtonsMethodTutor, PointInterpolation, RiemannSum, RollesTheorem, Roots, Rule, Show,

rectangles

of

equal

width whose heights were determined by ShowIncomplete, ShowSolution, ShowSteps, Summand, SurfaceOfRevolution, SurfaceOfRevolutionTutor, Tangent, TangentSecantTutor, TangentTutor,

the

function

value

at the midpoint of each resulting subinterval. TaylorApproximation, TaylorApproximationTutor, Understand, Undo, VolumeOfRevolution, VolumeOfRevolutionTutor, WhatProblem

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

3 5 3 7

K0.2

8

4

8

2

8

4

8

x K0.4

Using these rectangles,

1 + cos2 x dx 1 + 7

3

0 A midpoint Riemann sum approximation of f x dx, where

0

f x = 1 Ccos x 2 and the partition is uniform. The approximate

9. Recall the definitions of Hyperbolic Sine & Hyperbolic Cosine value of the integral is 3.817821847. Number of subintervals used: 3. from Math 121:

Hyperbolic Sine ex - e-x

sinh x = 2

Hyperbolic Cosine ex + e-x

cosh 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)

ex + e-x ex - e-x

=

+

2

2

= (ex)(e-x)

ex + e-x ex - e-x

-

2

2

=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 ex - e-x =-

2 = - sinh x

Thus, f (x) = sinh x is odd.

4

d (c) Show that (sinh x) = cosh x and deduce that cosh x dx = sinh x + C.

dx

d

d ex - e-x

(sinh x) =

dx

dx

2

d =

1 ex - 1 e-x

dx 2 2

= 1 ex + 1 e-x 22 ex + e-x

= 2

= cosh x

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.)

d (d) Show that (cosh x) = sinh x and deduce that sinh x dx = cosh x + C.

dx

d

d ex + e-x

(cosh x) =

dx

dx

2

d =

1 ex + 1 e-x

dx 2 2

= 1 ex - 1 e-x 22 ex - e-x

= 2

= sinh x

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

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