[PDF] Section 81: Arc Length - MATH 134 Calculus 2 with FUNdamentals

40130_6ArcLength.pdf

MATH 134 Calculus 2 with FUNdamentals

Section 8.1: Arc Length

In this section we will learn how to nd the length of a curve, specically, the length of the graph of a

function. This is known as thearc length. The resulting integral is typically impossible to compute by hand so numerical integration techniques (e.g., Midpoint Rule) are often necessary to compute the arc length. For this worksheet (and on homework), we choose functions where the integrals are possible to do by hand or by using an integration table.

Arc Length Formula

To nd the arc length of a dierentiable functionf(x) fromx=atox=b, we useArc Length = Z b ap1 + [f0(x)]2dx:(1) The basic idea behind the formula is to draw small tangent vectors (line segments) at successive points along the graph, and then sum up the lengths of these vectors. Since the slope toy=f(x) is given by the derivativedy=dx, a right triangle with length
xfor the base and heightf0(x)

xwill have a hypotenuse that approximates the curve. Using the Pythagorean Theorem, the length of the hypotenuse of this triangle is p(
x)2+ (f0(x)
x)2=p(
x)2(1 + [f0(x)]2) =
xp1 + [f0(x)]2: Taking the limit as
xapproaches 0 and summing over the curve fromx=atox=bgives the integral in formula (1). Example 1:Compute the arc length of the graph off(x) = 3x1 fromx= 1 tox= 5 using formula (1). Then check your answer using the distance formula. Answer:We havef0(x) = 3 for anyx, so formula (1) simply becomes L=Z 5

1p1 + 3

2dx=Z 5

1p10dx=p10

Z 5 1

1dx=p10x5

1= 4p10:

On the other hand, note thatf(x) is a linear function so its graph is just a line. Sincef(1) = 2 andf(5) = 14, the arc length fromx= 1 tox= 5 is the length of the line segment between the points (1;2) and (5;14). Using the distance formulas (essentially the Pythagorean Theorem), we have

L=p(51)2+ (142)2=p16 + 144 =

p160 = 4 p10; which agrees with the answer obtained using formula (1). Example 2:Compute the arc length of the graph off(x) =18 x2lnxfromx= 1 tox=e2. Answer:The key to this problem is to write 1 + [f0(x)]2as a perfect square. We computef0(x) = 14 xx1. Then,

1 + [f0(x)]2= 1 +116

x212 +x2=116 x2+12 +x2=14 x+x1 2 :

This implies that the arc length is

Z e2 1s 14 x+x1 2 dx=Z e2 114
x+x1dx=18 x2+ lnxe 2 1 =18 e4+ 218 + 0 =18 e4+ 15
: Exercises:For #1{2, compute the arc length of the graph of each function over the given interval.

1.f(x) =x3=2fromx= 0 tox= 4.

2.f(x) =x3+112

x1fromx= 1 tox= 2.

3. Verify that the circumference of the unit circle is 2by computing the arc length of the curve

y=p1x2fromx=1 tox= 1.