Calculus II MAT 146 Integration Applications: Arc Length

Again we use a definite integral to sum an infinite number of measures, each infinitesimally small. We seek to determine the length of a curve that represents the graph of some real-valued function f, measuring from the point (a, f (a)) on the curve to the point (b, f (b)) on the curve.

The graph of y = f is shown. We seek to determine the length of the curve, known as arc length, from the point (a, f (a)) on the curve to the point (b, f (b)) .

Together with endpoints (a, f (a)) and (b, f (b)) , we select additional points (xi, f (xi )) on the curve. Here, we’ve created n line segments. The length of the ith line segment is denoted by Li, with 1 ≤ i ≤ n.

Concentrate on the segment whose length is Li. Determine the slope of that segment: slope of segment whose length is Li = mi f x ! f x = ( i ) ( i!1) xi ! xi!1 "y = i "xi

"yi # mi = "xi

# mi $ "xi = "yi

We know by the Mean Value Theorem that there exists an xi*, with xi–1 ≤ xi* ≤ xi-1, such that mi = f !(xi* ) . We use this fact, and the distance formula, to calculate the typical segment length Li: 2 2 Li = (xi ! xi!1) +( f (xi ) ! f (xi!1))

2 2 = ("xi ) +("yi )

2 2 = ("xi ) +("yi )

2 2 = ("xi ) +(mi"xi )

2 2 = ("xi ) +( f #(xi* )"xi )

2 2 = "x $1+ f # x ' ( i ) %& ( ( i* )) ()

2 = "xi 1+( f #(xi* )) We now let the number of segments between (a, f(a)) and (b, f(b)) grow without bound.

That is, we let n—> ∞. We can now write a definite integral to add all the lengths Li: total length of the curve from x = a to x = b :

b 2 & 1+#" f !(x)%$ dx a Example #1: Determine the length of the curve y = x2 from x = 0 to x = 3. b 2 We use the arc length formula & 1+#" f !(x)%$ dx . Here, a = 0 and b = 3. The function a y = f (x) = x2 has derivative f !(x) = 2x , so b 2 arc length = & 1+#" f !(x)%$ dx a 3 2 = & 1+ [2x] dx 0 3 = & 1+ 4x2 dx 0 1 3 = ln 37 + 6 + 37 4 ( ) 2 ' 9.747089

x3 1 1 Example #2: What is the arc length for y = + , ! x !1 ? 6 2x 2 b 2 Again, use the arc length formula & 1+#" f !(x)%$ dx . Here, a = ½ and b = 1, with a dy x2 1 = ! . dx 2 2x2 b 2 arc length = & 1+#" f !(x)%$ dx a 2 1 " x2 1 $ = & 1+( ' 2 ) dx 1 2 # 2 2x % 31 = 48

Example #3: For x = ln(1! y2 ) , determine the length of arc x(y) for 0 ≤ y ≤ ½. d 2 Here, we use the arc length formula with y as the independent variable: & 1+#"h!(y)%$ dy . c dx " 1 % !2y Here, we have c = 0 and d = ½, with = $ 2 '(!2y) = 2 . dy #1! y & 1! y

d 2 arc length = & 1+#" f !(y)%$ dy c

1 2 2 " '2y $ = & 1+( 2 ) dy 0 #1' y % * 0.5986123