Calculus II MAT 146 Integration Applications: Arc Length
[Pages:3]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 (xi ) ! f (xi!1)
xi ! xi!1
= "yi "xi
#
mi
=
"yi "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:
( ) Li = (xi ! ) xi!1 2 + f (xi ) ! f (xi!1) 2
= ("xi )2 + ("yi )2
= ("xi )2 + ("yi )2
= ("xi )2 + (mi"xi )2
= ("xi )2 + ( f #(xi*)"xi )2
=
("xi
)2
$%&1
+
(
f
#(
xi*
))2
' ()
= "xi 1+ ( f #(xi*))2
We now let the number of segments between (a, f(a)) and (b, f(b)) grow without bound.
That is, we let n--> . We can now write a definite integral to add all the lengths Li:
total length of the curve from x = a to x = b :
b
&
1+"# f !(x)$%2 dx
a
Example #1: Determine the length of the curve y = x2 from x = 0 to x = 3.
b
We use the arc length formula &
1+"# f !(x)$%2 dx . Here, a = 0 and b = 3. The function
a
y = f (x) = x2 has derivative f !(x) = 2x , so
b
arc length = &
1+"# f !(x)$%2 dx
a
3
=&
1+ [2x]2 dx
0
3
= & 1+ 4x2 dx
0
( ) = 1 ln 37 + 6 + 3 37
4
2
' 9.747089
Example #2: What is the arc length for y = x3 + 1 , 1 ! x !1 ?
6 2x 2
b
Again, use the arc length formula &
1+"# f !(x)$%2 dx . Here, a = ? and b = 1, with
a
dy dx
=
x2 2
!
1 2x2
.
b
arc length = &
1+"# f !(x)$%2 dx
a
1
=&
1 2
" 1+(
#
x2 2
'
1 2x2
$2 ) %
dx
= 31 48
Example #3: For x = ln(1! y2 ) , determine the length of arc x(y) for 0 y ?.
d
Here, we use the arc length formula with y as the independent variable: &
1+"#h!(y)$%2 dy .
c
Here, we have c = 0 and d = ?, with
dx dy
=
"1 #$1! y2
%
' (!2 y)
&
=
!2 y 1! y2
.
d
arc length = &
1+"# f !(y)$%2 dy
c
1 2
=&
0
1
+
" '2y #(1' y2
$2 ) %
dy
* 0.5986123
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