MathMentorArcLength

Length of a Plane Curve - Arc Length - Classwork
%
71 B)3< 45 /0025A)*/41 4<1 (18C4< 5: 4<1 ';2@1 y # x
:25* A = , 45 A = % ;3)8C - ()813# \2/B 4<1 - ()813 /89
:)89 4<1)2 (18C4<3 /89 3;* 4<1*#
, 45 #$
]52*;(/ ^^^^^^^^^^^^ &18C4< ^^^^^^^^^
#$ 45 !
]52*;(/ ^^^^^^^^^^^^ &18C4< ^^^^^^^^^
! 45 !#$ ]52*;(/ ^^^^^^^^^^^^ &18C4< ^^^^^^^^^
!#$ 45 % ]52*;(/ ^^^^^^^^^^^^ &18C4< ^^^^^^^^^
=54/( &18C4< ^^^^^^^^^^^^^^^^^^^^^
We can now make this statement: A curve between two points has length L #
lim
;x *0, ;y *0
? ;x + ;y
? f (c ) ;x .
2
2
.
The limit of a chord length can be found exactly by transforming it to a Riemann sum:
Here is the technique used. It is some involved algebra. Do not worry why steps are done.
5 ;y 2 8 2 5 , ;y / 2 8 2
;x + ;y # 71 + 2 : ;x # 71 + . 1 : ;x
76 - ;x 0 :9
6 ;x 9
2
2
So a chord length can be written as
?
5 , ;y / 2 8
71 + . 1 : ;x 2 # ?
76 - ;x 0 :9
5 , ;y / 2 8
71 + . 1 : ;x
76 - ;x 0 :9
;y
# f 3( x )
;x
So the exact length of the curve between x # a and x # b is
As ;x approaches 0, you know that
b
L#
>
2
1 + ( f 3( x )) dx
NOTE : very few problems can be solved using the FTC.
a
Example 1) Find the exact length of the curve y # - x from x = 0 to x = 3. Verify geometrically.
MasterMathMentor.com
- 57 -
Stu Schwartz
Example 2) Find the exact length of the curve y # 5 x ! 2 from x = 0 to x = 2. No calculators.
Example 3) Find the exact length of the curve y # x
3
2
from x = 0 to x = 4. No calculators.
Example 4) At time = 0, a bug starts crawling along the path y #
the bug crawl in 2 seconds? (no calculators)
32
1 2
t + 2) , y in feet, t in seconds.. How far does
(
3
Example 5) Find the length of the given curves from between the indicated x-values (show setup - calculator
allowed)
5 48
a. y # e x q0, 2r
b. y # tan x 70, :
6 49
c. y # 4 x ! x 2
MasterMathMentor.com
q1, 4r
d. y # x sin x q0, 2#r
- 58 -
Stu Schwartz
Arc Length - Homework
Find the exact length of the curves between the indicated x values. No calculators.
1)
2) y # 4 x
(0,2)
y # 10 x ! 1
3
2
(0,4)
x3 1
+ from x = 1 to x = 2. No calculators. The algebra is
12 x
intricate but straightforward. Take your time and do it neatly.
3) Find the exact length of the curve y #
Find the length of the curves between the x-values. Calculators allowed.
4) y # x 2 ! 5 x + 3
5) y # x 3 ! 9 x 2 + 5 x + 50
(1,6)
6) y # sec x
5 48
760, 4 :9
MasterMathMentor.com
7) y # (ln x )
- 59 -
2
(-1,9)
(0.1, e)
Stu Schwartz
L) =<1 0)'4;21 58 4<1 2)C<4 3<5B3
4<1 R18 ]2/8>()8 R2)9C1 /'2533
4<1 \1(/B/21 S)@12 614B118 F/
/89 _`## =<1 '18412 30/8 5: 4<1
62)9C1 )3 /65;4 -%,, :114 (58C#
=<1 3;30183)583 '/6(13 </8C )8
0/2/65()' /2'3 :25* 45B123 M$, :4
/65@1 4<1 B/412a3 3;2:/'1# =<131
'/6(13 '5*1 /3 '(531 /3 %%, :4 45
4<1 B/412 /4 4<1 '18412 5: 4<1
30/8# D31 4<)3 )8:5*/4)58 45
B2)41 /8 1b;/4)58 5: 4<1 b;/92/4)'
:;8'4)58 1A02133)8C 4<1 <1)C<4 5:
4<1 '/6(13 :25* 4<1 B/412 /3 /
:;8'4)58 /3 4<1 <52)X584/(
9)34/8'1 :25* 4<1 '18412 30/8
(21*1*612 <5B 45 C1812/41 /
0/2/65(/ :25* + 05)843)# D31 4<1
1b;/4)58 4<18 45 '/(';(/41 4<1
(18C4< 5: 4<1 0/2/65()' '/6(1#
-%,, :4
M$, :4
M$, :4
%%, :4
9. When a chain hangs under its own weight, its shape is a catenary (from the latin for chain). The equation of a
catenary with a vertex on the y-axis is y # e x + e! x . Find the length of the chain from x = -2 to x = 2. Then use
the same technique you used in v 8 to find the length of the parabola with the same vertex and endpoints.
MasterMathMentor.com
- 60 -
Stu Schwartz

Download Report