Section 7.4: Arc Length
Arc Length
The arch length s of the graph of f(x) over [a,b] is
simply the length of the curve.
a
b
Find the arch length s of the graph of f(x) = -3x + 12
over [1,3].
1,9 
No Calculator
White Board Challenge
 3  1   3  9 
2
40
 3,3
2 10
2
Linear Arc Length
Find the arch length s of the graph of f(x) = mx + b
f b  f  a 
over [a,b].
MVT : f '  c  
ba

a, f  a 
b  a    f b   f  a 
2
2
b  a    b  a   f ' c  

b  a 
 b, f  b  
a
2
2
b
x
b  a 
2
1  f '  c  
2
1 f 'c
1  f '  c   x
2
2
Arc Length as a Riemann Sum
Find the arch length s of the graph of f(x) over [a,b].
Approximate the arch
length with chords
Arc Length 

x
i 1 , f  xi 1 

N
lim
max xk  0

b
a
 x , f x  
i
a
i
b

i1
1  f '  ci   xi
2
1   f '  x   dx
2
Arc Length Formula
Assume that f '(x) exists and is continuous
on [a,b]. Then the arc length s of y = f (x)
over [a,b] is equal to:
s
b
a
1   f '  x   dx
2
Example 1
Find the arc length s of the graph
f (x) = 1/12 x3 + x-1 over [1,3].
f ' x  x  x
Use the formula:
Find the
derivative:
s
3
1
2
1
4
2
3
The derivative is not
defined at 0 but 0 is
not in our interval.
Thus we can use
the arc length
formula.
1   x  x  dx   1  161 x 4  12  x 4 dx
1
1
4
2
2
2

3

3
1
1
3

1

1
16
x 4  12  x 4 dx

x x
1
4
1
4
2
x 2  x 2  dx
 x x
1
12
 dx
2 2
3
1

3
1
17

6
Example 2
Find the arc length s of the graph y = x1/3 over
[-8,8].
The derivative is
y'  x
Find the
derivative:
Instead, try
solving for x:
1
3
x y
Find the new
derivative:
Find the new
limits:
Use the formula:
2 3
3
x '  3y
3
8  y
y  2
2
s
2
2

1
3 x2 3
not defined at 0
and 0 is in our
interval. Thus we
can not use the
arc length
formula.
Now the
derivative is
defined
everywhere.
3
Right now, we can NOT
evaluate this integral
without a calculator.
2 2
8 y
y2
1  3 y  dy
 17.26
Find the arc length of the curve of y = x2 – 4│x│ – x
over [-4,4].
 x  3x if
y 2
 x  5 x if
x0
2
s
0
4
x0
1   2 x  3 dx  
 19.56
2
Calculator
White Board Challenge
4
0
1   2 x  5  dx
2