12.6 Arc Length
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Let C be a curve joining points A and B. See Fig. 1.1. Choose the points P0, P1, P2, …, Pn–1, Pn along C such that A =
P0 and B = Pn. Denote the length of the chord (line segment) P0 P1 by |P0 P1|, that of the chord P1P2 by |P1P2|, etc, ie,
| Fig. 1.1 Length of curve or arc C is limit of sum of lengths of chords |
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| 2. Lengths Of Graphs Of Functions |
The derivation of the formula for the length of the graph of a function f requires that f is differentiable and f ‘ is
continuous, as seen below. A function f is said to be continuously differentiable if it’s differentiable on its domain and
its derivative f ‘ is continuous there. Let f be a continuously differentiable function and [a, b] a sub-interval of dom( f ).
See Fig. 2.1. We wish to find the length of the portion of the graph of f over [a, b].
Let s be the length of the portion of the graph of f over [a, b]. Form a regular partition of order n of [a, b]. We select
n = 5 as an example in Fig. 2.1. The partition points are a = x0 < x1 < x2 < … < xn–1 < xn = b. So the sub-intervals [xi–1,
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| The length s of the graph of continuously differentiable f over [a, b] is:
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Let y = f(x). Since f ‘(x) = dy/dx, Eq. [2.1] is equivalent to:
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Remark 2.1
If 2 functions differ by a constant on [a, b], as seen in Fig. 2.2, then clearly the portions of their graphs there have an
equal length. This fact is confirmed by Eq. [2.1]. As the functions differ by a constant on [a, b], they have the same
derivative there, so, by Eq. [2.1], the portions of their graphs there have an equal length.
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| If functions differ by a constant on [a, b] then portions of |
Example 2.1
Find the length of the graph of f(x) = x2/3 over [1, 8]. Give the answer in approximate decimal format.
Solution
EOS
Note that we don’t have to sketch the arc or curve in question. When finding areas and volumes, the sketch is needed to correctly set up the integral. Here, it’s not. However, it may be necessary to determine the characteristics of the shape of
the curve; see Problem & Solution 5.
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| 3. Differential Of Arc Lengths |
The graph of f and the regular n-order partition of [a, b] in Fig. 2.1 are re-produced in Fig. 3.1. Consider the Riemann

Fig. 3.1 as an example for [xi–1, xi]. Recall from Section 4.3 Definitions 2.1 that, at xi–1, the differential of x equals the
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Remark that here we’re dealing with segments of tangent lines at xi–1, while in Part 2 we’re dealing with chords Pi–1Pi.. The arc length element is the length of a segment of a tangent line, not the length of a chord. As the arc length element
is obtained by operations on differentials (of x and y), it’s also called the differential of arc length.
Definition 3.1
| Suppose f is a continuously differentiable function and let y = f(x). Then the differential ds of arc length for f is: |
We’ve seen above that:
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The arc length is the definite integral of the differential of arc length.
The small right triangle on the graph of f in Fig. 3.1 has all its 3 sides expressed in differential terms, and thus is called a
differential triangle. We can manipulate Eq. [3.1] as follows:
| The differential ds of arc length for continuously differential f where y = f(x) is also given by: |
From Eqs. [3.2] and [3.3] we obtain:
But this is exactly Eq. [2.2]. We observe that the differential triangle with its equation:
serves as a useful mnemonic device to obtain Eqs. [2.1], [2.2], or [3.4].
Example 3.1
Calculate the length of the graph of:
from x = 1 to x = 2, using differential of arc length in forming the integral.
Solution
The differential of arc length is:
The required arc length is:
EOS
When asked to use the differential of arc length in forming the integral, we use Eqs. [3.3] and [3.4]. For the function in
this example, we factor 1 + (dy/dx)2 into the square of something, and so the radical (square root symbol) in the
expression for ds is removed.
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We can do this reversing because g is a continuously differentiable function of the independent variable y and the
dependent variable is x where x = g( y), and because the formulas for arc length discussed above are valid for any
continuously differentiable function, regardless of the variable. Eq. [3.1] can also be manipulated as follows:
So:
Example 4.1
Compute the length of the graph of:
from y = 1 to y = 2.
Solution
The differential of arc length is:
The desired arc length is:
EOS
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To avoid the possible confusion as to which differential dx or dy to use in the equation for arc length, we make the
following observation. For arc length calculated along the x-axis from x = a to x = b, the function is a function of the
variable x, so it’s the derivative with respect to x and the differential of x, ie ( f ‘(x) or dy/dx) and dx. For arc length
calculated along the y-axis from y = c to y = d, the function is a function of the variable y, so it’s the derivative with
respect to y and the differential of y, ie ( g ‘( y) or dx/dy) and dy.
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1. Find the length of the graph of the function f(x) = x3/2 on [0, 1].
Solution
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2. Calculate the length of the curve y = f(x) = x4 + 1/(32x2) on [1, 2]. Give the answer in approximate decimal format.
Solution
The desired arc length is:
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3. Compute the length of the curve:
Solution
The differential of arc length is:
The required arc length is:
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4. Evaluate the length of the graph of y = x2 from the origin to the point (2, 4). Give an approximate value in decimal
format.
Solution
Let f(x) = y = x2. Then f ‘(x) = 2x. The required arc length is:
Hence:
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5. Find the circumference of the closed curve x2/3 + y2/3 = a2/3, where a > 0 is a constant.
Note
| Graph Of x2/3 + y2/3 = 32/3. |
Solution
Note
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