Calc 2- 6.5 Arc Length

Arc Length via Line Segment Approximation

• Real-world curves are often not straight lines (e.g., planetary orbits are ellipses). To find their length, we approximate the curve with straight line segments and then pass to a limit.
• Idea: If you have a function f on an interval [a, b], you can approximate the curve y = f(x) by a sequence of straight segments connecting points (x{i-1}, f(x{i-1})) to (xi, f(xi)). This is analogous to using a Riemann sum for area under a curve.
• Notation for a single segment:
  • Δx = xi - x{i-1}
  • Δy = f(xi) - f(x{i-1})
  • Length of the i-th segment ≈
    ( \sqrt{(Δx)^2 + (Δy)^2} )
    (by the Pythagorean theorem).
• If we use N segments to approximate the curve, the total approximate length is
  $L<em>N = \sum</em>{i=1}^N \sqrt{(Δx<em>i)^2 + (Δy</em>i)^2}.$
• The goal is to take the limit as the maximum Δx_i → 0 (i.e., as the partition gets finer) to obtain the arc length of the curve.

From Straight-Line Segments to an Integral

• Algebraic rearrangement of a single segment length:
  • Start with
    $\sqrt{(Δx)^2 + (Δy)^2}.$
  • Factor out the square of Δx:
    $\sqrt{(Δx)^2 + (Δy)^2} = |Δx| \sqrt{1 + \left(\frac{Δy}{Δx}\right)^2}.$
  • If the partition is taken in the direction of increasing x, then Δx > 0 and
    $\sqrt{(Δx)^2 + (Δy)^2} = Δx \sqrt{1 + \left(\frac{Δy}{Δx}\right)^2}.$
• Relate Δy/Δx to the derivative: as the partition gets finer (Δx → 0),
  $\frac{Δy}{Δx} \rightarrow f'(x)$ for some x in [x{i-1}, xi].
• Thus, each segment length can be approximated by
  $Δx<em>i \sqrt{1 + \left(\frac{Δy</em>i}{Δx<em>i}\right)^2} \approx Δx</em>i \sqrt{1 + \bigl(f'(x<em>i^<em>)\bigr)^2},$ where xi^ is a point in [x{i-1}, xi].
• In the limit, the sum becomes a Riemann sum for the integral
  $L = \int_a^b \sqrt{1 + \bigl(f'(x)\bigr)^2} \, dx.$
• Therefore, the arc length of the curve y = f(x) from x = a to x = b is
  
  L = \int_a^b \sqrt{1 + \bigl(f'(x)\bigr)^2} \, dx.
  

Example: y = 2x + 1 on [2, 3]

• Consider the straight-line function f(x) = 2x + 1 on the interval [2, 3].
• The derivative is a constant:
  $f'(x) = 2.$
• Arc length via the integral:
  $L = \int<em>2^3 \sqrt{1 + \bigl(2\bigr)^2} \, dx = \int</em>2^3 \sqrt{5} \, dx = \sqrt{5} \cdot (3 - 2) = \sqrt{5}.$
• Check via endpoints and a single segment (distance between endpoints):
  • Endpoints: x = 2 → y = f(2) = 5; x = 3 → y = f(3) = 7.
  • Δx = 3 - 2 = 1; Δy = 7 - 5 = 2.
  • Length of the straight-line segment joining endpoints:
    $\sqrt{(Δx)^2 + (Δy)^2} = \sqrt{1^2 + 2^2} = \sqrt{5}.$
• This matches the arc length computed via the integral, as expected for a straight-line (linear) function.
• The diagram intuition: the left endpoint is (2, 5) and the right endpoint is (3, 7); the actual curved path would have length approximated by many short segments, but for a straight line the length equals the distance between endpoints, which is (\sqrt{5}).
• Summary check: differentiating y = 2x + 1 gives a constant slope of 2, and the arc length integral reduces to a constant times the interval length, yielding (L = (\sqrt{5})(3-2) = \sqrt{5}).

Check for consistency and why this works

• The premise is that mathematics should be consistent: the same problem solved in different but equivalent ways should yield the same answer.
• In this example, the two methods (arc length integral and direct endpoint distance for a straight line) agree, illustrating consistency.
• The approach generalizes beyond straight lines to any differentiable curve, by taking finer partitions and using the derivative to approximate small arc segments.

Connections to prior concepts and real-world relevance

• Connection to Riemann sums: the arc length integral is obtained by taking a Riemann-sum-like limit of segment lengths.
• Connection to the Pythagorean theorem: the length of each short segment is given by the hypotenuse of a right triangle with legs Δx and Δy.
• Connection to derivatives: the local slope Δy/Δx tends to f'(x) as the segment width shrinks, leading to the integrand (\sqrt{1 + (f'(x))^2}).
• Real-world relevance: planetary orbits are elliptical, not straight lines, so arc length formulas are essential for calculating distances along curved paths in physics and astronomy.

Assumptions, conditions, and practical notes

• Assumptions:
  • f is differentiable on [a, b], and (\sqrt{1 + (f'(x))^2}) is integrable over [a, b].
• Practical notes:
  • The derivation uses the limit of sums of segment lengths, mirroring the logic used for areas under curves (
    Riemann sums).
  • When the function is linear, the arc length reduces to the distance between endpoints; for nonlinear functions, the integral captures the accumulated length of the curved path.

Key formulas to remember

• Arc length of a curve y = f(x) from x = a to x = b:
  $L = \int_a^b \sqrt{1 + \bigl(f'(x)\bigr)^2} \, dx.$
• Length of a small segment between x{i-1} and xi:
  \sqrt{(Δx)^2 + (Δy)^2} = |Δx| \sqrt{1 + \left(\frac{Δy}{Δx}\right)^2}. $$
• In the limit as Δx → 0, Δy/Δx → f'(x), giving the integral form.

Quick recap

• Curves are approximated by line segments; use Pythagoras to get segment lengths.
• Factor and relate Δy/Δx to the derivative; take limits to obtain the arc length integral.
• Example with a linear function confirms the method and yields the same result via both the integral and the endpoint-distance approach.
• The arc length formula generalizes to any differentiable curve and is a foundational tool in physics, astronomy, and geometry.