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Arc Length and Curve Length · Tutorial 418 of 998

Common Setup Errors in Arc-Length Problems

Use the correct arc-length formula and bounds by checking how the curve is described, differentiated, and traversed.

Intermediate 10 min read

What You'll Learn

  • Distinguish graph and parametric arc-length formulas
  • Check derivatives before building the integrand
  • Match integration bounds to the chosen variable
  • Preserve absolute values when simplifying square roots
  • Verify a setup by relating it to the curve’s motion

Why Arc-Length Setups Go Wrong

Arc-length calculations can become difficult because the integrand is complicated. But many errors occur before integration begins: the square root is omitted, the derivative is wrong, or the bounds belong to a different variable than the one being integrated. These mistakes can produce a neat-looking calculation for the wrong quantity.

The safest approach is to identify how the curve is described, write the matching formula, and check each part of the setup before evaluating the integral. This tutorial focuses on those checks rather than on methods for evaluating difficult arc-length integrals.

Graph formula: If a curve is the graph \(y=f(x)\) for \(a\le x\le b\), and \(f\) is continuously differentiable on that interval, its length is \(L=\int_a^b\sqrt{1+[f'(x)]^2}\,dx\).
Parametric formula: If a curve is given by \(\mathbf r(t)=(x(t),y(t))\) for \(\alpha\le t\le\beta\), where \(x\) and \(y\) are piecewise continuously differentiable, its length, counted along the stated traversal, is \(L=\int_\alpha^\beta\sqrt{[x'(t)]^2+[y'(t)]^2}\,dt\).

These formulas are related, but their variables and bounds have different meanings. For a graph, the independent variable is \(x\). For a parametrization, the variable is \(t\), and \(x(t)\) tells us which horizontal coordinate the moving point has at time \(t\).

Why the Square Root Is Essential

For a graph, a small horizontal change \(dx\) and vertical change \(dy\) form the legs of a right triangle. The corresponding small segment of the curve has length \(ds\), so the Pythagorean theorem gives \(ds^2=dx^2+dy^2\). Since \(dy=f'(x)\,dx\), for an increasing horizontal coordinate the length element is \(ds=\sqrt{1+[f'(x)]^2}\,dx\). Adding these elements along the interval gives the graph formula.

For a parametrized curve, the coordinate changes over a small parameter increment \(dt\) are approximately \(x'(t)\,dt\) and \(y'(t)\,dt\). The corresponding speed, or distance travelled per unit parameter, is therefore \(\sqrt{[x'(t)]^2+[y'(t)]^2}\). The length formula integrates this speed over the parameter interval.

In either formula, the square root comes from combining perpendicular coordinate changes with the Pythagorean theorem. Adding the squared derivatives without taking the square root does not give a length element.

Worked Example: The Missing Square Root

Worked Example: The Missing Square Root

Find the length of the graph \(y=x^2\) from \(x=0\) to \(x=1\), and identify the error in the setup that integrates \(1+4x^2\).

First differentiate the graph:

$$ f(x)=x^2,\qquad f'(x)=2x. $$

Use the graph formula. The squared derivative is \(4x^2\), but it belongs inside the square root:

$$ L=\int_0^1\sqrt{1+(2x)^2}\,dx =\int_0^1\sqrt{1+4x^2}\,dx. $$

An antiderivative is \(F(x)=\frac{x}{2}\sqrt{1+4x^2}+\frac14\operatorname{asinh}(2x)\), where \(\operatorname{asinh}(u)=\ln(u+\sqrt{1+u^2})\). To check it, differentiate:

$$ \begin{aligned} F'(x) &=\frac12\sqrt{1+4x^2} +\frac{2x^2}{\sqrt{1+4x^2}} +\frac{1}{2\sqrt{1+4x^2}}\\ &=\frac{1+4x^2}{\sqrt{1+4x^2}} =\sqrt{1+4x^2}. \end{aligned} $$

Now evaluate at the bounds:

$$ \begin{aligned} L &=F(1)-F(0)\\ &=\frac{\sqrt{5}}{2}+\frac14\operatorname{asinh}(2). \end{aligned} $$

The incorrect setup \(\int_0^1(1+4x^2)\,dx\) gives \(1+\frac43=\frac73\). Its integrand is not the curve’s length element: it omits the square root. The corrected answer has been checked by differentiating the antiderivative and substituting both bounds.

Check the Derivative Before Building the Integrand

A correct arc-length formula cannot compensate for an incorrect derivative. Differentiate the function as written, including any constant factors, before squaring the result. Squaring a mistaken derivative can hide the original mistake because the final integrand may still look plausible.

Worked Example: A Derivative Factor Error

Find the length of \(y=\frac{x^2}{4}\) on \(0\le x\le2\). A common error is to differentiate it as \(x\), forgetting the factor \(\frac14\).

Differentiate carefully using the constant multiple rule:

$$ f'(x)=\frac14(2x)=\frac{x}{2}. $$

Substitute that derivative in the graph formula:

$$ L=\int_0^2\sqrt{1+\left(\frac{x}{2}\right)^2}\,dx =\int_0^2\sqrt{1+\frac{x^2}{4}}\,dx. $$

An antiderivative is \(G(x)=\frac{x}{2}\sqrt{1+\frac{x^2}{4}}+\operatorname{asinh}(x/2)\). Its derivative is:

$$ \begin{aligned} G'(x) &=\frac12\sqrt{1+\frac{x^2}{4}} +\frac{x^2}{8\sqrt{1+x^2/4}} +\frac{1}{2\sqrt{1+x^2/4}}\\ &=\frac{1+x^2/4}{\sqrt{1+x^2/4}} =\sqrt{1+\frac{x^2}{4}}. \end{aligned} $$

Thus, evaluating at \(0\) and \(2\) gives:

$$ L=G(2)-G(0)=\sqrt{2}+\operatorname{asinh}(1). $$

If the derivative were incorrectly taken to be \(x\), the resulting integral would be \(\int_0^2\sqrt{1+x^2}\,dx\), a different length calculation. The derivative check prevents that error: the derivative of \(x^2/4\) is \(x/2\), not \(x\). Differentiating \(G\) and substituting the bounds verifies the corrected calculation.

Match the Bounds to the Integration Variable

The limits in a definite integral must be values of the variable of integration. If the curve is parametrized by \(t\), the integral’s limits are parameter values. If it is written as a graph \(y=f(x)\), the limits are horizontal coordinates. The same geometric endpoints can have different numerical coordinates in these two descriptions.

Worked Example: Converting Bounds Between a Graph and a Parameter

Consider \(\mathbf r(t)=(t^2,t^3)\) for \(1\le t\le2\). Find its length, then confirm that the graph description uses different bounds.

In the parametrization, differentiate both coordinates and form the speed:

$$ x'(t)=2t,\qquad y'(t)=3t^2,\qquad \sqrt{[x'(t)]^2+[y'(t)]^2} =\sqrt{4t^2+9t^4}. $$

Since \(t\ge1\), we have \(t>0\), so \(\sqrt{4t^2+9t^4}=t\sqrt{4+9t^2}\). The length is therefore:

$$ \begin{aligned} L &=\int_1^2 t\sqrt{4+9t^2}\,dt\\ &=\left.\frac{(4+9t^2)^{3/2}}{27}\right|_1^2\\ &=\frac{40^{3/2}-13^{3/2}}{27} =\frac{80\sqrt{10}-13\sqrt{13}}{27}. \end{aligned} $$

For the graph description, \(x=t^2\), and \(t\) is positive on the given interval. Thus \(t=\sqrt{x}\), \(y=t^3=x^{3/2}\), and the parameter bounds \(t=1\) and \(t=2\) correspond to \(x=1\) and \(x=4\). The graph and its derivative are:

$$ y=x^{3/2},\qquad \frac{dy}{dx}=\frac32\sqrt{x}. $$

Using the graph formula with the corresponding \(x\)-bounds gives:

$$ \begin{aligned} L &=\int_1^4\sqrt{1+\frac94x}\,dx\\ &=\left.\frac{8}{27}\left(1+\frac94x\right)^{3/2}\right|_1^4\\ &=\frac{80\sqrt{10}-13\sqrt{13}}{27}. \end{aligned} $$

Both descriptions give the same length. If using the graph formula, the bounds \(1\) and \(2\) would be incorrect: those are parameter values, not the graph’s \(x\)-coordinates. The conversion \(x=t^2\) sends the parameter interval \([1,2]\) to the graph interval \([1,4]\). Matching the bounds this way checks that the two computations describe the same segment.

Keep Absolute Values When Simplifying

A square root is nonnegative. In particular, \(\sqrt{u^2}=|u|\), not \(u\) in general. Replacing \(\sqrt{(2t)^2}\) by \(2t\) is valid only where \(t\ge0\). When a parameter interval includes negative values, the sign must be handled on the appropriate subintervals.

Worked Example: Losing the Sign of the Parameter

Find the length of \(\mathbf r(t)=(t^2,0)\) for \(-1\le t\le2\). The curve travels left from \(x=1\) to \(x=0\), then right from \(x=0\) to \(x=4\).

Differentiate and calculate the speed:

$$ \mathbf r'(t)=(2t,0),\qquad \|\mathbf r'(t)\|=\sqrt{(2t)^2}=2|t|. $$

The sign changes at \(t=0\), so split the integral there. On \([-1,0]\), \(|t|=-t\); on \([0,2]\), \(|t|=t\):

$$ \begin{aligned} L &=\int_{-1}^0(-2t)\,dt+\int_0^2 2t\,dt\\ &=\left[-t^2\right]_{-1}^0+\left[t^2\right]_0^2\\ &=1+4=5. \end{aligned} $$

If one incorrectly replaces \(2|t|\) by \(2t\) across the entire interval, the result is \(\int_{-1}^2 2t\,dt=[t^2]_{-1}^2=3\). That is wrong because the speed cannot be negative. The corrected piecewise calculation gives positive lengths \(1\) and \(4\), matching the two distances travelled along the line.

A Practical Setup Check

Before evaluating an arc-length integral, check the following items in order. Each one catches a distinct setup error.

1
Identify the description.
For \(y=f(x)\), use the graph formula and integrate with respect to \(x\). For \(\mathbf r(t)=(x(t),y(t))\), use the parametric formula and integrate with respect to \(t\).
2
Differentiate each coordinate or function.
Check constants, powers, and variables before substituting derivatives into the formula.
3
Keep the square root.
For a graph, the integrand is \(\sqrt{1+[f'(x)]^2}\). For a parametrization, it is \(\sqrt{[x'(t)]^2+[y'(t)]^2}\).
4
Match bounds to the variable.
Use \(x\)-coordinates for a graph integral and parameter values for a parametric integral. Convert endpoints when changing descriptions.
5
Check signs and traversal.
Use \(\sqrt{u^2}=|u|\). If the parameter interval crosses a point where the sign changes, split the integral or retain the absolute value.

These checks matter because an arc-length integral represents accumulated distance, not signed displacement. A negative velocity in one coordinate is allowed; a negative speed is not. Likewise, a correctly evaluated integral still answers the wrong question if its derivatives, variable, or bounds were set up incorrectly.

Check Your Understanding

For each question, identify the setup issue or state the correct expression before attempting any integration.

  1. For the graph \(y=3x^2\) on \(0\le x\le1\), what derivative and length integrand should be used?
  2. Why is \(\sqrt{(5t)^2}\) equal to \(5|t|\), and when can it be replaced by \(5t\)?
  3. A curve is parametrized for \(0\le t\le3\) by \(x(t)=2t\). What are the corresponding \(x\)-bounds for a graph integral over the same interval?
  4. For \(\mathbf r(t)=(\cos t,\sin t)\), write the speed integrand. Which derivatives must be squared?
  5. What is wrong with using \(\int_a^b(1+[f'(x)]^2)\,dx\) as the graph-length formula?

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