Comparing Graph and Parametric Length Setups

One Curve, Two Length Setups

A curve can often be described either as a graph \(y=f(x)\) or parametrically as \(x=x(t)\), \(y=y(t)\). Each description gives a valid way to set up its length, but the resulting integrals may differ in how simple they look, which endpoints they include, and how readily they can be evaluated.

The goal is not to prefer one formula in every situation. It is to identify which description represents the part of the curve accurately and produces the most useful integral. The limits must match the variable of integration, and any endpoint behavior must be understood.

Key idea: For the same curve segment, graph and parametric length integrals give the same length when their domains cover that segment exactly. Their usefulness can nevertheless differ substantially.

The Two Formulas and Their Domains

If a curve is the graph of a continuously differentiable function \(y=f(x)\) on \([a,b]\), its length is

L=\int_a^b\sqrt{1+\left(f'(x)\right)^2}\,dx.

If a curve is described by continuously differentiable coordinate functions \(x=x(t)\) and \(y=y(t)\) for \(t\in[\alpha,\beta]\), its length is

L=\int_\alpha^\beta \sqrt{\left(x'(t)\right)^2+\left(y'(t)\right)^2}\,dt.

For piecewise smooth curves, the length is computed by applying the appropriate formula on each smooth piece and adding the results. The graph formula assumes that the relevant piece is represented as a single-valued function of \(x\). A parametric description does not require this: it can represent curves that turn back in \(x\), including many curves that are not one graph over the entire segment.

In the graph formula, the limits \(a,b\) are \(x\)-coordinates. In the parametric formula, the limits \(\alpha,\beta\) are parameter values. They cannot be interchanged just because both describe the endpoints of the same curve.

Why the Formulas Agree

Suppose a graph segment is parametrized by \(x=x(t)\) and \(y=f(x(t))\), with \(x(t)\) increasing from \(a\) to \(b\). By the chain rule, \(y'(t)=f'(x(t))x'(t)\). Substituting this into the parametric speed gives

\sqrt{\left(x'(t)\right)^2+\left(y'(t)\right)^2} = \sqrt{\left(x'(t)\right)^2\left(1+\left(f'(x(t))\right)^2\right)} = x'(t)\sqrt{1+\left(f'(x(t))\right)^2},

where the last equality uses \(x'(t)\geq 0\). Thus the parametric length is

\int_\alpha^\beta \sqrt{1+\left(f'(x(t))\right)^2}\,x'(t)\,dt = \int_a^b\sqrt{1+\left(f'(x)\right)^2}\,dx.

The final equality is the change of variable \(x=x(t)\). If \(x(t)\) decreases, its derivative has the opposite sign, so the oriented parameter limits must be handled accordingly; the length itself remains nonnegative. If \(x(t)\) reverses direction, split the curve into monotone pieces or use the parametric formula directly.

Worked Example: A Parabola with the Same Integral Either Way

Worked Example: Comparing Two Setups for \(y=x^2\)

Find the length of the arc \(y=x^2\) from \(x=0\) to \(x=1\), first as a graph and then with \(x=t,\ y=t^2\).

Graph setup. Since \(f'(x)=2x\), the length is

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

An antiderivative is \(\frac{x}{2}\sqrt{1+4x^2}+\frac14\operatorname{asinh}(2x)\). Its derivative is

\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}.

Therefore

L= \left[\frac{x}{2}\sqrt{1+4x^2} +\frac14\operatorname{asinh}(2x)\right]_0^1 = \frac{\sqrt5}{2}+\frac14\operatorname{asinh}(2) = \frac{\sqrt5}{2}+\frac14\ln(2+\sqrt5).

Parametric setup. For \(x=t,\ y=t^2\), the matching parameter limits are \(t=0\) to \(t=1\). Since \(x'(t)=1\) and \(y'(t)=2t\),

L=\int_0^1\sqrt{1+4t^2}\,dt =\frac{\sqrt5}{2}+\frac14\ln(2+\sqrt5).

The two integrals are identical after renaming the integration variable. The graph form is direct; the parameterization adds no simplification here. As a check, differentiating the displayed antiderivative gives \(\sqrt{1+4x^2}\), the required integrand, and both evaluations use the same endpoints.

Worked Example: A Graph Endpoint Singularity, a Simple Parametric Speed

Worked Example: The Upper Semicircle

Compare the graph and parametric setups for the upper semicircle of radius \(1\). As a graph, it is \(y=\sqrt{1-x^2}\) for \(-1\leq x\leq 1\).

For \(-1<x<1\),

y'=-\frac{x}{\sqrt{1-x^2}}, \qquad \sqrt{1+(y')^2} = \sqrt{1+\frac{x^2}{1-x^2}} = \frac{1}{\sqrt{1-x^2}}.

The graph integral is improper at both endpoints, because its integrand is unbounded there. Its value is obtained by the substitution \(x=\sin u\), with \(-\frac{\pi}{2}<u<\frac{\pi}{2}\). On that interval, \(\cos u>0\), so \(dx=\cos u\,du\) and \(\sqrt{1-\sin^2u}=\cos u\). Consequently,

L=\int_{-1}^1\frac{dx}{\sqrt{1-x^2}} =\int_{-\pi/2}^{\pi/2}\frac{\cos u}{\cos u}\,du =\pi.

As a parametric description, take \(x=\cos t,\ y=\sin t\) for \(0\leq t\leq\pi\). Then \(x'=-\sin t\) and \(y'=\cos t\), so the speed is \(\sqrt{\sin^2t+\cos^2t}=1\). Hence

L=\int_0^\pi 1\,dt=\pi.

The parametric speed is constant and its endpoints cause no singularity in the integrand, so it is a simpler setup for evaluation. The graph description remains valid, but its length integral needs improper-integral handling. Both results check against the same semicircle length, \(\pi\).

Worked Example: Different Algebra, Equal Exact Length

Worked Example: A Semicubical Parabola Arc

Consider the arc \(y=x^{3/2}\) for \(0\leq x\leq1\). A convenient parametrization is \(x=t^2,\ y=t^3\), for \(0\leq t\leq1\). Compare the two setups and evaluate them.

For the graph, \(y'=\frac32\sqrt{x}\), giving

L=\int_0^1\sqrt{1+\frac94x}\,dx.

An antiderivative is \(\frac{8}{27}(1+\frac94x)^{3/2}\), since its derivative is \(\frac{8}{27}\cdot\frac32\cdot\frac94(1+\frac94x)^{1/2}=\sqrt{1+\frac94x}\). Thus

L=\frac{8}{27}\left[\left(\frac{13}{4}\right)^{3/2}-1\right] =\frac{13\sqrt{13}-8}{27}.

For the parametrization, \(x'=2t\) and \(y'=3t^2\), so

L=\int_0^1\sqrt{4t^2+9t^4}\,dt =\int_0^1 t\sqrt{4+9t^2}\,dt.

The last equality uses \(t\geq0\). With \(u=4+9t^2\), \(du=18t\,dt\), and the limits become \(u=4\) and \(u=13\). Therefore

L=\frac{1}{18}\int_4^{13}u^{1/2}\,du =\frac{1}{27}\left[u^{3/2}\right]_4^{13} =\frac{13\sqrt{13}-8}{27}.

The graph integrand has the simpler form, while the parameterized speed factors as \(t\sqrt{4+9t^2}\) and is also straightforward to integrate by substitution. The parameterization has zero speed at \(t=0\), but the length integral is still valid for this smooth parameterized arc. The matching exact values verify that the descriptions cover the same segment.

Choosing Between the Setups

When both descriptions are available, compare not only the appearance of the integrands but also their domains and endpoint behavior.

Feature	Graph setup	Parametric setup
Integration variable	The coordinate \(x\)	The parameter \(t\)
Best suited to	A segment that is one graph \(y=f(x)\)	A segment naturally described by coordinate functions
Potential difficulty	Large or unbounded \(f'(x)\); graph may not cover a turning curve	Speed may involve complicated derivatives or radicals
Possible advantage	Short derivative and simple graph integrand	Simple speed, convenient parameter interval, or coverage beyond one graph

A useful comparison follows these checks:

Confirm the curve segment.
Match the endpoints and ensure each description traces the intended portion exactly once, unless repeated tracing is intended.

Write the correct limits.
Use \(x\)-coordinates for a graph integral and parameter values for a parametric integral.

Compare the integrands.
Look for simpler derivatives, factorizations, substitutions, or endpoint behavior.

Choose a workable form.
A setup with a shorter-looking integrand is not automatically better if its domain or endpoint behavior makes evaluation harder.

A Common Variable-Confusion Error

A frequent mistake is to use the derivative of a graph as though it were the derivative with respect to a parameter. For \(x=t^2,\ y=t^3\), the graph derivative is \(dy/dx=\frac32\sqrt{x}\). Substituting \(x=t^2\) gives \(\frac32t\), but that is still \(dy/dx\), not \(dy/dt\). Using \(\sqrt{1+(\frac32t)^2}\) as the parametric speed is incorrect.

For instance, at \(t=\frac12\), that mistaken expression is \(\sqrt{1+\frac{9}{16}}=\frac54\), whereas the actual parametric speed is

\sqrt{(2t)^2+(3t^2)^2}\bigg|_{t=1/2} = \sqrt{1+\frac{9}{16}} = \frac54.

Here the numerical values happen to agree because the mistaken calculation has accidentally used the correct speed value at this parameter? It has not: the graph derivative at \(x=t^2\) is \(\frac32t\), which at \(t=\frac12\) is \(\frac34\), giving \(\sqrt{1+\frac{9}{16}}=\frac54\); this equality alone does not make it a speed formula. At \(t=1\), the mistaken value is \(\frac{\sqrt{13}}{2}\), while the actual speed is \(\sqrt{13}\). The error is the missing factor \(dx/dt\).

The correct relation is \(dy/dt=(dy/dx)(dx/dt)\). A graph length integrand can be converted to parameter form only with the corresponding change in \(x\): when \(x'(t)\geq0\), the parametric speed is \(\sqrt{1+(dy/dx)^2}\,x'(t)\). Keeping both the derivative and the integration variable consistent prevents this mistake.

Why Comparing Setups Matters

The same geometric length can appear as an elementary integral in one description and an awkward one in another. A graph may expose a simple derivative, while a parametrization may remove a graph endpoint singularity or represent a curve that reverses direction in \(x\). Comparing the domains also catches errors that algebra alone cannot reveal: an integral can be computed correctly and still measure the wrong arc if its limits trace a different segment.

Before evaluating, check that the integrand is the appropriate speed, the variable matches the limits, and the chosen description covers exactly the desired curve segment. These checks make the comparison useful whether the final integral is evaluated exactly or by numerical methods.

Check Your Understanding

For each question, identify both the relevant formula and the domain issue involved.

For the graph \(y=f(x)\), what variable determines the limits of the graph length integral?
For \(x=t^2,\ y=t^3\), write the parametric speed and state why \(dy/dx\) alone is not that speed.
Why is the graph length integral for the upper unit semicircle improper, and why is its length nevertheless finite?
Give one reason the parametric description of the upper semicircle is more convenient for evaluation.
If \(x(t)\) reverses direction on an interval, what must be considered before using the graph formula on the whole curve?