Why the Choice of Parameter Matters
A curve can have more than one description. A graph \(y=f(x)\) may also be written parametrically as \(x=x(t)\), \(y=y(t)\). Both descriptions trace the same geometric curve, but they need not make its length equally easy to calculate.
The parameter is not part of the curve itself: it tells us how the curve is traversed. A good choice can simplify derivatives, turn a difficult interval into a convenient one, or make the square root in the length formula easier to integrate. The goal is not to find a parameter that is always best, but to compare valid descriptions of the same segment and use the one that leads to the clearest calculation.
These formulas presume that the specified interval traces the curve segment being measured. If a parametrization retraces part of the curve, its integral counts that travel again. If the graph derivative or parametric speed is unbounded at an endpoint, the length integral is interpreted as an improper integral, provided the corresponding limit exists.
How a Change of Parameter Preserves Length
Suppose a curve has a continuously differentiable parametrization \(\mathbf r(t)\), \(a\le t\le b\). Let \(t=\phi(u)\), where \(\phi\) is continuously differentiable and strictly increasing on \([\alpha,\beta]\), with \(\phi(\alpha)=a\) and \(\phi(\beta)=b\). The new parametrization is \(\mathbf r(\phi(u))\). The chain rule gives its velocity as \(\mathbf r'(\phi(u))\phi'(u)\). Since \(\phi\) is increasing, \(\phi'(u)\ge 0\), so its speed is \(\|\mathbf r'(\phi(u))\|\phi'(u)\). Therefore:
The equality is the change-of-variables formula for integrals, with \(t=\phi(u)\). Thus, a valid increasing change of parameter preserves length. A decreasing change also preserves length when the reversed limits are handled correctly: the speed contains \(|\phi'(u)|\). The parametrization must trace the intended segment once; a parameter that reverses direction partway through or retraces the curve can count additional travel.
This gives a practical test for a proposed parameter: check that its interval covers the intended segment, that it does not add unwanted retracing, and that the derivatives and resulting integral are manageable. The parameter that produces the shortest-looking expression is useful only if it still describes the right curve segment.
Worked Example: A Quarter Circle
Worked Example: Simplifying a Quarter-Circle Length
Find the length of the quarter-circle \(x^2+y^2=1\) in the first quadrant, from \((0,1)\) to \((1,0)\). Compare its graph and parametric descriptions.
As a graph, the segment is \(y=\sqrt{1-x^2}\), for \(0\le x\le1\). For \(0\le x<1\),
The graph length is improper at \(x=1\). Taking the endpoint limit gives
Now choose \(x=\sin t\), \(y=\cos t\), for \(0\le t\le\pi/2\). This interval traces from \((0,1)\) to \((1,0)\), as required. The derivatives are \(x'=\cos t\) and \(y'=-\sin t\), so
As a check, the parametric antiderivative is \(t\), whose derivative is \(1\), and its endpoint difference is \(\pi/2-0=\pi/2\). The unit-circle equation also holds throughout: \(\sin^2t+\cos^2t=1\). The parameter avoids the graph’s improper endpoint integrand and makes the speed constant.
Worked Example: A Power Graph with a Useful Substitution
Worked Example: Reparameterizing \(y=x^{3/2}\)
Find the length of \(y=x^{3/2}\) from \(x=1\) to \(x=4\). Compare the graph calculation with the parameter \(x=t^2\).
For the graph, \(y'=\frac{3}{2}\sqrt{x}\). Hence
An antiderivative is \(\frac{8}{27}(1+\frac{9x}{4})^{3/2}\): differentiating it gives \(\frac{8}{27}\cdot\frac{3}{2}\cdot\frac{9}{4}(1+\frac{9x}{4})^{1/2}=\sqrt{1+\frac{9x}{4}}\). Evaluating at the endpoints gives
For the parametric description, set \(x=t^2\). Since \(x\) ranges from \(1\) to \(4\), take \(1\le t\le2\); then \(y=x^{3/2}=t^3\). The derivatives and speed are
The last equality uses \(t>0\) on this interval. An antiderivative is \(\frac{(4+9t^2)^{3/2}}{27}\), since its derivative is \(t\sqrt{4+9t^2}\). Therefore
The result agrees with the graph calculation. The parameter change is valid because \(x=t^2\) is strictly increasing from \(1\) to \(4\), so it traces the same segment once. In this case it turns the speed integral into a direct substitution, although the graph integral is also manageable.
Worked Example: When the Graph Is Simpler
Worked Example: A Line Segment with Two Parameters
Find the length of the line segment \(y=2x+1\) from \(x=0\) to \(x=1\). Then check the answer using a nonlinear parameter.
For the graph, \(y'=2\), so the speed factor is constant:
For another description, let \(x=(t+t^2)/2\) and \(y=2x+1\), with \(0\le t\le1\). The function \(x(t)\) increases from \(0\) to \(1\), because \(x'=(1+2t)/2>0\) throughout the interval. The segment is therefore traversed once. The derivatives are \(x'=(1+2t)/2\) and \(y'=1+2t\), so the speed and length are
The parametric antiderivative is \(\frac{\sqrt{5}}{2}(t+t^2)\); its derivative is the displayed speed, and its endpoint difference is \(\frac{\sqrt{5}}{2}(2-0)=\sqrt{5}\). Both descriptions give the same length, but the graph form has the simpler constant integrand.
A Common Mistake: Adding Coordinate Derivatives
Parametric speed is not \(x'(t)+y'(t)\). It is the magnitude of the velocity vector, \(\sqrt{(x')^2+(y')^2}\). For the quarter circle above, an incorrect calculation would use \(\cos t-\sin t\) as the speed. That expression is not even always nonnegative, so it cannot represent distance travelled per unit parameter. The correct calculation is
The distinction matters whenever coordinate derivatives have different sizes or signs. The length formula uses their squares because it measures the magnitude of motion, not the signed sum of coordinate changes.
A Practical Choice of Parameter
When several descriptions are available, compare them before committing to an integral. A useful choice often comes from a relation already visible in the curve: a square root in \(x\) may suggest \(x=t^2\); trigonometric coordinates on a circle may suggest sine and cosine; or the graph variable may already give a constant or simple speed. Then verify the interval and the traced segment before simplifying the integrand.
Record its endpoints and the part of the curve whose length is wanted.
Write a graph formula when possible, and consider whether another parameter removes a square root or simplifies the derivatives.
Confirm that it traces the desired segment once, in either direction, without adding retraced travel.
Use the graph speed factor or the magnitude of the parametric velocity, then choose the integral that is easiest to evaluate reliably.
Choosing a parameter does not change the geometric length when the same segment is traced once. It changes the calculation used to obtain that length. This is especially useful when the graph description has a difficult endpoint or an awkward square root, but it is equally important to recognize when a change of parameter adds work rather than removing it.
Check Your Understanding
For each question, explain both the calculation and what the chosen parameter describes.
- For a parametric curve, why is speed \(\sqrt{(x')^2+(y')^2}\) rather than \(x'+y'\)?
- What conditions on an increasing change of parameter ensure that it preserves the length of the same curve segment?
- For \(y=x^{3/2}\) on \(1\le x\le4\), what interval for \(t\) corresponds to \(x=t^2\), and why?
- Why does the graph description of the line \(y=2x+1\) give a simpler length integral than the nonlinear parameter in the example?
- What can go wrong if a parameter retraces part of a curve?