From Polynomial Coordinates to Arc Length
A polynomial parametrization describes a curve by giving polynomial formulas for its coordinates as functions of a parameter. For example, \(x=t\) and \(y=t^2\) trace a parabola as \(t\) varies. To find the length of the traced curve over a specified parameter interval, the key step is to calculate how quickly the point moves along it.
If the coordinate functions are \(x(t)\) and \(y(t)\), then their derivatives \(x'(t)\) and \(y'(t)\) are the velocity components. The speed is the magnitude of that velocity. Arc length is the accumulated speed over the parameter interval. Polynomial coordinates make the derivatives straightforward, but the square root in the speed can still lead to integrals that require careful simplification.
The expression under the square root is a sum of squares, so the speed is nonnegative. That fact matters: when a simplification produces \(\sqrt{g(t)^2}\), the result is \(|g(t)|\), not automatically \(g(t)\). On an interval where \(g(t)\geq 0\), the absolute value can be removed; on an interval where \(g(t)\) changes sign, the integral may need to be split.
A Parabola with an Elementary Arc-Length Integral
Consider the standard parametrization \(x=t,\ y=t^2\). It traces the parabola \(y=x^2\), since \(x=t\) and thus \(y=x^2\). Its speed involves \(\sqrt{1+4t^2}\), which is not a polynomial, but it has an elementary antiderivative. This example illustrates why finding the speed is only the setup: the remaining integral must still be evaluated correctly.
Worked Example: Length of \(x=t,\ y=t^2\) from \(t=0\) to \(t=1\)
Differentiate the coordinate functions: $$ x'(t)=1,\qquad y'(t)=2t. $$ Therefore the speed is $$ \sqrt{(1)^2+(2t)^2}=\sqrt{1+4t^2}. $$ The arc length is $$ L=\int_0^1\sqrt{1+4t^2}\,dt. $$ An antiderivative on this interval is $$ F(t)=\frac{t}{2}\sqrt{1+4t^2} +\frac14\ln\!\left(2t+\sqrt{1+4t^2}\right). $$ To check it, differentiate the first term: $$ \frac{d}{dt}\left(\frac{t}{2}\sqrt{1+4t^2}\right) =\frac12\sqrt{1+4t^2}+\frac{2t^2}{\sqrt{1+4t^2}}. $$ The derivative of the logarithm term is $$ \frac{d}{dt}\left(\frac14\ln\!\left(2t+\sqrt{1+4t^2}\right)\right) =\frac{1}{2\sqrt{1+4t^2}}. $$ Adding these derivatives gives \(\sqrt{1+4t^2}\), as required. Evaluating at the endpoints, $$ F(1)=\frac{\sqrt5}{2}+\frac14\ln(2+\sqrt5), \qquad F(0)=0. $$ Thus $$ L=\frac{\sqrt5}{2}+\frac14\ln(2+\sqrt5). $$ The result is positive, as a length must be. Differentiating the antiderivative recovers the speed, confirming the calculation.
Why the Speed Is the Integrand
The arc-length formula uses speed because speed measures distance traveled per unit change in the parameter. Over a short parameter change \(\Delta t\), the coordinate changes are approximately \(x'(t)\Delta t\) and \(y'(t)\Delta t\). The straight-line distance between the two nearby points is therefore approximately $$ \sqrt{\bigl(x'(t)\Delta t\bigr)^2+\bigl(y'(t)\Delta t\bigr)^2} = \sqrt{\bigl(x'(t)\bigr)^2+\bigl(y'(t)\bigr)^2}\,|\Delta t|. $$ For increasing parameter values, \(|\Delta t|=\Delta t\). Adding these short distances over the interval and taking the limit gives the integral of speed. This is why the coordinates’ derivatives enter as a sum of squares under a square root: they are the components of the velocity vector, whose magnitude is the speed.
For polynomial parametrizations, the main opportunity is to simplify that sum of squares. Sometimes the result remains a square root such as \(\sqrt{1+4t^2}\), which can still be integrated. In other examples, the sum is itself a perfect square, and the speed reduces to a polynomial on the interval.
A Polynomial Identity That Makes Speed Simple
The next example uses a polynomial identity rather than an inverse-trigonometric or logarithmic antiderivative. The identity is an instance of the algebraic relation $$ (1-t^2)^2+(2t)^2=(1+t^2)^2. $$ Because \(1+t^2\) is positive for every real \(t\), its square root is exactly \(1+t^2\). The speed therefore becomes a polynomial, and the length integral is especially direct.
Worked Example: A Polynomial Parametrization with Polynomial Speed
Let $$ x(t)=t-\frac{t^3}{3},\qquad y(t)=t^2,\qquad -1\leq t\leq 1. $$ Differentiate: $$ x'(t)=1-t^2,\qquad y'(t)=2t. $$ The speed is $$ \begin{aligned} \sqrt{(1-t^2)^2+(2t)^2} &=\sqrt{1-2t^2+t^4+4t^2}\\ &=\sqrt{1+2t^2+t^4}\\ &=\sqrt{(1+t^2)^2}\\ &=1+t^2. \end{aligned} $$ The last equality holds because \(1+t^2>0\) throughout the interval. Thus $$ L=\int_{-1}^{1}(1+t^2)\,dt =\left[t+\frac{t^3}{3}\right]_{-1}^{1}. $$ At the upper endpoint the antiderivative is \(1+\frac13=\frac43\). At the lower endpoint it is \(-1-\frac13=-\frac43\). Hence $$ L=\frac43-\left(-\frac43\right)=\frac83. $$ As a check, differentiating \(t+\frac{t^3}{3}\) gives \(1+t^2\), exactly the speed. The positive result is also consistent with arc length being a distance.
A Second Parabolic Example
A parabola need not be parametrized with \(x=t\). For \(x=t^2,\ y=2t\), eliminating \(t\) gives \(x=y^2/4\), another parabola. The coordinate derivatives are different, but the speed still simplifies to an elementary integrand.
Worked Example: Length of \(x=t^2,\ y=2t\) from \(t=0\) to \(t=1\)
Differentiate: $$ x'(t)=2t,\qquad y'(t)=2. $$ Then $$ \sqrt{(2t)^2+2^2}=2\sqrt{1+t^2}, $$ where the positive factor \(2\) is taken outside the square root. Therefore $$ L=\int_0^1 2\sqrt{1+t^2}\,dt. $$ An antiderivative is $$ G(t)=t\sqrt{1+t^2}+\ln\!\left(t+\sqrt{1+t^2}\right). $$ To verify it, $$ \frac{d}{dt}\left(t\sqrt{1+t^2}\right) =\sqrt{1+t^2}+\frac{t^2}{\sqrt{1+t^2}}, $$ and $$ \frac{d}{dt}\ln\!\left(t+\sqrt{1+t^2}\right) =\frac{1}{\sqrt{1+t^2}}. $$ Their sum is $$ \frac{1+2t^2+1}{\sqrt{1+t^2}} =2\sqrt{1+t^2}. $$ Now evaluate: $$ G(1)=\sqrt2+\ln(1+\sqrt2),\qquad G(0)=0. $$ Thus $$ L=\sqrt2+\ln(1+\sqrt2). $$ The derivative check confirms the antiderivative matches the speed, and the endpoint subtraction gives a positive length.
Absolute Values Cannot Be Ignored
A common error is to take the square root of a square without considering the sign. For example, \(\sqrt{t^2}=|t|\), not \(t\) for all real \(t\). This distinction can affect an arc length calculation when the parameter interval crosses zero.
For a concrete correction, take \(x=t^2,\ y=\frac23t^3\) on \([-1,1]\). Then \(x'=2t\), \(y'=2t^2\), and $$ \sqrt{(2t)^2+(2t^2)^2}=2|t|\sqrt{1+t^2}. $$ A wrong calculation would replace \(2|t|\sqrt{1+t^2}\) by \(2t\sqrt{1+t^2}\) over the entire interval. That expression is negative for \(t<0\), which cannot be speed. Instead split at zero: $$ L=2\int_{-1}^{0}(-t)\sqrt{1+t^2}\,dt +2\int_0^1t\sqrt{1+t^2}\,dt. $$ Using \(\frac{d}{dt}(1+t^2)^{3/2}=3t\sqrt{1+t^2}\), the second integral is $$ 2\int_0^1t\sqrt{1+t^2}\,dt =\frac23\left[(1+t^2)^{3/2}\right]_0^1 =\frac23(2\sqrt2-1). $$ The first integral has the same value, by the substitution \(u=-t\). Consequently, $$ L=\frac43(2\sqrt2-1). $$ The integrand is nonnegative on both pieces, as speed must be. This example shows why a sign check is part of simplification, not an optional detail.
When Polynomial Coordinates Do Not Mean a Polynomial Integral
Even when both coordinate functions are polynomials, the speed is generally the square root of a polynomial: $$ \sqrt{\bigl(x'(t)\bigr)^2+\bigl(y'(t)\bigr)^2}. $$ That square root may or may not simplify to a polynomial or have an elementary antiderivative. The examples above were chosen because their integrals can be evaluated exactly. One should not assume every polynomial parametrization has a simple exact arc-length formula. In a problem where the speed integral has no elementary antiderivative, an exact elementary evaluation may not be available; numerical arc length is a separate approach.
A reliable workflow keeps the geometric and algebraic steps distinct. First identify the parameter interval that traces the requested part of the curve. Next differentiate both coordinate functions and form the nonnegative speed. Then simplify the square root, checking signs whenever an absolute value could appear. Finally integrate over the stated interval and verify the antiderivative by differentiating it. This process works for parabolas and for more general polynomial parametrizations.
Check Your Understanding
For each question, begin by finding the velocity components and then form the speed.
- For \(x=t,\ y=3t^2\) on \(0\leq t\leq 1\), what definite integral gives the arc length?
- For \(x=t-\frac{t^3}{3},\ y=t^2\), why can the square root in the speed be replaced by \(1+t^2\) on \([-1,1]\)?
- What is the correct value of \(\sqrt{(3t)^2}\) when \(t<0\), and why?
- For \(x=t^2,\ y=2t\) on \([0,1]\), what are the two velocity components and the speed?
- Why does polynomial form of \(x(t)\) and \(y(t)\) not guarantee that the speed is a polynomial?
