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Parametric Arc Length and Motion · Tutorial 941 of 998

Polynomial Paths: An Arc-Length Workshop

Find the speed of polynomially parametrized paths, simplify the resulting integral, and evaluate lengths exactly on specified intervals.

Intermediate 9 min read

What You'll Learn

  • Compute speed from the derivatives of polynomial coordinate functions.
  • Set up arc length on a specified parameter interval.
  • Simplify speed when the sum of squared derivatives is a perfect square.
  • Evaluate a polynomial-path length using substitution.
  • Handle absolute values in speed when a derivative changes sign.

From Polynomial Coordinates to Length

A polynomial parametrization describes a path by giving each coordinate as a polynomial in a parameter \(t\). For example, \(x=t\) and \(y=t^2\) trace a parabola as \(t\) varies. The length of the path over a specified interval is not generally the difference between its endpoint coordinates: the path may curve, and its distance accumulates along the curve.

The practical method is to differentiate both coordinate functions, combine their derivatives to find speed, and integrate that speed over the parameter interval. Polynomial derivatives are usually straightforward; the main work is often simplifying and evaluating the resulting square root.

Definition: For a differentiable parametrized plane curve \(x=x(t)\), \(y=y(t)\), its speed at parameter value \(t\) is \(v(t)=\sqrt{(x'(t))^2+(y'(t))^2}\). If the curve is continuously differentiable on \([a,b]\), its arc length on that interval is \(L=\int_a^b v(t)\,dt\).

The speed is nonnegative. The bounds \(a\) and \(b\) specify which portion of the parametrized path is measured. In this workshop, we will use this formula to work through several polynomial paths, including cases where the square root becomes a perfect square and cases where it does not.

Why the Speed Integral Gives Length

Over a small parameter change \(\Delta t\), the position changes by approximately \((x'(t)\Delta t,y'(t)\Delta t)\). The length of this small displacement is approximately \(\sqrt{(x'(t)\Delta t)^2+(y'(t)\Delta t)^2}=v(t)|\Delta t|\). Adding these displacement lengths over a partition of \([a,b]\) gives an approximation to the curve’s length. As the partition is refined, the approximations lead to the integral of speed.

More explicitly, for a continuously differentiable path \(\mathbf r(t)=(x(t),y(t))\), the displacement over each subinterval \([t_{i-1},t_i]\) is \(\mathbf r(t_i)-\mathbf r(t_{i-1})\). By the fundamental theorem of calculus, this is \(\int_{t_{i-1}}^{t_i}\mathbf r'(t)\,dt\). Its norm is no greater than \(\int_{t_{i-1}}^{t_i}\|\mathbf r'(t)\|\,dt\). Summing over subintervals shows that polygonal approximations cannot exceed \(\int_a^b\|\mathbf r'(t)\|\,dt\); refining the partition recovers the curve length. Since \(\|\mathbf r'(t)\|=\sqrt{(x'(t))^2+(y'(t))^2}\), this is precisely the speed integral.

Key formula: For polynomial coordinate functions on a parameter interval \([a,b]\), $$ L=\int_a^b\sqrt{(x'(t))^2+(y'(t))^2}\,dt. $$ The square root must be taken before integrating, and the result is the length traced as \(t\) runs from \(a\) to \(b\).

Worked Example: The Parabola \(x=t,\ y=t^2\)

Worked Example: Finding the Length of a Parabola Segment

Find the arc length of \(x=t,\ y=t^2\) for \(0\le t\le1\).

First differentiate each coordinate:

$$ x'(t)=1,\qquad y'(t)=2t. $$

Thus the speed is

$$ v(t)=\sqrt{1^2+(2t)^2}=\sqrt{1+4t^2}. $$

Set up the length integral over the given interval:

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

To evaluate it, substitute \(u=2t\), so \(du=2\,dt\), \(dt=\frac12\,du\), and the bounds change from \(t=0,1\) to \(u=0,2\). Then

$$ L=\frac12\int_0^2\sqrt{1+u^2}\,du. $$

An antiderivative of \(\sqrt{1+u^2}\) is \(\frac12\bigl(u\sqrt{1+u^2}+\ln(u+\sqrt{1+u^2})\bigr)\). To check it, differentiate the expression inside the parentheses:

$$ \frac{d}{du}\left(u\sqrt{1+u^2}+\ln(u+\sqrt{1+u^2})\right) =\sqrt{1+u^2}+\frac{u^2}{\sqrt{1+u^2}}+\frac{1}{\sqrt{1+u^2}} =2\sqrt{1+u^2}. $$

Therefore, evaluating the integral gives

$$ \begin{aligned} L &=\frac14\left[u\sqrt{1+u^2}+\ln(u+\sqrt{1+u^2})\right]_0^2\\ &=\frac14\left(2\sqrt5+\ln(2+\sqrt5)-\ln(1)\right)\\ &=\frac{\sqrt5}{2}+\frac14\ln(2+\sqrt5). \end{aligned} $$

Since \(\ln(1)=0\), this is the exact length. Differentiating the antiderivative used for the original variable gives \(\frac{d}{dt}\left(\frac14\left(2t\sqrt{1+4t^2}+\ln(2t+\sqrt{1+4t^2})\right)\right)=\sqrt{1+4t^2}\), confirming the integrand and evaluation.

When the Speed Is a Perfect Square

For some polynomial paths, the expression under the square root is itself the square of a polynomial. Recognizing this can turn a radical integral into a polynomial integral. The identity must be checked, and the sign matters: \(\sqrt{q(t)^2}=|q(t)|\), not always \(q(t)\). If \(q(t)\ge0\) throughout the interval, then the speed simplifies to \(q(t)\) there.

Worked Example: A Polynomial Speed on a Curved Path

Find the arc length of \(x=t-\frac{t^3}{3}\), \(y=t^2\) for \(0\le t\le1\).

Differentiate the coordinates:

$$ x'(t)=1-t^2,\qquad y'(t)=2t. $$

Compute the sum of their squares, showing the terms before simplifying:

$$ \begin{aligned} (x'(t))^2+(y'(t))^2 &=(1-t^2)^2+(2t)^2\\ &=1-2t^2+t^4+4t^2\\ &=1+2t^2+t^4\\ &=(1+t^2)^2. \end{aligned} $$

On the entire interval, \(1+t^2>0\), so the speed is \(v(t)=1+t^2\). Thus

$$ L=\int_0^1(1+t^2)\,dt =\left[t+\frac{t^3}{3}\right]_0^1 =1+\frac13 =\frac43. $$

Checking the antiderivative, \(\frac{d}{dt}(t+t^3/3)=1+t^2\), which is the speed. Hence the length is \(\frac43\) units.

When Substitution Helps

Not every polynomial parametrization produces a perfect-square speed. A useful next step is to look for a factor in the integrand that is proportional to the derivative of the expression inside the square root. This can make a substitution effective even though the speed itself is not a polynomial.

Worked Example: A Substitution for a Polynomial Path

Find the arc length of \(x=t^2,\ y=t^3\) for \(0\le t\le1\).

Differentiate the coordinates and form the speed:

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

Because \(t\ge0\) on this interval, factor \(t^2\) out of the square root as \(t^2\), rather than as an unsigned \(t\):

$$ v(t)=t\sqrt{4+9t^2}. $$

The length is therefore \(L=\int_0^1t\sqrt{4+9t^2}\,dt\). Use \(u=4+9t^2\), so \(du=18t\,dt\), \(t\,dt=\frac1{18}\,du\), and the bounds become \(u=4\) and \(u=13\):

$$ \begin{aligned} L &=\frac1{18}\int_4^{13}\sqrt{u}\,du\\ &=\frac1{18}\left[\frac23u^{3/2}\right]_4^{13}\\ &=\frac1{27}\left(13^{3/2}-4^{3/2}\right)\\ &=\frac{13\sqrt{13}-8}{27}. \end{aligned} $$

To check, the corresponding antiderivative in \(t\) is \(\frac{(4+9t^2)^{3/2}-8}{27}\). Its derivative is \(\frac1{27}\cdot\frac32(4+9t^2)^{1/2}\cdot18t=t\sqrt{4+9t^2}\), the speed. The evaluated length is therefore \(\frac{13\sqrt{13}-8}{27}\) units.

A Sign Pitfall: Speed Is Never Negative

A common error is to replace the square root of a square by the expression inside the square without checking its sign. This can give a negative “length” on part of an interval. Consider the polynomial path \(x=t^2,\ y=0\) on \([-1,1]\).

Common mistake: Since \(x'(t)=2t\) and \(y'(t)=0\), one might write \(v(t)=\sqrt{(2t)^2}=2t\). This is wrong for \(t<0\), where \(2t<0\). The correct identity is \(\sqrt{(2t)^2}=|2t|=2|t|\).

Using the incorrect expression would give \(\int_{-1}^1 2t\,dt=0\), which cannot be the length of a path that moves from \(x=1\) to \(x=0\) and then back to \(x=1\). Correcting the speed gives

$$ L=\int_{-1}^1 2|t|\,dt =\int_{-1}^0(-2t)\,dt+\int_0^1 2t\,dt =1+1=2. $$

In general, factor squares from under a radical only after checking the parameter interval. For example, \(\sqrt{t^2}=|t|\); it equals \(t\) when \(t\ge0\), and \(-t\) when \(t\le0\).

A Reliable Workflow

For each polynomial path, keep the computation organized in this order:

1
Record the interval.
Identify the parameter bounds that determine the segment whose length is required.
2
Differentiate both coordinates.
Find \(x'(t)\) and \(y'(t)\) before combining them.
3
Build the speed.
Use \(v(t)=\sqrt{(x'(t))^2+(y'(t))^2}\), keeping the square root nonnegative.
4
Simplify with the interval in mind.
Check for a perfect square or a useful substitution, and preserve absolute values when signs can change.
5
Integrate and check.
Evaluate between the original bounds, or convert both bounds consistently after substitution; differentiate the antiderivative to verify the integrand.

Why These Examples Matter

The examples illustrate three different outcomes of the same arc-length method. For \(x=t,\ y=t^2\), speed leads to a radical integral and a logarithmic expression. In the second path, the sum of squared derivatives factors as a perfect square. In the third, a substitution reduces the radical integral to a power integral. The procedure is stable even when the algebra changes: calculate the speed first, then choose a valid way to integrate it.

Be careful not to confuse endpoint displacement with arc length. The displacement vector is \((x(b)-x(a),y(b)-y(a))\), while the arc length accumulates the nonnegative speed over the whole interval. A curved or reversing path can have positive length even when its endpoints coincide. For these polynomial examples, the key checks are that the interval is correct, the speed is nonnegative, and the final antiderivative differentiates back to that speed.

Check Your Understanding

Use the speed formula and the interval information carefully in each question.

  1. For \(x=t,\ y=3t^2\) on \(0\le t\le1\), write the speed and set up the arc-length integral. Which substitution is natural for evaluating it?
  2. For \(x=t-\frac{t^3}{3},\ y=t^2\), why is \(\sqrt{(1+t^2)^2}=1+t^2\) on \([0,1]\)?
  3. For \(x=t^2,\ y=t^3\) on \([0,1]\), identify the factor in the speed that makes \(u=4+9t^2\) useful.
  4. What is the correct value of \(\sqrt{(3t)^2}\) when \(t<0\), and why?
  5. How can differentiating a proposed antiderivative check an arc-length calculation?

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