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Arc Length and Curve Length · Tutorial 413 of 998

Arc Length of Polynomial Parametric Curves

Use polynomial coordinate functions to find a curve’s speed, simplify the length integrand, and evaluate exact arc lengths on specified intervals.

Intermediate 11 min read

What You'll Learn

  • State the arc-length formula for a continuously differentiable parametrization on a closed interval.
  • Differentiate polynomial coordinate functions to form the speed.
  • Simplify a speed expression by recognizing a perfect square.
  • Evaluate exact length integrals, including the length of a parabolic arc.
  • Check an antiderivative by differentiating it and verify the final endpoint calculation.
  • Recognize errors involving square roots and the difference between distance and arc length.

From Polynomial Coordinates to Curve Length

A curve can be described by giving each coordinate as a function of a parameter. When those coordinate functions are polynomials, their derivatives are also polynomials, so the speed is found by straightforward differentiation. The main challenge is then to simplify the square root in the length integral and evaluate the resulting integral correctly.

For a parametrization \(\mathbf r(t)=(x(t),y(t))\), the speed at parameter value \(t\) is the magnitude of the velocity vector \(\mathbf r'(t)\). Arc length adds up these speeds over the parameter interval. The integral may simplify to a polynomial, or it may require an exact antiderivative involving a logarithm.

Definition: If \(x(t)\) and \(y(t)\) are continuously differentiable on \([a,b]\), the length of the parametrized curve over that interval is $$ L=\int_a^b \sqrt{(x'(t))^2+(y'(t))^2}\,dt. $$ The integrand is the speed. The formula also applies when the velocity is zero at isolated points.

The direction in which the parameter traverses the curve does not make length negative: speed is a magnitude, so the integrand is nonnegative. In the examples below, the parameter interval is stated explicitly; changing that interval can change which portion of the curve is measured.

Why the Speed Integral Gives Length

The formula follows from approximating a curve by short chords. On a small parameter interval, the change in position is approximately velocity times the change in parameter. The length of that change is therefore approximately speed times the interval width. Adding these approximations and taking the limit gives the integral.

1
Partition the parameter interval.
Divide \([a,b]\) into subintervals and consider the curve’s displacement on each one.
2
Approximate each displacement.
For a subinterval of width \(\Delta t\) beginning at \(t_i\), continuous differentiability gives \(\mathbf r(t_i+\Delta t)-\mathbf r(t_i)\approx \mathbf r'(t_i)\Delta t\).
3
Take chord lengths and add.
The chord length is approximately \(\lVert\mathbf r'(t_i)\rVert\Delta t\), so the sum of chord lengths approximates a Riemann sum for the integral of speed.

For a precise error justification, write the displacement on one subinterval as \(\int_{t_i}^{t_i+\Delta t}\mathbf r'(u)\,du\). If \(\mathbf r'\) varies by at most \(\omega\) on that subinterval, its difference from \(\mathbf r'(t_i)\Delta t\) has magnitude at most \(\omega\Delta t\). Because \(\mathbf r'\) is uniformly continuous on the closed interval, the maximum such variation tends to zero as the partition is refined. The total difference between the chord sum and the Riemann sum for speed therefore tends to zero. The limiting length is \(\int_a^b\lVert\mathbf r'(t)\rVert\,dt\), which in two dimensions is the stated formula.

Worked Example: A Parabolic Arc

Worked Example: A Parabolic Arc

Find the exact length of the arc \(x=t,\ y=t^2\) for \(0\le t\le 1\). This traces the portion of the parabola \(y=x^2\) from \((0,0)\) to \((1,1)\).

Differentiate each coordinate:

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

Thus the speed is

$$ \sqrt{(x'(t))^2+(y'(t))^2} =\sqrt{1+4t^2}. $$

For \(t\ge 0\), an antiderivative 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 and the logarithmic term separately:

$$ \frac{d}{dt}\left(\frac{t}{2}\sqrt{1+4t^2}\right) =\frac12\sqrt{1+4t^2}+\frac{2t^2}{\sqrt{1+4t^2}}, $$ $$ \frac{d}{dt}\left(\frac14\ln(2t+\sqrt{1+4t^2})\right) =\frac{1}{2\sqrt{1+4t^2}}. $$

Adding these derivatives gives \(\sqrt{1+4t^2}\), as required. Evaluate at the endpoints:

$$ F(1)=\frac{\sqrt5}{2}+\frac14\ln(2+\sqrt5), \qquad F(0)=0+\frac14\ln(1)=0. $$

Therefore the arc length is

$$ L=F(1)-F(0) =\frac{\sqrt5}{2}+\frac14\ln(2+\sqrt5). $$

As a final check, the integrand used in the evaluation was obtained directly from the two coordinate derivatives, and \(F'(t)=\sqrt{1+4t^2}\). The endpoint subtraction is positive, as a length must be.

When the Speed Becomes a Perfect Square

A particularly convenient situation occurs when the expression under the square root is a perfect square. For polynomial parametrizations, expanding the squares can reveal this structure. The resulting speed still requires attention to sign: \(\sqrt{q(t)^2}=|q(t)|\), not automatically \(q(t)\). The sign of \(q(t)\) on the whole interval determines which expression to use.

Worked Example: A Polynomial Curve with a Simple Speed

Find the length of \(x=t-\frac{t^3}{3},\ y=t^2\) for \(0\le t\le 1\).

Differentiate:

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

Form the speed and expand the expression under the square root:

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

The radicand is \((1+t^2)^2\). Since \(1+t^2>0\), its nonnegative square root is \(1+t^2\). Hence

$$ L=\int_0^1(1+t^2)\,dt =\left[t+\frac{t^3}{3}\right]_0^1 =\left(1+\frac13\right)-0 =\frac43. $$

Differentiate the antiderivative \(t+\frac{t^3}{3}\) to check that it gives \(1+t^2\). Substituting \(t=1\) and \(t=0\) gives the displayed endpoint values, so the exact length is \(\frac43\).

Worked Example: A Polynomial Parametrization with a Zero-Speed Endpoint

Worked Example: A Polynomial Parametrization with a Zero-Speed Endpoint

Find the length of \(x=t^2,\ y=\frac23t^3\) for \(0\le t\le 1\).

The coordinate derivatives and speed are

$$ x'(t)=2t,\qquad y'(t)=2t^2, $$ $$ \sqrt{(2t)^2+(2t^2)^2} =\sqrt{4t^2+4t^4} =2t\sqrt{1+t^2}, $$

where \(2t\ge0\) on this interval. At \(t=0\), the speed is zero, but the continuous speed function still has a well-defined integral over the interval. Evaluate using the substitution \(u=1+t^2\), so \(du=2t\,dt\):

$$ L=\int_0^1 2t\sqrt{1+t^2}\,dt =\int_1^2\sqrt{u}\,du =\left[\frac23u^{3/2}\right]_1^2 =\frac23\left(2\sqrt2-1\right). $$

The antiderivative in the original parameter is \(\frac23(1+t^2)^{3/2}\); differentiating it gives \(2t\sqrt{1+t^2}\). Its values at \(1\) and \(0\) are \(\frac43\sqrt2\) and \(\frac23\), respectively, confirming the result.

A Common Square-Root Error

In the second example, a tempting but incorrect step would be to simplify the speed as \(1-t^2\), because that term appears in \(x'(t)\). This ignores the contribution from \(y'(t)\). The incorrect calculation would give

$$ \sqrt{(1-t^2)^2+(2t)^2} \overset{\text{wrong}}{=}1-t^2. $$

But squaring the proposed answer gives \((1-t^2)^2=1-2t^2+t^4\), whereas the actual radicand is \(1+2t^2+t^4\). They are not equal except at special parameter values. The correct simplification is

$$ 1+2t^2+t^4=(1+t^2)^2, \qquad \sqrt{(1+t^2)^2}=|1+t^2|=1+t^2. $$

Always combine and simplify the full sum of squared coordinate derivatives before taking its square root. Then check the sign when replacing \(\sqrt{q(t)^2}\) by \(q(t)\).

A Reliable Calculation Sequence

For a polynomially parametrized curve, a consistent sequence keeps the geometry and algebra connected:

  1. Write the coordinate functions and the exact parameter interval.
  2. Differentiate each coordinate with respect to the parameter.
  3. Square both derivatives, add them, and simplify the radicand.
  4. Take the nonnegative square root, accounting for signs on the interval.
  5. Evaluate the resulting definite integral and check the antiderivative and endpoint values.

The chord between the curve’s endpoints is generally not the arc length. For example, in the parabolic example the endpoint distance is \(\sqrt{(1-0)^2+(1-0)^2}=\sqrt2\), while the length is the integral of speed along the entire arc. The chord measures only the straight-line separation, not the distance traced along the curve.

Key takeaway: Polynomial coordinates make the velocity algebraic, but the speed is the square root of a sum of squares. Simplify that entire expression before integrating, and verify the sign and the parameter limits.

Check Your Understanding

For each question, identify the derivatives and the speed before attempting the integral.

  1. For \(x=t,\ y=3t^2\) on \(0\le t\le1\), what is the speed integrand?
  2. For \(x=t-\frac{t^3}{3},\ y=t^2\) on \(0\le t\le1\), why is the speed \(1+t^2\), rather than \(1-t^2\)?
  3. What exact integral gives the length of \(x=t^2,\ y=\frac23t^3\) on \(0\le t\le1\)?
  4. Why can the speed be zero at an endpoint without making the length integral undefined?
  5. What does the endpoint chord measure, and why is it generally different from arc length?

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