Arc Lengths and Surfaces for Parametric Curves, calculus lesson

A bike odometer does not know the shape of the road. It adds up speed times time, and that sum is the distance travelled. Length along a parametric curve works the same way. Integrate the speed of the moving point and you get the length of its path, and a small change to that integral gives the area of a surface of revolution.

Key terms

Parametric speed

Definition Let $x = x(t)$ and $y = y(t)$ be differentiable at $t$ . The velocity of the point at time $t$ is the vector $\langle x'(t), y'(t) \rangle$ , and its speed is the length of that vector,

\frac{ds}{dt} = \sqrt{x'(t)^2 + y'(t)^2}.

In plain words In a short time $dt$ the point moves about $x'\,dt$ sideways and $y'\,dt$ up, so by the Pythagorean theorem it covers about $\sqrt{x'^2 + y'^2}\,dt$ of path.

Example For $x = t^2$ , $y = t^3$ the speed is

\frac{ds}{dt} = \sqrt{4t^2 + 9t^4} = \lvert t \rvert \sqrt{4 + 9t^2}.

At $t = 1$ it equals $\sqrt{13} \approx 3.61$ , and at $t = 0$ it equals $0$ .

Regularity

Definition A parametrization $x = x(t)$ , $y = y(t)$ on an interval $I$ is smooth, or regular, if $x'$ and $y'$ are continuous on $I$ and are never both zero there, that is, the speed is positive at every $t$ in $I$ .

In plain words A regular parametrization never stops. Where the point comes to rest, the curve can form a sharp corner or cusp even though $x(t)$ and $y(t)$ are perfectly smooth functions.

Example For $x = t^2$ , $y = t^3$ both derivatives $2t$ and $3t^2$ vanish at $t = 0$ . The curve is $y = \pm x^{3/2}$ , which has a cusp at the origin. On any interval that avoids $t = 0$ the parametrization is regular.

Arc length

Definition Let a curve $C$ be given by $x = x(t)$ , $y = y(t)$ for $\alpha \le t \le \beta$ , where $x'$ and $y'$ are continuous on $[\alpha, \beta]$ and $C$ is traversed exactly once as $t$ increases from $\alpha$ to $\beta$ . Then the length of $C$ is

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

In plain words Length is the integral of speed over the time interval, as long as no part of the path is run over twice.

Example One arch of the cycloid $x = r(\theta - \sin\theta)$ , $y = r(1 - \cos\theta)$ with $r > 0$ and $0 \le \theta \le 2\pi$ has

x'^2 + y'^2 = r^2(1 - \cos\theta)^2 + r^2 \sin^2\theta = 2r^2(1 - \cos\theta) = 4r^2 \sin^2\frac{\theta}{2}.

Since $\sin(\theta/2) \ge 0$ on $[0, 2\pi]$ , the square root is $2r\sin(\theta/2)$ and the length is

L = \int_0^{2\pi} 2r\sin\frac{\theta}{2}\, d\theta = \left[-4r\cos\frac{\theta}{2}\right]_0^{2\pi} = 8r.

Piecewise smooth curves

Definition A curve $x = x(t)$ , $y = y(t)$ , $\alpha \le t \le \beta$ , is piecewise smooth if $x$ and $y$ are continuous on $[\alpha, \beta]$ and there is a partition $\alpha = t_0 < t_1 < \dots < t_n = \beta$ such that on each closed piece $[t_{k-1}, t_k]$ the functions $x$ and $y$ have continuous derivatives (one-sided at the ends) that are never both zero. So each piece is smooth in the sense above, and corners or cusps can occur only at the points $t_k$ . If the curve is traced exactly once, its length is the sum of the lengths of the pieces,

L = \sum_{k=1}^{n} \int_{t_{k-1}}^{t_k} \sqrt{x'(t)^2 + y'(t)^2}\, dt.

In plain words Corners are allowed. Measure each smooth stretch between corners and add.

Example The curve $x = t$ , $y = \lvert t \rvert$ for $-1 \le t \le 1$ is a V with a corner at $t = 0$ . On each half the speed is $\sqrt{1 + 1} = \sqrt{2}$ , so $L = \sqrt{2} + \sqrt{2} = 2\sqrt{2}$ , the two legs of length $\sqrt{2}$ each.

Repeated tracing

Definition Let $x'$ and $y'$ be continuous on $[\alpha, \beta]$ . If the parametrization runs over some part of the curve more than once, the integral

\int_{\alpha}^{\beta} \sqrt{x'(t)^2 + y'(t)^2}\, dt

is the total distance travelled by the point, not the length of the curve. To get the length, restrict to an interval on which the curve is traced exactly once.

In plain words The integral is an odometer. Drive the same road back and forth and it keeps counting.

Example For $x = \sin^2 t$ , $y = \cos^2 t$ we have $x + y = 1$ with $0 \le x \le 1$ , a segment of length $\sqrt{2}$ . The speed is $\sqrt{2}\,\lvert \sin 2t \rvert$ , and

\int_0^{3\pi} \sqrt{2}\,\lvert \sin 2t \rvert\, dt = \sqrt{2} \cdot 6 = 6\sqrt{2},

because the point runs along the segment six times. The interval $[0, \frac{\pi}{2}]$ traces it once and gives $\sqrt{2}$ .

Surface area of revolution

Definition Let $C$ be given by $x = x(t)$ , $y = y(t)$ for $\alpha \le t \le \beta$ , with $x'$ and $y'$ continuous on $[\alpha, \beta]$ and $C$ traversed exactly once. If $y(t) \ge 0$ on $[\alpha, \beta]$ , the surface obtained by rotating $C$ about the $x$ -axis has area

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

If instead $x(t) \ge 0$ on $[\alpha, \beta]$ , rotating about the $y$ -axis gives

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

In plain words Each tiny piece of curve of length $ds$ sweeps out a thin band. The band’s area is its circumference $2\pi \times \text{radius}$ times its width $ds$ .

Example The upper half circle $x = r\cos t$ , $y = r\sin t$ , $0 \le t \le \pi$ , with $r > 0$ has speed $r$ . Rotating it about the $x$ -axis gives a sphere with

S = \int_0^{\pi} 2\pi\, r\sin t \cdot r\, dt = 2\pi r^2 \left[-\cos t\right]_0^{\pi} = 4\pi r^2.

Distance to the axis

Definition In the surface area integral the factor in front of $ds$ is $2\pi\rho(t)$ , where $\rho(t)$ is the distance from the point $(x(t), y(t))$ to the axis of rotation. For the $x$ -axis $\rho = \lvert y \rvert$ , for the $y$ -axis $\rho = \lvert x \rvert$ , for the line $y = k$ it is $\lvert y - k \rvert$ and for the line $x = h$ it is $\lvert x - h \rvert$ . The curve must not cross the axis, because pieces on opposite sides can sweep over the same surface, which would then be counted twice.

In plain words The radius of each band is how far that bit of curve is from the axis, whatever the axis is.

Example Rotate the circle $x = \cos t$ , $y = 2 + \sin t$ , $0 \le t \le 2\pi$ , about the $x$ -axis. Its distance to the axis is $2 + \sin t > 0$ and its speed is $1$ , so the torus has area

S = \int_0^{2\pi} 2\pi (2 + \sin t)\, dt = 2\pi \cdot 4\pi = 8\pi^2.

Check yourself

Find the length of $x = e^t \cos t$ , $y = e^t \sin t$ for $0 \le t \le 1$ .

Show the answer

The derivatives are $x' = e^t(\cos t - \sin t)$ and $y' = e^t(\sin t + \cos t)$ , so

x'^2 + y'^2 = e^{2t}\left((\cos t - \sin t)^2 + (\sin t + \cos t)^2\right) = 2e^{2t}.

The length is

L = \int_0^1 \sqrt{2}\, e^t\, dt = \sqrt{2}(e - 1) \approx 2.43.

Rotate the segment $x = 3t$ , $y = 4t$ , $0 \le t \le 1$ , about the $x$ -axis. Find the surface area.

Show the answer

The speed is $5$ and the distance to the axis is $4t$ , so

S = \int_0^1 2\pi \cdot 4t \cdot 5\, dt = 20\pi.

This is the side of a cone with radius $4$ and slant height $5$ , and $\pi \cdot 4 \cdot 5 = 20\pi$ .

For $x = \cos t$ , $y = \sin t$ , what does the speed integral over $0 \le t \le 6\pi$ give, and what is the length of the circle?

Show the answer

The integral is $6\pi$ , the distance travelled in three laps. The circle has length $2\pi$ .

Arc Lengths and Surfaces for Parametric Curves

Key terms

Parametric speed

Regularity

Arc length

Piecewise smooth curves

Repeated tracing

Surface area of revolution

Distance to the axis

Check yourself

How this lesson connects

Builds on