周次: 2 备注: Attendance check 日期: March 2, 2022 节次: 1

Outline

We knew what is meant by the length of a straight line segment, but without calculus, we have no precise notion of the length of a general winding curve. The idea of approximating the length of a curve has no difference with the idea of approximating an area or approximating a volume. By subdividing the curve into many pieces and joining successive points of division by straight line segments, we can then pass through a limit process to find the “real” length of a curve. In this lecture, we show this process and discuss some examples and consequences.

Parametric Plane Curves

Let $C$ be a curve given parametrically by the equation

$$ x =f(t),\quad \text{and}\quad y = g(t),\quad a\le t\le b. $$

Definition. (Smooth (Plane) Curve)

If the functions $f$ and $g$ have continuous derivatives on $[a,b]$ ( that is, $f'(t)$ and $g'(t)$ are both continuous functions, such functions are called continuously differentiable) and $f'$ and $g'$ are not simultaneously zero. Then the curve $C$ is called a smooth curve.

Remark

It is helpful to imagine the curve as the path of a particle moving from point $A = (f(a),g(a))$ at time $t=a$ to point $B=(f(b),g(b))$.

Example. (Why $f'$ and $g'$ cannot be simultaneously zero?)

A plane curve $x=t^2$ and $y=t^3$ ($t\in[-1,1]$) are defined by functions with continuous derivatives. However, the plot shows there is a cusp when $t=0$, the reason being $f'(0)= g'(0)=0$.

Length of a Parametrically Defined Curve

For a plane curve $C: (x=f(t), y=g(t))$ , ($a\le t\le b$). Suppose $f$ and $g$ are continuously differentiable. Set points $A = (f(a),g(a))$ and $B = (f(b),g(b))$. In sequel, we always assume the path from $A$ to $B$ is traversed exactly once as $t$ increases from $t=a$ to $t=b$.

1. Partition

We subdivide the closed interval $[a,b]$ into $n$ pieces

$$ P: a=t_0 < t_1 < \cdots < t_n = b. $$

The partition $P$ induces a partition on the curve, as

$$ A = P_0, P_1, \cdots, P_n = B,\quad \text{where}\quad P_k = (f(t_k),g(t_k)) $$

2. Length for the straight line segments

Each straight line segment joining the successive points of the partition has length

$$ L_k = \sqrt{(\Delta x_k)^2 + (\Delta y_k)^2} \ = \sqrt{[f(t_k)-f(t_{k-1}]^2 + [g(t_k) - g(t_{k-1})]^2} $$

3. Approximating the curve length

The summation of all the lengths for straight line segment is assumed to be a good approximation for the length of the curve.

$$ \text{Length of the curve }C\approx \sum_{k=1}^n L_k $$

By the Lagrange Mean Value Theorem , there are numbers $t_k^*$ and $t_k^{**}$ $\in[t_{k-1},t_k]$ such that

$$ \Delta x_k = f(t_k)-f(t_{k-1}) = f'(t_k^*), \Delta t_k,\ \Delta y_k = g(t_k)- g(t_{k-1}) = g'(t_k^{**}) , \Delta t_k. $$

Then

$$ \sum_{k=1}^n L_k = \sum_{k=1}^n \sqrt{(\Delta x_k)^2 + (\Delta y_k)^2}\ = \sum_{k=1}^n \sqrt{[f'(t_k^)]^2 + [g'(t_k^{**})]^2},\Delta t_k\=\sum_{k=1}^n \sqrt{[f'(t_k^)]^2 + [g'(t_k^{*})]^2},\Delta t_k + \text{error} $$

4. Pass through the limit

The above summation is not exactly a Riemann sum (It is a Riemann summation up to an “error term”). However, a theorem in Advanced Calculus shows the error approaches $0$ as $||P||\to 0$. Therefore by passing through the limit process as $||P||\to 0$, we obtain the following formula for the length of the curve $C$.

Definition. (Length of a Parametric Curve)

If a curve is defined parametrically by $x=f(t)$ and $y=g(t)$, $a\le t\le b$, where $f'$ and $g'$are continuous on $[a,b]$, and $C$ is traversed exactly once as $t$ increases from $t=a$ to $t=b$, then the length of $C$ is the definite integral

$$ L = \int_a^b \sqrt{[f'(t)]^2 + [g'(t)]^2} dt $$

Remark

The different choices for the parametrization of the curve $C$ doesn’t affect the value $L$ at all. As long as the parametrization meets the conditions stated in the definition of the length of $C$.

Example 1. The Circumference of a Circle

A circle with radius $r$ can be defined parametrically by

$$ x=r\cos(t),\quad\text{and}\quad y=r\sin(t),\quad 0\le t\le 2\pi. $$

Therefore by using the definite integral we can see the length is $L=2\pi r$.

Example 2. The length of an astroid

An astroid is defined by $x=\cos^3(t)$ and $y=\sin^3(t)$, with $0\le t\le 2\pi.$

The length of the curve in the first-quadrant is $L=\dfrac32.$

Special Case: The Length of a Curve $y=f(x)$

Given a continuously differentiable function $y=f(x)$, $a\le x\le b$, we can assign $x=t$ as a parameter. The graph of the function $f$ is then the curve $C$ defined parametrically by

$$ x=t,\quad \text{and} \quad y=f(t),\quad a\le t\le b. $$

Then as a special case of parametric curve, the length of the graph $C$ is defined by

Definition. The length of curve $y=f(x)$

If $f$ is continuously differentiable on the closed interval $[a,b]$, then the length of the curve $y=f(x)$ from $x=a$ to $x=b$ is

$$ L = \int_a^b \sqrt{1+\left(\frac{dy}{dx}\right)^2}dx = \int_a^b \sqrt{1+[f'(x)]^2} dx. $$

Example 3. The arc-length of a graph

A graph defined by

$$ y=\frac{4\sqrt2}{3}x^{3/2} -1,\quad 0\le x\le 1. $$

is $L = \dfrac{13}{6}.$

Example 4. The length of a catenary

A catenary is typically expressed as

$$ y = c \cosh(x/c),\quad \text a\le x \le b. $$

Dealing with Discontinuities in $dy/dx$

As a case when $dy/dx$ fails to exist, $dx/dy$ may exist. Therefore, we may be able to find the curve length by expressing $x$ as a function of $y$ instead.

Example 5. Inverse the expression $y=f(x)$ to $x=g(y)$

Find the length of the curve $y=(x/2)^{2/3}$ from $x=0$ to $x=2.$ By express $x$ as a function of $y$ we find the length is approximately $L\approx 2.27.$

Refined Differential Formula for Length

The length is frequently written in terms of differentials in place of derivatives. Recall that for function $x=f(t)$, the differential $dx$ is defined as

$$ dx = df = f'(t) dt = \frac{d x}{dt} dt $$

So the equation for length is written by

$$ L = \int_a^b \sqrt{(dx)^2+(dy)^2}. $$

It is customary to eliminate the parentheses in $(dx)^2$ and write $dx^2$ instead. So

$$ L = \int_a^b \sqrt{dx^2+dy^2} $$

This is a unified formula for length. For different cases, one must express $x$ and $y$ in suitable parametrization and restore the formula $L = \int_a^b \sqrt{dx^2+dy^2} = \int_a^b \sqrt{g'(t)^2+g'(t)^2}dt$ to compute.

A useful way is to write

$$ ds = \sqrt{dx^2+dy^2} $$

and treat $ds$ as the differential of arc-length (弧长微元), which can be integrated between appropriate limits to give the total length of a curve. The geometric meaning of $ds$ is shown below, as an approximation of length for tiny arc-segment. Then to memorize, first recall the shortest expression

$$ L = \int ds $$

Then recall $ds = \sqrt{dx^2+dy^2}$, finally choose appropriate parametrization to expand $dx^2+dy^2$, as well as specify the the integral limits $a$ and $b$.