Section 2.4 Arc Length of a Curve and Surface Area
References.
- Calculus, Early Transcendentals by Stewart, Section 8.1.
Note: We only address the topic of arc length here; surface areas are better handled in the context of parametric curves, in Section 7.2.
Introduction.
Just as calculus was used to compute the areas of regions with curved boundaries, starting with an approximation by the area of a collection of polygons, we will now make sense of the length of a curve through approximating a curve by a collection of short straight line segments, and so approximate its area by the total length of that collection.
In this section we do this for curves in the plane that are the graph of a function: a curve \(C\) that is the set of points \((x,y)\) given by \(a \leq x \leq b\text{,}\) \(y=f(x)\text{.}\)
Later, in Section 7.2, we will see how to consider the length of other curves, like circles and spirals.
Approximation by Polygonal Curves.
The first step is familiar: divide the range of \(x\) values into \(n\) small intervals with values at spacing \(\Delta x = (b-a)/n\text{:}\) \(x_0, x_1, \dots x_n\) with \(x_i = a+i\Delta x\) so \(x_0=a\text{,}\) \(x_n=b\text{.}\)
This then gives a sequence of points along the curve: \(P_i(x_i,y_i)\) with \(y_i=f(x_i)\text{,}\) \(0 \leq i \leq n\text{.}\) The point \(P_0(a,f(a))\) is called the initial point of the curve, the point \(P_n(b,f(b))\) is called the terminal point, and both are called endpoints.
Joining these points with line segments gives a “polygonal curve”, of length
\begin{equation*}
L_n = \sum_{i=1}^n \left| P_{i-1} P_i \right|
= \sum_{i=1}^n \sqrt{(x_i-x_{i-1})^2 + (y_i-y_{i-1})^2}\
= \sum_{i=1}^n \sqrt{(\Delta x)^2 + (\Delta y_i)^2}
\end{equation*}
where \(\Delta y_i = y_i-y_{i-1}\) is the vertical step on the \(i\)-th edge.
Curve Length as a Limit.
The exact length of the curve \(C\) is then defined as the limit of these approximations:
\begin{equation*}
L = \lim_{n \to \infty} L_n = \lim_{n \to \infty} \sum_{i=1}^n \sqrt{(\Delta x)^2 + (\Delta y_i)^2}
\end{equation*}
As with areas and volumes, we want to turn this into a definite integral with the Fundamental Theorem of Calculus (FTC).
To get the factor \(\Delta x\) needed, rearrange as
\begin{equation*}
L = \lim_{n \to \infty} \sum_{i=1}^n \sqrt{1 + \left[\frac{\Delta y_i}{\Delta x}\right]^2} \Delta x
\end{equation*}
Each difference quotient is approximately a value of \(f'\text{:}\) using the Mean Value Theorem, we can get
\begin{equation*}
L = \lim_{n \to \infty} \sum_{i=1}^n \sqrt{1 + \left[ f'(x_i^*)\right]^2} \; \Delta x
\quad \mbox{[for suitable values ]}
\end{equation*}
Finally, the Fundamental Theorem of Calculus (Part 2) gives:
The Arc Length of a Curve.
If \(f'(x)\) is continuous on \([a,b]\) then the curve \(C\) of points \(P(x,y)\) given by \(a \leq x \leq b\text{,}\) \(y=f(x)\) has arc length
\begin{equation}
L = \int_a^b \sqrt{1+ \left[ f'(x)\right]^2} \; dx
= \int_a^b\sqrt{1+ \left( \frac{dy}{dx} \right)^2} \; dx\tag{2.4.1}
\end{equation}
Sometimes we are interested not just in the total length, but the length of the part of a curve from the initial point up to a certain \(x\) value.
This will be useful for example when the curve is describing motion in time, and we want to compute distance traveled as a function of time.
Changing the dummy variable of integration from \(x\) to \(t\text{,}\) the length of the curve from the initial point where \(t=a\) to the point where \(t=x\) is a function of \(x\text{,}\) the arc length function
\begin{equation}
s(x) = \int_a^x \sqrt{1+ \left[ f'(t)\right]^2} \; dt\tag{2.4.2}
\end{equation}
The rate of change of arc length as \(x\) increases is given by the Fundamental Theorem of Calculus (Part 1) as
\begin{equation*}
\frac{ds}{dx} = \sqrt{1+ \left[ f'(x)\right]^2} = \sqrt{1+ \left( \frac{dy}{dx} \right)^2}
\end{equation*}
The differential of the arc length function can be useful: it is
\begin{equation*}
ds = \sqrt{1+ \left[ f'(x) \right]^2} \; dx = \sqrt{1+ \left( \frac{dy}{dx} \right)^2} \; dx
\end{equation*}
which has a nicely mnemonic form if we square both sides:
\begin{equation*}
(ds)^2 = \left[ 1+ \left( \frac{dy}{dx} \right)^2 \right] (dx)^2 = (dx)^2 + (dy)^2
\end{equation*}
which is intuitively what the Pythagorean Theorem gives for the length of a small piece of the curve of horizontal extent \(dx\text{,}\) vertical extent \(dy\text{.}\)
Perhaps the best way to remember the arc length formulas is to first remember this intuitive differential formula and then integrate:
\begin{equation*}
s(x) = \int_a^x ds = \int_a^x \sqrt{1+ \left[ f'(t) \right]^2} \; dt,
\qquad
L = s(b) = \int_a^b ds = \int_a^b \sqrt{1+ \left( \frac{dy}{dx} \right)^2} \; dx.
\end{equation*}
Study Guide.
Study the part of Calculus Volume 2, Section 2.4 2 on Arc Length of a Curve; that is to Example 2.20 and Checkpoint 2.20; we omit the second topic of Area of a Surface of Revolution.
Do one or several exercises from each of the ranges 171–174 and 176–179, but if you cannot evaluate it explicitly, simply leave the result as a definite integral expresssion.
openstax.org/books/calculus-volume-2/pages/2-4-arc-length-of-a-curve-and-surface-areaopenstax.org/books/calculus-volume-2/pages/2-4-arc-length-of-a-curve-and-surface-area
