2.4 Arc Length

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We find the length of a curved segment.

Arc Length

Arc Length The length of the graph of the differentiable function, $y = f(x)$ , from $x = a$ to $x = b$ , is given by $L = \int _a^b \sqrt {1+ f'(x)^2} \; dx.$

example 1 Find the length of the curve $y = 3x + 2$ from $x = 0$ to $x = 4$ .
Since the graph of this function is a line, calculus is not required to find the length of the indicated segment. Instead, we can use the distance formula, $d = \sqrt {(x_2 - x_1)^2 + (y_2-y_1)^2},$ with the points $(0,2)$ and $(4, 14)$ to get $d = \sqrt {(4-0)^2 + (14-2)^2} = \sqrt {4^2 + 12^2} = \sqrt {160} = 4\sqrt {10}.$ We will also use the formula from the above theorem as a verification that the theorem is correct. Since $f(x) = 3x+2$ , we have $f'(x) = 3$ and the length of the segment is $L = \int _0^4 \sqrt {1 + 3^2} \;dx = \int _0^4 \sqrt {10} \; dx = \sqrt {10}x\bigg |_0^4 = 4\sqrt {10},$ as expected.

(problem 1a) Find the length of the curve $y = 2x - 5$ from $x = 2$ to $x = 5$ using the arc length formula.

$f'(x) = \answer {2}$

$\sqrt {1+f'(x)^2} = \answer {\sqrt {5}}$

Arc length = $\answer {3\sqrt {5}}$ .

(problem 1b) Find the length of the curve $y = 2x - 5$ from $x = 2$ to $x = 5$ using the distance formula.

The endpoints of the line segment are $(2, \answer {-1})$ and $(5, \answer {5})$

$d = \sqrt {(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt {(\answer {3})^2 + (\answer {6})^2}$

Arc length = $\answer {3\sqrt {5}}$ .

example 2 Find the length of the graph of $y = 1+ x^{3/2}$ from $x = 0$ to $x = 4$ .
First, $f(x) = x^{3/2} + 1$ and so $f'(x) = \frac 32 x^{1/2}$ . Then the arc length is $L = \int _0^4 \sqrt {1 + \left (\frac 32 x^{1/2}\right )^2} \; dx = \int _0^4 \sqrt {1 + \frac 94 x} \;dx.$ To compute this integral, we will use a $u$ -substitution with $u = 1+\frac 94 x$ and $du = \frac 94 \; dx$ . Thus, the anti-derivative is, $\int \sqrt {1+\frac 94 x} \; dx = \int \frac 49 \sqrt u \; du = \frac 49 \cdot \frac 23 u^{3/2} + C = \frac {8}{27} \left (1 + \frac 94 x \right )^{3/2} + C.$

Returning to our computation of arc length, $L = \frac {8}{27} \left (1 + \frac 94 x \right )^{3/2} \bigg |_0^4 = \frac {8}{27} (10^{3/2} - 1).$

(problem 2a) Find the length of the curve $y = 2 + 4x^{3/2}$ from $x = 0$ to $x = 1$

$f'(x) = \answer {6x^{1/2}}$

$\sqrt {1+f'(x)^2} = \answer {\sqrt {1+ 36x}}$

Arc length = $\answer {\frac {1}{54} (37^{3/2} -1)}$

(problem 2b) Find the length of the curve $y = 3 + 2x^{3/2}$ from $x = 7$ to $x = 11$

$f'(x) = \answer {3x^{1/2}}$

$\sqrt {1+f'(x)^2} = \answer {\sqrt {1+ 9x}}$

Arc length = $\answer {\frac {2}{27} (488)}$

example 3 Find the length of the graph of $y = \frac {x^3}{12} + \frac {1}{x}$ from $x = 1$ to $x = 2$ .
In this example, computing a simplified form of the integrand is laborious, so we will proceed slowly. First, $f(x) = \frac {x^3}{12} + x^{-1}$ , so $f'(x) = \frac {x^2}{4} - \frac {1}{x^2}$ and $f'(x)^2 = \left ( \frac {x^2}{4} - \frac {1}{x^2} \right )^2 = \frac {x^4}{16} - \frac 12 + \frac {1}{x^4}.$ Next, we add one: $1 + f'(x)^2 = \frac {x^4}{16} + \frac 12 + \frac {1}{x^4},$ and the key to this problem is that this expression is a perfect square: $\frac {x^4}{16} + \frac 12 + \frac {1}{x^4} = \left ( \frac {x^2}{4} + \frac {1}{x^2}\right )^2.$ Now, we can compute the square root: $\sqrt {1+f'(x)^2} = \sqrt {\left ( \frac {x^2}{4} + \frac {1}{x^2}\right )^2 } = \frac {x^2}{4} + \frac {1}{x^2}.$ Finally, we can integrate to get the arc length: $\begin{align*} L &= \int_1^2 \sqrt{1+f'(x)^2} \; dx \\ &= \int_1^2 \left(\frac{x^2}{4} + \frac{1}{x^2} \right) \; dx \\ &= \int_1^2 \left(\frac{x^2}{4} + x^{-2} \right) \; dx \\ &= \left(\frac{x^3}{12} + \frac{x^{-1}}{-1} \right) \bigg|_1^2 \\ &= \left(\frac{x^3}{12} - \frac{1}{x} \right) \bigg|_1^2 \\ &= \left(\frac23 - \frac12\right) - \left(\frac{1}{12} - 1\right) \\ &= \frac{13}{12}. \end{align*}$

(problem 3a) Find the length of the curve $\displaystyle {y = \frac {x^4}{4} + \frac {1}{8x^2}}$ from $x = 1$ to $x = 2$

$f'(x) = \answer {x^3 - 1/(4x^3)}$

$f'(x)^2 =$

$\displaystyle {x^6 + \frac {1}{16x^6}}$ $\displaystyle {x^6 -\frac 14 + \frac {1}{16x^6}}$ $\displaystyle {x^6 -\frac 12 + \frac {1}{16x^6}}$

$1+f'(x)^2 = \left (\answer {x^3 + \frac {1}{4x^3}}\right )^2$

Arc length = $\answer {123/32}$

(problem 3b) Find the length of the curve $\displaystyle {y = \frac {x^2}{8} - \ln (x)}$ from $x = 2$ to $x = 4$

$f'(x) = \answer {x/4 - 1/x}$

$f'(x)^2 =$

$\displaystyle {\frac {x^2}{16} + \frac {1}{x^2}}$ $\displaystyle {\frac {x^2}{16} -\frac 14 + \frac {1}{x^2}}$ $\displaystyle {\frac {x^2}{16} -\frac 12 + \frac {1}{x^2}}$

$1+f'(x)^2 = \left (\answer {x/4 + 1/x}\right )^2$

Arc length = $\answer {3/2 + \ln 2}$

(problem 3c) Find the length of the curve $y = \ln (\sec (x))$ from $x = 0$ to $x = \pi /4$ .

$f'(x) = \answer {\tan (x)}$

$1 + f'(x)^2 =$

$\sec ^2(x)$ $\tan ^2(x)$ $\cos ^2(x)$

Arc length = $\answer {\ln (1 + \sqrt 2)}$

(problem 3d) Find the length of the curve $y = \frac {e^x + e^{-x}}{2}$ from $x = 0$ to $x = 1$ .

$f'(x) = \answer {\frac {e^x - e^{-x}}{2}}$

$1 + f'(x)^2 =$

$\left (\frac {e^{x} - e^{-x}}{2}\right )^2$ $\left (\frac {e^{x} + e^{-x}}{2}\right )^2$ $\left (\frac {e^{x} + e^{-x}}{4}\right )^2$

Arc length = $\answer {\frac {e^2 - 1}{2e}}$

Video Lesson

Here is a detailed, lecture style video on arc length:

Theoretical Justifications

In this section, we derive the arc length formula. Suppose $f(x)$ is a differentiable function on the open interval $(a,b)$ and continuous on the closed interval $[a,b]$ . Let $\Delta x = \frac {b-a}{n}$ and let $x_i = a + i\cdot \Delta x$ for $i = 0, 1, 2, 3, \ldots , n$ . Note that $x_0 = a$ and $x_n = a + n \cdot \Delta x = b$ . Next, to approximate the arc length, connect the adjacent points $(x_{i-1}, f(x_{i-1}))$ and $(x_i, f(x_i))$ with straight line segments for $i = 1, 2, \ldots n.$ From the distance formula, the length of the $i^{th}$ segment is $L_i = \sqrt {(x_i - x_{i-1})^2 + \left [f(x_i) -f(x_{i-1})\right ]^2}$

Now, we apply the Mean Value Theorem to $f(x)$ on the interval $[x_{i-1}, x_i]$ to conclude that $f(x_i) -f(x_{i-1}) = f'(x_i^*)(x_i - x_{i-1}) = f'(x_i^*) \cdot \Delta x$ for some number $x_i^*$ in the interval $(x_{i -1}, x_i)$ . Thus,

$L_i = \sqrt { (\Delta x)^2 + f'(x_i^*)^2 \cdot (\Delta x)^2} = \sqrt { 1 + f'(x_i^*)^2 }\cdot \Delta x, \; i = 1, 2, \ldots n.$

Returning to the arc length, $L$ , of $y = f(x)$ on $[a,b]$ , we have $L \approx \sum _{i=1}^n L_i = \sum _{i=1}^n \sqrt {1 + f'(x_i^*)^2} \cdot \Delta x.$ Taking the limit as $n \to \infty$ , we define the arc length as $L = \lim _{n \to \infty } \sum _{i=1}^n \sqrt {1 + f'(x_i^*)^2} \cdot \Delta x = \int _a^b \sqrt {1 + f'(x)^2} \; dx.$