Skip to main content  Contents  
 Prev  Up Next  \(\newcommand{\markedPivot}[1]{\boxed{#1}}
\newcommand{\IR}{\mathbb{R}}
\newcommand{\IC}{\mathbb{C}}
\renewcommand{\P}{\mathcal{P}}
\renewcommand{\Im}{\operatorname{Im}}
\newcommand{\RREF}{\operatorname{RREF}}
\newcommand{\vspan}{\operatorname{span}}
\newcommand{\setList}[1]{\left\{#1\right\}}
\newcommand{\setBuilder}[2]{\left\{#1\,\middle|\,#2\right\}}
\newcommand{\unknown}{\,{\color{gray}?}\,}
\newcommand{\drawtruss}[2][1]{
\begin{tikzpicture}[scale=#1, every node/.style={scale=#1}]
\draw (0,0) node[left,magenta]{C} -- 
      (1,1.71) node[left,magenta]{A} -- 
      (2,0) node[above,magenta]{D} -- cycle;
\draw (2,0) -- 
      (3,1.71) node[right,magenta]{B} -- 
      (1,1.71) -- cycle;
\draw (3,1.71) -- (4,0) node[right,magenta]{E} -- (2,0) -- cycle;
\draw[blue] (0,0) -- (0.25,-0.425) -- (-0.25,-0.425) -- cycle;
\draw[blue] (4,0) -- (4.25,-0.425) -- (3.75,-0.425) -- cycle;
\draw[thick,red,->] (2,0) -- (2,-0.75);
#2
\end{tikzpicture}
}
\newcommand{\trussNormalForces}{
\draw [thick, blue,->] (0,0) -- (0.5,0.5);
\draw [thick, blue,->] (4,0) -- (3.5,0.5);
}
\newcommand{\trussCompletion}{
\trussNormalForces
\draw [thick, magenta,<->] (0.4,0.684) -- (0.6,1.026);
\draw [thick, magenta,<->] (3.4,1.026) -- (3.6,0.684);
\draw [thick, magenta,<->] (1.8,1.71) -- (2.2,1.71);
\draw [thick, magenta,->] (1.6,0.684) -- (1.5,0.855);
\draw [thick, magenta,<-] (1.5,0.855) -- (1.4,1.026);
\draw [thick, magenta,->] (2.4,0.684) -- (2.5,0.855);
\draw [thick, magenta,<-] (2.5,0.855) -- (2.6,1.026);
}
\newcommand{\trussCForces}{
\draw [thick, blue,->] (0,0) -- (0.5,0.5);
\draw [thick, magenta,->] (0,0) -- (0.4,0.684);
\draw [thick, magenta,->] (0,0) -- (0.5,0);
}
\newcommand{\trussStrutVariables}{
\node[above] at (2,1.71) {\(x_1\)};
\node[left] at (0.5,0.866) {\(x_2\)};
\node[left] at (1.5,0.866) {\(x_3\)};
\node[right] at (2.5,0.866) {\(x_4\)};
\node[right] at (3.5,0.866) {\(x_5\)};
\node[below] at (1,0) {\(x_6\)};
\node[below] at (3,0) {\(x_7\)};
}
\newcommand{\N}{\mathbb N}
\newcommand{\Z}{\mathbb Z}
\newcommand{\Q}{\mathbb Q}
\newcommand{\R}{\mathbb R}
\DeclareMathOperator{\arcsec}{arcsec}
\DeclareMathOperator{\arccot}{arccot}
\DeclareMathOperator{\arccsc}{arccsc}
\newcommand{\tuple}[1]{\left\langle#1\right\rangle}
\newcommand{\lt}{<}
\newcommand{\gt}{>}
\newcommand{\amp}{&}
\definecolor{fillinmathshade}{gray}{0.9}
\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle     \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle        \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle      \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
\) 
Section   6.2   Arclength (AI2) 
 
Learning Outcomes  
Subsection   6.2.1   Activities 
 
Activity   6.2.1 . 
 
Suppose we wanted to find the arclength of the parabola 
\(y=-x^2+6x\)  over the interval 
\([0,4]\text{.}\)  
 Plot of \(y=-x^2+6x\)  over \([0,4]\text{.}\) 
 
 
Figure   121.    Plot of \(y=-x^2+6x\)  over \([0,4]\text{.}\)  
 
(a)  
Suppose we wished to estimate this length with two line segments where 
\(\Delta x=2\text{.}\)  
 Plot of \(y=-x^2+6x\)  over \([0,4]\text{.}\) 
 
 
Figure   122.    Plot of \(y=-x^2+6x\)  over \([0,4]\)  with two line segments where \(\Delta x=2\text{.}\)  
 
Which of  the following expressions represents the sum of the lengths of the line segments with endpoints \((0,0)\text{,}\)  \((2,8)\)  and  \((4,8)\text{?}\) 
\(\displaystyle \sqrt{4+8}\) 
 
\(\displaystyle \sqrt{2^2+8^2}+\sqrt{(4-2)^2+(8-8)^2}\) 
 
\(\displaystyle \sqrt{4^2+8^2}\) 
 
\(\displaystyle \sqrt{2^2+8^2}+\sqrt{4^2+8^2}\) 
 
 (b)  
Suppose we wished to estimate this length with four line segments where 
\(\Delta x=1\text{.}\)  
 Plot of \(y=-x^2+6x\)  over \([0,4]\text{.}\) 
 
 
Figure   123.    Plot of \(y=-x^2+6x\)  over \([0,4]\)  with four line segments where \(\Delta x=1\text{.}\)  
 
Which of  the following expressions represents the sum of the lengths of the line segments with endpoints \((0,0)\text{,}\)  \((1,5)\text{,}\)  \((2,8)\text{,}\)  \((3,9)\)  and \((4,8)\text{?}\) 
\(\displaystyle \sqrt{4^2+8^2}\) 
 
\(\displaystyle \sqrt{1^2+(5-0)^2}+\sqrt{1^2+(8-5)^2}+\sqrt{1^2+(9-8)^2}+\sqrt{1^2+(8-9)^2}\) 
 
\(\displaystyle \sqrt{1^2+5^2}+\sqrt{2^2+8^2}+\sqrt{3^2+9^2}+\sqrt{4^2+8^2}\) 
 
 (c)  
Suppose we wished to estimate this length with 
\(n\)  line segments where 
\(\displaystyle \Delta x=\frac{4}{n}\text{.}\)   Let 
\(f(x)=-x^2+6x\text{.}\)  
 Plot of \(y=-x^2+6x\)  over \([0,4]\text{.}\) 
 
 
Figure   124.    Plot of \(y=-x^2+6x\)  over \([0,4]\)  with \(n\)  line segments where \(\displaystyle \Delta x=\frac{4}{n}\text{.}\)  
 
Which of  the following expressions represents the length of the line segment from \((x_0, f(x_0))\)  to \((x_0+\Delta x, f(x_0+\Delta x))\text{?}\) 
\(\displaystyle \sqrt{x_0^2+f(x_0)^2}\) 
 
\(\displaystyle \sqrt{(x_0+\Delta x)^2+f(x_0+\Delta x)^2}\) 
 
\(\displaystyle \sqrt{(\Delta x)^2+f(\Delta x)^2}\) 
 
\(\displaystyle \sqrt{(\Delta x)^2+(f(x_0+\Delta x)-f(x_0))^2}\) 
 
 (d)  
Which of  the following Riemann sums best estimates the arclength of the parabola \(y=-x^2+6x\)  over the interval \([0,4]\text{?}\)   Let \(f(x)=-x^2+6x\text{.}\) 
\(\displaystyle \displaystyle \sum \sqrt{(\Delta x)^2+f(\Delta x)^2}\) 
 
\(\displaystyle \displaystyle \sum \sqrt{(x_i+\Delta x)^2+f(x_i+\Delta x)^2}\) 
 
\(\displaystyle \displaystyle \sum \sqrt{x_i^2+f(x_i)^2}\) 
 
\(\displaystyle \displaystyle \sum \sqrt{(\Delta x)^2+(f(x_i+\Delta x)-f(x_i))^2}\) 
 
 (e)  
Note that
\begin{align*}
\sqrt{(\Delta x)^2+(f(x_i+\Delta x)-f(x_i))^2} & = \sqrt{(\Delta x)^2\left(1+\left(\frac{f(x_i+\Delta x)-f(x_i)}{\Delta x} \right)^2\right)}\\
&=\sqrt{1+\left(\frac{f(x_i+\Delta x)-f(x_i)}{\Delta x} \right)^2}\Delta x\text{.}
\end{align*}
 
Which of the following best describes \(\displaystyle\lim_{\Delta x\to 0} \frac{f(x_i+\Delta x)-f(x_i)}{\Delta x}\text{?}\) 
\(\displaystyle 0\) 
 
\(\displaystyle 1\) 
 
\(\displaystyle f'(x_i)\) 
 
This limit is undefined.
 
 
 
Fact   6.2.2 . 
 
Given a differentiable function \(f(x)\text{,}\)  the arclength  of \(y=f(x)\)  defined on \([a,b]\)  is computed by the integral
\begin{align*}
\lim_{n\to \infty}\sum \sqrt{(\Delta x)^2+(f(x_i+\Delta)-f(x_i))^2} & =\lim_{n\to \infty}\sum \sqrt{1+\left(\frac{f(x_i+\Delta x)-f(x_i)}{\Delta x} \right)^2}\Delta x\\
& = \int_a^b \sqrt{1+(f'(x))^2}dx\text{.}
\end{align*}
 
Activity   6.2.3 . 
 
Use 
Fact 6.2.2  to find an integral which measures the arclength of the parabola 
\(y=-x^2+6x\)  over the interval 
\([0,4]\text{.}\) 
Activity   6.2.4 . 
 
Consider the curve \(y=2^x-1\)  defined on \([1,5]\text{.}\) 
 
(a)  
Estimate the arclength of this curve with two line segments where \(\Delta x=2\text{.}\) 
\(x_i\)  
\((x_i,f(x_i))\)  
\((x_i+\Delta x,f(x_i+\Delta x))\)  
Length of segment 
 
\(1\)  
 
 
 
 
\(3\)  
 
 
 
 
(b)  
Estimate the arclength of this curve with four line segments where \(\Delta x=1\text{.}\) 
\(x_i\)  
\((x_i,f(x_i))\)  
\((x_i+\Delta x,f(x_i+\Delta x))\)  
Length of segment 
 
\(1\)  
 
 
 
 
\(2\)  
 
 
 
 
\(3\)  
 
 
 
 
\(4\)  
 
 
 
 
(c)  
Find an integral which computes the arclength of the curve \(y=2^x-1\)  defined on \([1,5]\text{.}\) 
Activity   6.2.5 . 
 
Consider the curve \(y=5e^{-x^2}\)  over the interval \([-1,4]\text{.}\) 
 
(a)  
Estimate this arclength with 5 line segments where \(\Delta x=1\text{.}\) 
(b)  
Find an integral which computes this arclength.
Subsection   6.2.2   Videos 
 
Figure   125.    Video: Estimate the arclength of a curve with Riemann sums and find an integral which computes the arclength
Subsection   6.2.3   Exercises