Objectives
- To come
Section 8.1 Arc Length
Subsection 8.1.1 Before Class
Subsubsection 8.1.1.1 The Formula
Example 8.1.1.
- Find the distance between the points \((0,-2)\) and \((5,0)\text{.}\)
- Now, find the length of the curve \(y = \dfrac{2}{5}x - 2\) on the interval \([0,5]\text{.}\)
Solution.
- The distance is \(\sqrt{29}\)
- The length is \(\sqrt{29}\) as well.
Example 8.1.2.
Consider \(f(x) = x-x^2\text{.}\)
Arc Length Formula.
If \(f'\) is continuous on \([a,b]\text{,}\) then the length of the curve \(y = f(x)\) on \([a,b]\) is given by
\begin{equation*}
L = \int_a^b \sqrt{1 + [f'(x)]^2}\, dx
\end{equation*}
Proof.
Let \(f\) be a curve on the interval \([a,b]\) with the property that \(f'\) is continuous on \([a,b]\) as well.
Subdivide \([a,b]\) into smaller intervals of length \(\Delta x\text{.}\) Then, the length of each subdivision, \(L_i\text{,}\) can be approximated by the straight-line distance between the points \(f(x_{i-1})\) and \(f(x_i)\text{:}\)
\begin{equation*}
L_i = \sqrt{(x_i-x_{i-1})^2 + (y_i-y_{i-1})^2}
\end{equation*}
Since \(x_i-x_{i-1} = \Delta x\text{,}\) we can rewrite \(L_i\) as
\begin{equation*}
L_i = \sqrt{(\Delta x)^2 + (\Delta y_i)^2}
\end{equation*}
where \(\Delta y_i\) represents \(y_i-y_{i-1}\text{.}\)
From the Mean Value Theorem (see Section 3.2), we know that
\begin{equation*}
f'(x_i) = \dfrac{f(x_i)-f(x_{i-1})}{x_i-x_{i-1}}
\end{equation*}
which can be rearranged to say
\begin{equation*}
f(x_i) - f(x_{i-1}) = f'(x_i)\cdot (x_i-x_{i-1})
\end{equation*}
Since \(f(x_i) - f(x_{i-1}) = \Delta y_i\) because it precisely measures the change in the output values and \(x_i-x_{i-1} = \Delta x\text{,}\) we can rewrite the Mean Value Theorem expression as
\begin{equation*}
\Delta y_i = f'(x_i)\cdot \Delta x
\end{equation*}
Replacing this in the expression for \(L_i\text{,}\) we have
\begin{equation*}
L_i = \sqrt{(\Delta x)^2 + (f'(x_i)\Delta x)^2}
\end{equation*}
Now we may simplify:
\begin{align*}
L_i \amp = \sqrt{(\Delta x)^2 + (f'(x_i)\Delta x)^2}\\
\amp = \sqrt{(\Delta x)^2[1 + (f(x_i))^2]} \\
\amp = \sqrt{1 + (f(x_i))^2)}\cdot \Delta x
\end{align*}
The approximation for the total length of the curve is given by
\begin{align*}
L \amp \approx \sum_{i=1}^n L_i\\
\amp \approx \sum_{i=1}^n \sqrt{1 + (f(x_i))^2)}\cdot \Delta x
\end{align*}
Taking the limit will gives an expression for the precise length of the curve:
\begin{align*}
L \amp = \lim_{n\to\infty}\sum_{i=1}^n \sqrt{1 + (f(x_i))^2)}\cdot \Delta x\\
\amp = \int_a^b \sqrt{1 + (f'(x))^2}\, dx
\end{align*}
Example 8.1.3.
Find the length of the arc of the curve \(y^2 = x^3\) between \((1,1)\) and \((4,8)\text{.}\)
Solution.
\(\dfrac{1}{27}(80\sqrt{10}-13\sqrt{13})\)
Example 8.1.4.
Find the length of the curve \(y = \dfrac{1}{2}x^2\) between \(x = -1\) and \(x = 1\text{.}\)
Solution.
\(\sqrt{2} + \ln(1+\sqrt{2})\)
Subsection 8.1.2 Pre-Class Activities
Example 8.1.5.
Use this space to write any questions you might have from the videos.
Solution.
Answers vary
Example 8.1.6.
Set up (but do not evaluate!) a definite integral that would find the length of the curve \(f(x) = x-x^2\) between \(x = 0\) and \(x = 1\text{.}\)
Solution.
\(L = \ds\int_0^1 \sqrt{1 + (1-2x)^2}\, dx\)
Example 8.1.7.
Find the exact length of the curve \(f(x) = 1 + 6x^{3/2}\) on the interval \(0\leq x \leq 1\text{.}\)
Solution.
\(\dfrac{2}{243}(82\sqrt{82}-1)\)
Subsection 8.1.3 In Class
Subsubsection 8.1.3.1 Some Examples
Example 8.1.8.
Find the length of the curve \(f(x) =x-x^2\) between \(x = 0\) and \(x =1\text{.}\)
Solution.
\(\dfrac{1}{\sqrt{2}} + \dfrac{1}{2}\ln (1+\sqrt{2})\)
Example 8.1.9.
If \(g(x) = \ln |\sec x|\text{,}\) find the length of \(g(x)\) between \(x = \pi/6\) and \(x = \pi/4\text{.}\)
Solution.
\(\ln (\sqrt{2}+1)-\ln \lrpar{\dfrac{3}{\sqrt{3}}}\)
Example 8.1.10.
Find the length of the arc of the parabola \(x = y^2\) from \((1,1)\) to \((4,2)\)
Solution.
\(\sqrt{17} + \dfrac{1}{4}\ln (\sqrt{17}+4) - \dfrac{\sqrt{5}}{2}-\ln(\sqrt{5} + 2)\)
Example 8.1.11.
Find the length of the curve \(x = \dfrac{1}{8}y^4 + \dfrac{1}{4y^2}\) for \(1\leq y \leq 2\text{.}\)
Solution.
\(\dfrac{49}{16}\)
Example 8.1.12.
Determine the length of the arc of the curve \(x^2 = (y-4)^3\) between \(y = 5\) and \(y = 8\)
Solution.
\(\dfrac{1}{27}(80\sqrt{10}-13\sqrt{13})\)
Subsubsection 8.1.3.2 Arc Length Function
Arc Length Function.
If \(C\) is a smooth curve with equation \(y = f(x)\) on the interval \([a,b]\text{,}\) then the function
\begin{equation*}
s(x) = \int_a^x \sqrt{1 + (f'(t))^2}\, dt
\end{equation*}
gives the length of \(f(x)\) between the starting point \((a,f(a))\) and the endpoint \((x,f(x))\text{.}\)
Example 8.1.13.
Find the arc length function for the curve \(y = x^2 - \dfrac{1}{8}\ln x\text{,}\) starting at the point \((1,1)\text{.}\)
Solution.
\(s(x) = x^2 + \dfrac{1}{8}\ln x - 1\)
Example 8.1.14.
Find the arc length function for the curve \(y = \inv{\sin}(x) + \sqrt{1-x^2}\text{,}\) starting at \(x = 0\text{.}\)
Solution.
\(s(x) = 2\sqrt{2} x\)
Example 8.1.15.
Let \(f(x) = \dfrac{1}{4}e^x + e^{-x}\text{.}\) Show that on any interval \([a,b]\text{,}\) the arc length of \(f(x)\) is exactly the area under the curve.
Solution.
First compute the arc length of \(f(x)\) on \([a,b]\text{.}\) \(f'(x) = \dfrac{1}{4}e^x - e^{-x}\text{,}\) so that \((f'(x))^2 = \dfrac{1}{16}e^{2x} - \dfrac{1}{2} + e^{-2x}\text{.}\) When we add 1, we have \(1 + (f'(x))^2 = \dfrac{1}{16}e^{2x} + \dfrac{1}{2} + e^{-2x}\text{,}\) which can be simplified to \(\dfrac{1}{4}\lrpar{\dfrac{1}{2}e^x + 2e^{-x}}^2\text{.}\)
Taking the root, we get the following expression for arc length:
\begin{equation*}
L = \int_a^b \dfrac{1}{4}e^x + e^{-x}\, dx
\end{equation*}
which is exactly the area under the curve on the same interval.
Subsection 8.1.4 After Class Activities
Example 8.1.16.
Find the length of the curve \(y = \dfrac{1}{3}x^3 + \dfrac{1}{4x}\) on the interval \([1,2]\)
Solution.
\(\dfrac{59}{24}\)
Example 8.1.17.
Find the length of the curve \(y = \ln(\cos x)\) on \(0\leq x\leq \pi/3\)
Solution.
\(ln\lrpar{1+\dfrac{1}{\sqrt{3}}} - ln\lrpar{1-\dfrac{1}{\sqrt{3}}}\)
Example 8.1.18.
Find the length of the curve \(y = \dfrac{1}{4}x^2 -\dfrac{1}{2}\ln x\) on the interval \([1,2]\text{.}\)
Solution.
\(\dfrac{3}{4} + \dfrac{1}{4}\ln 4\)
Example 8.1.19.
Find the arc length function for the following curves:
- \(y = 2x^{3/2}\text{,}\) starting at \((1,2)\text{.}\)
- \(y = \ln (\sin x)\) (on \((0,\pi)\)), with the starting point of \((\pi/2,0)\text{.}\)
Solution.
- \(s(x) = \dfrac{2}{27}\lrpar{(1+9x)^{3/2}-1}\text{.}\)
- \(s(x) = \ln (\csc x - \cot x)\text{.}\)
