MATH 221

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 6 September 2014

Lecture 32

Lengths of curves

Idea: Use the grid to slice up the curve into little pieces. $$\begin{array}{c} \end{array}$$ Each little piece $\begin{array}{c} dy dx ds \end{array}$ has length $$ds=\sqrt{{\left(dx\right)}^2+{\left(dy\right)}^2}\text{.}$$ Add up the lengths of the little pieces with an integral.

Use integration to find the length of a circle of radius $r\text{.}$ $$\begin{array}{c} - r r x - r r y \end{array}$$ The length of the whole circle is 4 times the length of $$\begin{array}{c} r x r y \end{array}$$ Divide this part of the curve into little pieces $\begin{array}{c} dy dx ds \end{array}\text{.}$ Each little piece has length $ds=\sqrt{{\left(dx\right)}^2+{\left(dy\right)}^2}$ Add up the lengths of the little pieces with an integral. $$\begin{array}{ccc}{\displaystyle {\int }_{x=0}^{x=r}ds}& =& {\displaystyle {\int }_{x=0}^{x=r}\sqrt{{\left(dx\right)}^2+{\left(dy\right)}^2}}\\ & =& {\displaystyle {\int }_{x=0}^{x=r}\frac{\sqrt{{\left(dx\right)}^2+{\left(dy\right)}^2}}{dx}dx}\\ & =& {\displaystyle {\int }_{x=0}^{x=r}\sqrt{\frac{{\left(dx\right)}^2+{\left(dy\right)}^2}{{\left(dx\right)}^2}}dx}\\ & =& {\displaystyle {\int }_{x=0}^{x=r}\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx}\\ & =& {\displaystyle {\int }_{x=0}^{x=r}\sqrt{1+{(-\frac{2x}{2y})}^2}dx,}\end{array}$$ since for $x^2+y^2=r^2,$ $2x+2y\frac{dy}{dx}=0,$ and so $\frac{dy}{dx}=-\frac{2x}{2y}\text{.}$ So $$\begin{array}{ccc}{\displaystyle {\int }_{x=0}^{x=r}ds}& =& {\displaystyle {\int }_{x=0}^{x=r}\sqrt{1+\frac{x^2}{y^2}}dx}\\ & =& {\displaystyle {\int }_{x=0}^{x=r}\sqrt{\frac{y^2+x^2}{y^2}}dx}\\ & =& {\displaystyle {\int }_{x=0}^{x=r}\sqrt{\frac{r^2}{r^2-x^2}}dx}\\ & =& {\displaystyle {\int }_{x=0}^{x=r}r\frac{1}{\sqrt{r^2-x^2}}dx}\\ & =& {\displaystyle {\int }_{x=0}^{x=r}r\frac{1}{\sqrt{r^2(1-\frac{x^2}{r^2})}}dx}\\ & =& {\displaystyle {\int }_{x=0}^{x=r}\frac{1}{\sqrt{1-{\left(\frac{x}{r}\right)}^2}}dx}\\ & =& {\displaystyle {\int }_{x=0}^{x=r}\frac{r·\frac{1}{r}}{\sqrt{1-{\left(\frac{x}{r}\right)}^2}}dx}\\ & =& {\displaystyle r{\text{sin}}^{-1}\left(\frac{x}{r}\right){|}_{x=0}^{x=r}}\\ & =& {\displaystyle r{\text{sin}}^{-1}1-r{\text{sin}}^{-1}\left(0\right)}\\ & =& {\displaystyle r\frac{\pi }{2}-0\text{.}}\end{array}$$ So the total length of the circle is $$4\left(\frac{r\pi }{2}\right)=2\pi r\text{.}$$

Find the length of the curve $x=t-\text{sin}\hspace{0.17em}t,$ $y=1-\text{cos}\hspace{0.17em}t,$ $0\le t\le 2\pi \text{.}$

Divide the curve into little pieces $\begin{array}{c} dy dx ds \end{array}$ Each little piece has length $ds=\sqrt{{\left(dx\right)}^2+{\left(dy\right)}^2}\text{.}$ Add up the lengths of the little pieces: $$\begin{array}{ccc}{\displaystyle {\int }_{t=0}^{t=2\pi }ds}& =& {\displaystyle {\int }_{t=0}^{t=2\pi }\sqrt{{\left(dx\right)}^2+{\left(dy\right)}^2}}\\ & =& {\displaystyle {\int }_{t=0}^{t=2\pi }\frac{\sqrt{{\left(dx\right)}^2+{\left(dy\right)}^2}}{dt}dt}\\ & =& {\displaystyle {\int }_{t=0}^{t=2\pi }\sqrt{\frac{{\left(dx\right)}^2+{\left(dy\right)}^2}{{\left(dt\right)}^2}}dt}\\ & =& {\displaystyle {\int }_{t=0}^{t=2\pi }\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt}\\ & =& {\displaystyle {\int }_{t=0}^{t=2\pi }\sqrt{{(1-\text{cos}\hspace{0.17em}t)}^2+{\left(\text{sin}\hspace{0.17em}t\right)}^2}dt}\\ & =& {\displaystyle {\int }_{t=0}^{t=2\pi }\sqrt{1-2\text{cos}\hspace{0.17em}t+{\text{cos}}^{2}t+{\text{sin}}^{2}t}dt}\\ & =& {\displaystyle {\int }_{t=0}^{t=2\pi }\sqrt{1-2\text{cos}\hspace{0.17em}t+1}dt}\\ & =& {\displaystyle {\int }_{t=0}^{t=2\pi }\sqrt{2-2\text{cos}\hspace{0.17em}t}dt}\\ & =& {\displaystyle {\int }_{t=0}^{t=2\pi }\sqrt{2-2\text{cos}(\frac{t}{2}+\frac{t}{2})}dt}\\ & =& {\displaystyle {\int }_{t=0}^{t=2\pi }\sqrt{2-2({\text{cos}}^2\left(\frac{t}{2}\right)-{\text{sin}}^2\left(\frac{t}{2}\right))}dt}\\ & =& {\displaystyle {\int }_{t=0}^{t=2\pi }\sqrt{2}\sqrt{1-{\text{cos}}^2\left(\frac{t}{2}\right)+{\text{sin}}^2()\frac{t}{2}}dt}\\ & =& {\displaystyle {\int }_{t=0}^{t=2\pi }\sqrt{2}\sqrt{{\text{sin}}^2\left(\frac{t}{2}\right)+{\text{sin}}^2\left(\frac{t}{2}\right)}dt}\\ & =& {\displaystyle {\int }_{t=0}^{t=2\pi }\sqrt{2}\sqrt{2{\text{sin}}^2\left(\frac{t}{2}\right)}dt}\\ & =& {\displaystyle {\int }_{t=0}^{t=2\pi }\sqrt{2}\sqrt{2}\text{sin}\left(\frac{t}{2}\right)dt}\\ & =& {\displaystyle 2(-\text{cos}\left(\frac{t}{2}\right))·2{|}_{t=0}^{t=2\pi }}\\ & =& {\displaystyle -4\text{cos}\left(\frac{2\pi }{2}\right)-(-4\text{cos}\hspace{0.17em}0)}\\ & =& {\displaystyle (-4)(-1)+4·1}\\ & =& {\displaystyle 4+4}\\ & =& {\displaystyle 8\text{.}}\end{array}$$

Find the length of the curve $x=\frac{3}{5}{y}^{\frac{5}{3}}-\frac{3}{4}{y}^{\frac{1}{3}}$ from $y=0$ to $y=1\text{.}$

Divide the curve into little pieces $\begin{array}{c} dy dx ds \end{array}\text{.}$ Each little piece has length $ds=\sqrt{{\left(dx\right)}^2+{\left(dy\right)}^2}\text{.}$ Add up the lengths of the little pieces $$\begin{array}{ccc}{\displaystyle {\int }_{y=0}^{y=1}ds}& =& {\displaystyle {\int }_{y=0}^{y=1}\sqrt{{\left(dx\right)}^2+{\left(dy\right)}^2}}\\ & =& {\displaystyle {\int }_{y=0}^{y=1}\frac{\sqrt{{\left(dx\right)}^2+{\left(dy\right)}^2}}{dy}dy}\\ & =& {\displaystyle {\int }_{y=0}^{y=1}\sqrt{\frac{{\left(dx\right)}^2+{\left(dy\right)}^2}{{\left(dy\right)}^2}}dy}\\ & =& {\displaystyle {\int }_{y=0}^{y=1}\sqrt{{\left(\frac{dx}{dy}\right)}^2+1}dy}\\ & =& {\displaystyle {\int }_{y=0}^{y=1}\sqrt{{(y^{\frac{2}{3}}-\frac{1}{4}{y}^{-\frac{2}{3}})}^2+1}dy}\\ & =& {\displaystyle {\int }_{y=0}^{y=1}\sqrt{y^{\frac{4}{3}}-\frac{1}{2}+\frac{1}{16}{y}^{-\frac{4}{3}}+1}dy}\\ & =& {\displaystyle {\int }_{y=0}^{y=1}\sqrt{y^{\frac{4}{3}}+\frac{1}{2}+\frac{1}{16}{y}^{-\frac{4}{3}}}dy}\\ & =& {\displaystyle {\int }_{y=0}^{y=1}\sqrt{{(y^{\frac{2}{3}}+\frac{1}{4}{y}^{-\frac{2}{3}})}^2}dy}\\ & =& {\displaystyle {\int }_{y=0}^{y=1}(y^{\frac{2}{3}}+\frac{1}{4}{y}^{-\frac{2}{3}})dy}\\ & =& {\displaystyle \frac{3}{5}{y}^{\frac{5}{3}}+\frac{1}{4}·3y^{\frac{1}{3}}{\int }_{y=0}^{y=1}}\\ & =& {\displaystyle \frac{3}{5}+\frac{3}{4}-(0+0)}\\ & =& {\displaystyle \frac{12}{20}+\frac{15}{20}}\\ & =& {\displaystyle \frac{27}{20}\text{.}}\end{array}$$

Notes and References

These are a typed copy of Lecture 32 from a series of handwritten lecture notes for the class MATH 221 given on November 27, 2000.

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