How To Find The Length Of An Irregular Arc

Question

How would you find out the length of an irregular arc. e.g. An arc with a base length of $10$cm and a height of $5$cm - what would be the length of that arc? Is there a specific formula I could use?

Answer 1

If y is given as function of x describing irregularity formula, its derivative with respect

to x is $ y^{'}(x) $, $ y(0) = 0, y(10) =5,$ then the length of arc is given by

$$ \int \sqrt{1+y^{'2}} \cdot dx $$

Answer 2

Let $y=f(x)$ describe the arc in $xy$ plane.
Consider an infinitesimally small part of the $x$ axis $dx$.
Since the part of the curve directly above will also be very small, we can consider it to be a straight line.
The slope of the function will be $$f'(x)= \tan \theta$$.
Acording to basic trigonometry, the length of hypotenuse will be
$$dx\sec\theta=dx\sqrt{1+\tan^2\theta}=\sqrt{1+(f'(x))^2}dx$$.
By integrating this, we can find out the arc length within any interval.
$$\int_{x_1}^{x_2}\sqrt{1+(f'(x))^2}dx$$

Answer 3

I assume you mean a circular arc, with chord length ("base length") of $w$ and distance along the perpendicular bisector of the base from the base chord to the arc of $h$ (the "height" of the arc).

Although by those definitions, the arc you describe is easy (it is in fact a semi-circle of a circle with radius 5 cm, so has a length of $5\pi$) the general case is not as easy but also tractable using elementary trigonometry:

Call the chord (the base) $AB$, and let $P$ be the midpoint of $AB$, $Q$ the meeting point on the arc, and $O$ be the circle center of the arc from $A$ to $B$. Then $OA = OQ$ and in right triangle $OPA$, $$ OP = OA - h \\ PA = w/2 \\ (OA)^2 = (OP)^2+(PA)^2 = (OA)^2-2h(OA) +h^2 + \frac{w^2}{4} \\ 2h(OA) = h^2 + \frac{w^2}{4}\\ OA = \frac{h}{2} + \frac{w^2}{8h} \\ \cos \angle POA = \frac{OA-h}{OA} = \frac{ \frac{w^2}{8h} - \frac{h}{2} } { \frac{w^2}{8h}+\frac{h}{2} } $$ and the length of the arc from $A$ to $Q$ and then from $Q$ to $B$ is $$ 2 (OA) \cos^{-1}\frac{w^2-4h^2}{w^2+4h^2} = 2 \frac{4h^2+w^2}{8h}\cos^{-1}\frac{w^2-4h^2}{w^2+4h^2} $$ In the example given, $w = 10, h=5$ and the arc length is $$ 2 \frac{4\cdot 25 +100}{8\cdot 5}\cos^{-1}\frac{100-4\cdot 25}{100-4\cdot 25} =2\cdot \frac{200}{40} \cos^{-1}(0) = 2\cdot 5 \cdot \frac{\pi}{2} = 5\pi $$ If, instead, the height were 3 cm then the circular arc length would be $$ \frac{34}{3}\cos^{-1}\frac{8}{17} \approx 12.25$$