how to find arc length of a circle

Distance along a curve

When rectified, the curve gives a directly line segment with the aforementioned length as the curve's arc length.

Arc length is the distance between two points along a section of a curve.

Determining the length of an irregular arc segment is likewise called rectification of a bend. If the rectification of a curve results in a finite number (then the bend has a finite length), and so the curve is said to be rectifiable. The advent of minute calculus led to a general formula that provides airtight-course solutions in some cases.

General approach [edit]

Approximation to a bend by multiple linear segments

A curve in the plane can be approximated by connecting a finite number of points on the curve using line segments to create a polygonal path. Since it is straightforward to calculate the length of each linear segment (using the Pythagorean theorem in Euclidean infinite, for example), the total length of the approximation can exist constitute by summation of the lengths of each linear segment; that approximation is known as the (cumulative) chordal altitude.^[ane]

If the curve is not already a polygonal path, using a progressively larger number of segments of smaller lengths will result in better approximations. The lengths of the successive approximations will not decrease and may keep increasing indefinitely, merely for polish curves they will tend to a finite limit equally the lengths of the segments get arbitrarily small.

For some curves there is a smallest number ${\displaystyle 50}$ that is an upper bound on the length of all polygonal approximations. These curves are chosen rectifiable and the arc length is defined as the number ${\displaystyle L}$ .

Formula for a smooth curve [edit]

Permit ${\displaystyle f\colon [a,b]\to \mathbb {R} ^{n}}$ be an injective and continuously differentiable (i.east., the derivative is a continuous part) function. The length of the curve defined by ${\displaystyle f}$ tin can be defined as the limit of the sum of linear segment lengths for a regular partition of ${\displaystyle [a,b]}$ as the number of segments approaches infinity. This means

$${\displaystyle L(f)=\lim _{N\to \infty }\sum _{i=1}^{Due north}{\bigg |}f(t_{i})-f(t_{i-1}){\bigg |}}$$

where ${\displaystyle t_{i}=a+i(b-a)/N=a+i\Delta t}$ with ${\displaystyle \Delta t={\frac {b-a}{N}}=t_{i}-t_{i-1}}$ for ${\displaystyle i=0,i,\dotsc ,Northward.}$ This definition is equivalent to the standard definition of arc length as an integral:

$${\displaystyle Fifty(f)=\lim _{Northward\to \infty }\sum _{i=i}^{N}{\bigg |}f(t_{i})-f(t_{i-1}){\bigg |}=\lim _{Due north\to \infty }\sum _{i=i}^{Due north}\left|{\frac {f(t_{i})-f(t_{i-1})}{\Delta t}}\right|\Delta t=\int _{a}^{b}{\Big |}f'(t){\Large |}\ dt.}$$

The last equality is proved by the post-obit steps:

1. The 2nd cardinal theorem of calculus shows ${\displaystyle f(t_{i})-f(t_{i-1})=\int _{t_{i-1}}^{t_{i}}f'(t)\ dt=\Delta t\int _{0}^{1}f'(t_{i-ane}+\theta (t_{i}-t_{i-ane}))\ d\theta }$ where ${\displaystyle t=t_{i-one}+\theta (t_{i}-t_{i-i})}$ over ${\displaystyle \theta \in [0,ane]}$ maps to ${\displaystyle [t_{i-1},t_{i}]}$ and ${\displaystyle dt=(t_{i}-t_{i-1})d\theta =\Delta td\theta }$ . In the beneath step, the following equivalent expression is used.

   $${\displaystyle {\frac {f(t_{i})-f(t_{i-1})}{\Delta t}}=\int _{0}^{1}f'(t_{i-ane}+\theta (t_{i}-t_{i-i}))\ d\theta .}$$

   

2. The function ${\displaystyle \left|f'\correct|}$ is a continuous function from a closed interval ${\displaystyle [a,b]}$ to the fix of real numbers, thus it is uniformly continuous co-ordinate to the Heine–Cantor theorem, so at that place is a positive real and monotonically non-decreasing function ${\displaystyle \delta (\varepsilon )}$ of positive real numbers ${\displaystyle \varepsilon }$ such that ${\displaystyle \Delta t<\delta (\varepsilon )}$ implies ${\displaystyle \left|\left|f'(t_{i-1}+\theta (t_{i}-t_{i-ane}))\right|-\left|f'(t_{i})\right|\right|<\varepsilon }$ where ${\displaystyle \Delta t=t_{i}-t_{i-ane}}$ and ${\displaystyle \theta \in [0,1]}$ . Let's consider the limit ${\displaystyle {\ce {N\to \infty }}}$ of the post-obit formula,

   $${\displaystyle \sum _{i=1}^{North}\left|{\frac {f(t_{i})-f(t_{i-i})}{\Delta t}}\correct|\Delta t-\sum _{i=1}^{N}\left|f'(t_{i})\right|\Delta t.}$$

   

   With the above step result, it becomes

   $${\displaystyle \sum _{i=one}^{Northward}\left|\int _{0}^{1}f'(t_{i-1}+\theta (t_{i}-t_{i-1}))\ d\theta \right|\Delta t-\sum _{i=i}^{North}\left|f'(t_{i})\right|\Delta t.}$$

   

   Terms are rearranged and so that it becomes

   $${\displaystyle \Delta t\sum _{i=1}^{Northward}\left(\left|\int _{0}^{1}f'(t_{i-one}+\theta (t_{i}-t_{i-i}))\ d\theta \right|-\int _{0}^{1}\left|f'(t_{i})\correct|d\theta \correct)\leqq \Delta t\sum _{i=1}^{N}\left(\int _{0}^{1}\left|f'(t_{i-1}+\theta (t_{i}-t_{i-one}))\correct|\ d\theta -\int _{0}^{1}\left|f'(t_{i})\right|d\theta \right)=\Delta t\sum _{i=1}^{N}\int _{0}^{1}\left|f'(t_{i-ane}+\theta (t_{i}-t_{i-i}))\right|-\left|f'(t_{i})\correct|\ d\theta }$$

   

   where in the leftmost side ${\displaystyle \left|f'(t_{i})\right|=\int _{0}^{1}\left|f'(t_{i})\right|d\theta }$ is used. By ${\displaystyle \left|\left|f'(t_{i-one}+\theta (t_{i}-t_{i-1}))\right|-\left|f'(t_{i})\right|\right|<\varepsilon }$ for ${\displaystyle N>(b-a)/\delta (\varepsilon )}$ so that ${\displaystyle \Delta t<\delta (\varepsilon )}$ , it becomes

   ${\displaystyle \Delta t\sum _{i=i}^{Northward}\left(\left|\int _{0}^{i}f'(t_{i-1}+\theta (t_{i}-t_{i-1}))\ d\theta \right|-\left|f'(t_{i})\right|\right)<\varepsilon Due north\Delta t}$

   

   with ${\displaystyle \left|f'(t_{i})\right|=\int _{0}^{1}\left|f'(t_{i})\right|d\theta }$ , ${\displaystyle \varepsilon N\Delta t=\varepsilon (b-a)}$ , and ${\displaystyle North>(b-a)/\delta (\varepsilon )}$ . In the limit ${\displaystyle Due north\to \infty ,}$ ${\displaystyle \delta (\varepsilon )\to 0}$ and then ${\displaystyle \varepsilon \to 0}$ thus the left side of ${\displaystyle <}$ approaches to ${\displaystyle 0}$ . In other words, ${\displaystyle \sum _{i=one}^{N}\left|{\frac {f(t_{i})-f(t_{i-1})}{\Delta t}}\right|\Delta t=\sum _{i=1}^{N}\left|f'(t_{i})\correct|\Delta t}$ in this limit, and the right side of this equality is but the Riemann integral of ${\displaystyle \left|f'(t)\correct|}$ on ${\displaystyle [a,b].}$ This definition of arc length shows that the length of a curve represented past a continuously differentiable office ${\displaystyle f:[a,b]\to \mathbb {R} ^{n}}$ on ${\displaystyle [a,b]}$ is always finite. In other words, the curve is e'er rectifiable.

The definition of arc length of a smoothen curve as the integral of the norm of the derivative is equivalent to the definition

$${\displaystyle Fifty(f)=\sup \sum _{i=1}^{N}{\bigg |}f(t_{i})-f(t_{i-1}){\bigg |}}$$

where the supremum is taken over all possible partitions ${\displaystyle a=t_{0}<t_{ane}<\dots <t_{Northward-1}<t_{Due north}=b}$ of ${\displaystyle [a,b].}$ ^[two] This definition is as well valid if ${\displaystyle f}$ is merely continuous, not differentiable.

A curve tin can be parameterized in infinitely many ways. Let ${\displaystyle \varphi :[a,b]\to [c,d]}$ be any continuously differentiable bijection. Then ${\displaystyle one thousand=f\circ \varphi ^{-1}:[c,d]\to \mathbb {R} ^{n}}$ is another continuously differentiable parameterization of the bend originally defined past ${\displaystyle f.}$ The arc length of the bend is the aforementioned regardless of the parameterization used to define the curve:

$${\displaystyle {\begin{aligned}L(f)&=\int _{a}^{b}{\Big |}f'(t){\Big |}\ dt=\int _{a}^{b}{\Big |}g'(\varphi (t))\varphi '(t){\Big |}\ dt\\&=\int _{a}^{b}{\Big |}1000'(\varphi (t)){\Big |}\varphi '(t)\ dt\quad {\text{in the case of }}\varphi {\text{ existence not-decreasing}}\\&=\int _{c}^{d}{\Big |}one thousand'(u){\Big |}\ du\quad {\text{using integration by exchange}}\\&=L(chiliad).\end{aligned}}}$$

Finding arc lengths by integration [edit]

If a planar bend in ${\displaystyle \mathbb {R} ^{2}}$ is divers by the equation ${\displaystyle y=f(x),}$ where ${\displaystyle f}$ is continuously differentiable, then it is simply a special case of a parametric equation where ${\displaystyle 10=t}$ and ${\displaystyle y=f(t).}$ The Euclidean distance of each minute segment of the arc can be given by:

$${\displaystyle {\sqrt {dx^{ii}+dy^{2}}}={\sqrt {1+\left({\frac {dy}{dx}}\right)^{ii}\,}}dx.}$$

The arc length is then given by:

$${\displaystyle south=\int _{a}^{b}{\sqrt {1+\left({\frac {dy}{dx}}\correct)^{2}\,}}dx.}$$

Curves with closed-form solutions for arc length include the catenary, circle, cycloid, logarithmic spiral, parabola, semicubical parabola and straight line. The lack of a closed form solution for the arc length of an elliptic and hyperbolic arc led to the development of the elliptic integrals.

Numerical integration [edit]

In nearly cases, including even simple curves, there are no airtight-course solutions for arc length and numerical integration is necessary. Numerical integration of the arc length integral is unremarkably very efficient. For instance, consider the problem of finding the length of a quarter of the unit circle by numerically integrating the arc length integral. The upper half of the unit circle can be parameterized as ${\displaystyle y={\sqrt {i-ten^{two}}}.}$ The interval ${\displaystyle x\in \left[-{\sqrt {2}}/ii,{\sqrt {ii}}/2\correct]}$ corresponds to a quarter of the circle. Since ${\textstyle dy/dx=-x/{\sqrt {1-x^{2}}}}$ and ${\displaystyle 1+(dy/dx)^{2}=i/\left(1-x^{2}\correct),}$ the length of a quarter of the unit circumvolve is

$${\displaystyle \int _{-{\sqrt {ii}}/2}^{{\sqrt {2}}/ii}{\frac {dx}{\sqrt {1-10^{ii}}}}\,.}$$

The fifteen-betoken Gauss–Kronrod rule estimate for this integral of 1.570796 326 808 177 differs from the true length of

$${\displaystyle {\Big [}\arcsin x{\Big ]}_{-{\sqrt {two}}/2}^{{\sqrt {ii}}/2}={\frac {\pi }{2}}}$$

by 1.3×ten⁻¹¹ and the 16-point Gaussian quadrature rule guess of 1.570796 326 794 727 differs from the truthful length by only 1.7×10⁻¹³ . This means it is possible to evaluate this integral to most automobile precision with simply xvi integrand evaluations.

Curve on a surface [edit]

Allow ${\displaystyle \mathbf {x} (u,v)}$ be a surface mapping and permit ${\displaystyle \mathbf {C} (t)=(u(t),v(t))}$ be a bend on this surface. The integrand of the arc length integral is ${\displaystyle \left|\left(\mathbf {10} \circ \mathbf {C} \right)'(t)\right|.}$ Evaluating the derivative requires the chain dominion for vector fields:

$${\displaystyle D(\mathbf {10} \circ \mathbf {C} )=(\mathbf {x} _{u}\ \mathbf {10} _{five}){\binom {u'}{v'}}=\mathbf {x} _{u}u'+\mathbf {x} _{five}v'.}$$

The squared norm of this vector is

$${\displaystyle \left(\mathbf {10} _{u}u'+\mathbf {x} _{v}5'\right)\cdot (\mathbf {x} _{u}u'+\mathbf {x} _{v}v')=g_{11}\left(u'\right)^{ii}+2g_{12}u'v'+g_{22}\left(v'\right)^{two}}$$

(where ${\displaystyle g_{ij}}$ is the first fundamental course coefficient), so the integrand of the arc length integral tin exist written as ${\displaystyle {\sqrt {g_{ab}\left(u^{a}\right)'\left(u^{b}\right)'\,}}}$ (where ${\displaystyle u^{1}=u}$ and ${\displaystyle u^{2}=v}$ ).

Other coordinate systems [edit]

Let ${\displaystyle \mathbf {C} (t)=(r(t),\theta (t))}$ be a curve expressed in polar coordinates. The mapping that transforms from polar coordinates to rectangular coordinates is

$${\displaystyle \mathbf {x} (r,\theta )=(r\cos \theta ,r\sin \theta ).}$$

The integrand of the arc length integral is ${\displaystyle \left|\left(\mathbf {x} \circ \mathbf {C} \right)'(t)\right|.}$ The concatenation dominion for vector fields shows that ${\displaystyle D(\mathbf {x} \circ \mathbf {C} )=\mathbf {x} _{r}r'+\mathbf {x} _{\theta }\theta '.}$ So the squared integrand of the arc length integral is

$${\displaystyle \left(\mathbf {x_{r}} \cdot \mathbf {x_{r}} \right)\left(r'\right)^{2}+2\left(\mathbf {ten} _{r}\cdot \mathbf {ten} _{\theta }\right)r'\theta '+\left(\mathbf {x} _{\theta }\cdot \mathbf {x} _{\theta }\right)\left(\theta '\correct)^{2}=\left(r'\right)^{2}+r^{2}\left(\theta '\right)^{two}.}$$

So for a curve expressed in polar coordinates, the arc length is

$${\displaystyle \int _{t_{1}}^{t_{2}}{\sqrt {\left({\frac {dr}{dt}}\right)^{2}+r^{2}\left({\frac {d\theta }{dt}}\right)^{ii}\,}}dt=\int _{\theta (t_{1})}^{\theta (t_{two})}{\sqrt {\left({\frac {dr}{d\theta }}\right)^{ii}+r^{two}\,}}d\theta .}$$

At present let ${\displaystyle \mathbf {C} (t)=(r(t),\theta (t),\phi (t))}$ be a bend expressed in spherical coordinates where ${\displaystyle \theta }$ is the polar angle measured from the positive ${\displaystyle z}$ -axis and ${\displaystyle \phi }$ is the azimuthal bending. The mapping that transforms from spherical coordinates to rectangular coordinates is

$${\displaystyle \mathbf {x} (r,\theta ,\phi )=(r\sin \theta \cos \phi ,r\sin \theta \sin \phi ,r\cos \theta ).}$$

Using the concatenation rule again shows that ${\displaystyle D(\mathbf {x} \circ \mathbf {C} )=\mathbf {ten} _{r}r'+\mathbf {x} _{\theta }\theta '+\mathbf {x} _{\phi }\phi '.}$ All dot products ${\displaystyle \mathbf {ten} _{i}\cdot \mathbf {ten} _{j}}$ where ${\displaystyle i}$ and ${\displaystyle j}$ differ are zero, so the squared norm of this vector is

$${\displaystyle \left(\mathbf {x} _{r}\cdot \mathbf {ten} _{r}\right)\left(r'^{2}\right)+\left(\mathbf {10} _{\theta }\cdot \mathbf {x} _{\theta }\right)\left(\theta '\correct)^{two}+\left(\mathbf {x} _{\phi }\cdot \mathbf {x} _{\phi }\right)\left(\phi '\right)^{2}=\left(r'\right)^{2}+r^{2}\left(\theta '\correct)^{two}+r^{2}\sin ^{2}\theta \left(\phi '\right)^{2}.}$$

So for a curve expressed in spherical coordinates, the arc length is

$${\displaystyle \int _{t_{one}}^{t_{2}}{\sqrt {\left({\frac {dr}{dt}}\correct)^{ii}+r^{2}\left({\frac {d\theta }{dt}}\correct)^{2}+r^{2}\sin ^{2}\theta \left({\frac {d\phi }{dt}}\right)^{two}\,}}dt.}$$

A very like calculation shows that the arc length of a curve expressed in cylindrical coordinates is

$${\displaystyle \int _{t_{1}}^{t_{2}}{\sqrt {\left({\frac {dr}{dt}}\right)^{2}+r^{2}\left({\frac {d\theta }{dt}}\right)^{2}+\left({\frac {dz}{dt}}\correct)^{2}\,}}dt.}$$

Unproblematic cases [edit]

Arcs of circles [edit]

Arc lengths are denoted past south, since the Latin give-and-take for length (or size) is spatium.

In the following lines, ${\displaystyle r}$ represents the radius of a circumvolve, ${\displaystyle d}$ is its diameter, ${\displaystyle C}$ is its circumference, ${\displaystyle s}$ is the length of an arc of the circle, and ${\displaystyle \theta }$ is the bending which the arc subtends at the middle of the circle. The distances ${\displaystyle r,d,C,}$ and ${\displaystyle s}$ are expressed in the same units.

• ${\displaystyle C=2\pi r,}$ which is the same as ${\displaystyle C=\pi d.}$ This equation is a definition of ${\displaystyle \pi .}$
• If the arc is a semicircle, then ${\displaystyle s=\pi r.}$
• For an arbitrary round arc:
  • If ${\displaystyle \theta }$ is in radians and so ${\displaystyle s=r\theta .}$ This is a definition of the radian.
  • If ${\displaystyle \theta }$ is in degrees, so ${\displaystyle due south={\frac {\pi r\theta }{180^{\circ }}},}$ which is the aforementioned equally ${\displaystyle southward={\frac {C\theta }{360^{\circ }}}.}$
  • If ${\displaystyle \theta }$ is in grads (100 grads, or grades, or gradians are one right-angle), then ${\displaystyle due south={\frac {\pi r\theta }{200{\text{ grad}}}},}$ which is the same as ${\displaystyle due south={\frac {C\theta }{400{\text{ grad}}}}.}$
  • If ${\displaystyle \theta }$ is in turns (i turn is a complete rotation, or 360°, or 400 grads, or ${\displaystyle 2\pi }$ radians), and so ${\displaystyle s=C\theta /1{\text{ plow}}}$ .

Great circles on World [edit]

Two units of length, the nautical mile and the metre (or kilometre), were originally defined so the lengths of arcs of great circles on the World'due south surface would be simply numerically related to the angles they subtend at its centre. The elementary equation ${\displaystyle s=\theta }$ applies in the following circumstances:

The lengths of the distance units were chosen to make the circumference of the Globe equal 40000 kilometres, or 21600 nautical miles. Those are the numbers of the corresponding angle units in one complete plow.

Those definitions of the metre and the nautical mile take been superseded by more precise ones, simply the original definitions are still accurate enough for conceptual purposes and some calculations. For instance, they imply that one kilometre is exactly 0.54 nautical miles. Using official modern definitions, one nautical mile is exactly 1.852 kilometres,^[3] which implies that one kilometre is about 0.539956 80 nautical miles.^[4] This modern ratio differs from the one calculated from the original definitions past less than 1 part in 10,000.

Other simple cases [edit]

• Archimedean spiral § Arc length
• Cycloid § Arc length
• Ellipse § Arc length
• Helix § Arc length
• Parabola § Arc length
• Sine and cosine § Arc length
• Triangle wave § Arc length

Historical methods [edit]

Antiquity [edit]

For much of the history of mathematics, even the greatest thinkers considered it incommunicable to compute the length of an irregular arc. Although Archimedes had pioneered a way of finding the surface area below a curve with his "method of exhaustion", few believed it was even possible for curves to have definite lengths, as exercise straight lines. The starting time ground was cleaved in this field, as information technology often has been in calculus, by approximation. People began to inscribe polygons within the curves and compute the length of the sides for a somewhat accurate measurement of the length. By using more segments, and by decreasing the length of each segment, they were able to obtain a more and more accurate approximation. In detail, by inscribing a polygon of many sides in a circle, they were able to find gauge values of π.^{[5] [6]}

17th century [edit]

In the 17th century, the method of exhaustion led to the rectification by geometrical methods of several transcendental curves: the logarithmic spiral by Evangelista Torricelli in 1645 (some sources say John Wallis in the 1650s), the cycloid by Christopher Wren in 1658, and the catenary past Gottfried Leibniz in 1691.

In 1659, Wallis credited William Neile'due south discovery of the commencement rectification of a nontrivial algebraic curve, the semicubical parabola.^[seven] The accompanying figures appear on page 145. On page 91, William Neile is mentioned every bit Gulielmus Nelius.

Integral form [edit]

Before the full formal development of calculus, the footing for the modern integral form for arc length was independently discovered by Hendrik van Heuraet and Pierre de Fermat.

In 1659 van Heuraet published a structure showing that the problem of determining arc length could be transformed into the trouble of determining the area nether a curve (i.eastward., an integral). Every bit an example of his method, he determined the arc length of a semicubical parabola, which required finding the expanse nether a parabola.^[8] In 1660, Fermat published a more full general theory containing the same result in his De linearum curvarum cum lineis rectis comparatione dissertatio geometrica (Geometric dissertation on curved lines in comparison with straight lines).^[nine]

Fermat'due south method of determining arc length

Building on his previous work with tangents, Fermat used the curve

${\displaystyle y=ten^{\frac {iii}{two}}\,}$

whose tangent at x = a had a slope of

${\displaystyle {3 \over ii}a^{\frac {1}{2}}}$

so the tangent line would have the equation

${\displaystyle y={three \over 2}a^{\frac {one}{2}}(x-a)+f(a).}$

Next, he increased a by a small amount to a + ε, making segment AC a relatively good approximation for the length of the curve from A to D. To find the length of the segment AC, he used the Pythagorean theorem:

${\displaystyle {\begin{aligned}AC^{2}&=AB^{two}+BC^{two}\\&=\varepsilon ^{2}+{ix \over 4}a\varepsilon ^{ii}\\&=\varepsilon ^{two}\left(ane+{9 \over 4}a\right)\terminate{aligned}}}$

which, when solved, yields

${\displaystyle AC=\varepsilon {\sqrt {1+{nine \over 4}a\,}}.}$

In order to approximate the length, Fermat would sum up a sequence of short segments.

Curves with infinite length [edit]

As mentioned above, some curves are non-rectifiable. That is, at that place is no upper bound on the lengths of polygonal approximations; the length can exist made arbitrarily large. Informally, such curves are said to have infinite length. There are continuous curves on which every arc (other than a single-bespeak arc) has infinite length. An instance of such a curve is the Koch bend. Another example of a curve with infinite length is the graph of the function defined by f(ten) =x sin(1/x) for any open set with 0 equally one of its delimiters and f(0) = 0. Sometimes the Hausdorff dimension and Hausdorff measure are used to quantify the size of such curves.

Generalization to (pseudo-)Riemannian manifolds [edit]

Permit ${\displaystyle M}$ exist a (pseudo-)Riemannian manifold, ${\displaystyle \gamma :[0,1]\rightarrow Grand}$ a curve in ${\displaystyle M}$ and ${\displaystyle g}$ the (pseudo-) metric tensor.

The length of ${\displaystyle \gamma }$ is defined to be

$${\displaystyle \ell (\gamma )=\int _{0}^{one}{\sqrt {\pm g\left(\gamma '(t),\gamma '(t)\correct)\,}}dt,}$$

where ${\displaystyle \gamma '(t)\in T_{\gamma (t)}K}$ is the tangent vector of ${\displaystyle \gamma }$ at ${\displaystyle t.}$ The sign in the square root is chosen one time for a given curve, to ensure that the square root is a real number. The positive sign is chosen for spacelike curves; in a pseudo-Riemannian manifold, the negative sign may exist called for timelike curves. Thus the length of a curve is a non-negative existent number. Usually no curves are considered which are partly spacelike and partly timelike.

In theory of relativity, arc length of timelike curves (world lines) is the proper time elapsed along the globe line, and arc length of a spacelike curve the proper distance along the curve.

See likewise [edit]

• Arc (geometry)
• Circumference
• Crofton formula
• Elliptic integral
• Geodesics
• Intrinsic equation
• Integral approximations
• Line integral
• Meridian arc
• Multivariable calculus
• Sinuosity

References [edit]

1. ^ Ahlberg; Nilson (1967). The Theory of Splines and Their Applications . Academic Press. p. 51. ISBN9780080955452.
2. ^ Rudin, Walter (1976). Principles of Mathematical Analysis . McGraw-Hill, Inc. pp. 137. ISBN978-0-07-054235-8.
3. ^ Suplee, Curt (ii July 2009). "Special Publication 811". nist.gov.
4. ^ CRC Handbook of Chemistry and Physics, p. F-254
5. ^ Richeson, David (May 2015). "Circular Reasoning: Who Start Proved That C Divided by d Is a Constant?". The College Mathematics Journal. 46 (three): 162–171. doi:10.4169/college.math.j.46.3.162. ISSN 0746-8342. S2CID 123757069.
6. ^ Coolidge, J. L. (February 1953). "The Lengths of Curves". The American Mathematical Monthly. 60 (2): 89–93. doi:x.2307/2308256. JSTOR 2308256.
7. ^ Wallis, John (1659). Tractatus Duo. Prior, De Cycloide et de Corporibus inde Genitis…. Oxford: University Press. pp. 91–96.
8. ^ van Heuraet, Hendrik (1659). "Epistola de transmutatione curvarum linearum in rectas [Letter on the transformation of curved lines into right ones]". Renati Des-Cartes Geometria (2nd ed.). Amsterdam: Louis & Daniel Elzevir. pp. 517–520.
9. ^ Thou.P.E.A.South. (pseudonym of Fermat) (1660). De Linearum Curvarum cum Lineis Rectis Comparatione Dissertatio Geometrica. Toulouse: Arnaud Colomer.

Sources [edit]

• Farouki, Rida T. (1999). "Curves from movement, motion from curves". In Laurent, P.-J.; Sablonniere, P.; Schumaker, L. L. (eds.). Curve and Surface Pattern: Saint-Malo 1999. Vanderbilt Univ. Press. pp. 63–90. ISBN978-0-8265-1356-4.

External links [edit]

Wikimedia Commons has media related to Arc length.

• "Rectifiable curve", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
• The History of Curvature
• Weisstein, Eric Due west. "Arc Length". MathWorld.
• Arc Length by Ed Pegg Jr., The Wolfram Demonstrations Project, 2007.
• Calculus Study Guide – Arc Length (Rectification)
• Famous Curves Index The MacTutor History of Mathematics annal
• Arc Length Approximation by Chad Pierson, Josh Fritz, and Angela Sharp, The Wolfram Demonstrations Project.
• Length of a Curve Experiment Illustrates numerical solution of finding length of a curve.

Source: https://en.wikipedia.org/wiki/Arc_length

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