*Notes to a video lecture on http://www.unizor.com*

__Definite Integrals -__

Length of Curve

Length of Curve

Our task is to calculate a length of a curve on a coordinate plane, defined parametrically as a set of points (

*), where*

**x(t),y(t)***and*

**x(t)***are smooth functions of parameter*

**y(t)***∈[*

**t***].*

**a,b**This is one of the typical integration problem and can be approached similarly to our approach to calculate the area under a curve.

Let's break down a segment [

*] into*

**a,b***intervals by points*

**N***, which results in corresponding breaking of our curve into smaller pieces by points*

**a=t**_{0}, t_{1},...t_{N}=b(

*)=(*

**x**_{0},y_{0}*),*

**x(t**_{0}),y(t_{0})(

*)=(*

**x**_{1},y_{1}*),...*

**x(t**_{1}),y(t_{1})...

(

*)=(*

**x**_{N},y_{N}*),*

**x(t**_{N}),y(t_{N})Each small piece of a curve can be approximated with a straight line from one of its ends to another, and this approximation will be better with the density of break points increasing.

So, the

*-th piece of a curve is approximated with a segment from point (*

**n***) to point (*

**x**_{n−1},y_{n−1}*).*

**x**_{n},y_{n}The length of this

*-th piece of a curve*

**n***can be calculated using the regular formula of a length of a segment between two points on a plane:*

**L**_{n}

**L**_{n}² = (x_{n}−x_{n−1})²+(y_{n}−y_{n−1})²Taking into account that both coordinates are functions of parameter

*, we can express it differently:*

**t**

**L**

+ (y(t_{n}² = (x(t_{n})−x(t_{n−1}))² ++ (y(t

_{n})−y(t_{n−1}))²Notice that

**x(t**_{n})−x(t_{n−1}) ≅ x^{I}(t_{n})·(t_{n}−t_{n−1})where

*is a derivative of function*

**x**^{I}(t)*by*

**x(t)***and approximation gets better and better as we increase the density of points*

**t***.*

**t**_{n}So, we can express the length of the

*-th piece of a curve as*

**n***[*

**L**_{n}² ≅*]*

**x**^{I}(t_{n})² + y^{I}(t_{n})²

**·**

· (t· (t

_{n}−t_{n−1})²From this we get

*SQRT[*

**L**_{n}≅*]*

**x**^{I}(t_{n})² + y^{I}(t_{n})²

**·**

· (t· (t

_{n}−t_{n−1})As before, we use Δ

*.*

**t**_{n}=t_{n}−t_{n−1}Now the length of a curve can be approximated by this sum:

*≅ Σ*

**L**_{n∈[1,N]}

*≅*

**L**_{n}≅ Σ

_{n∈[1,N]}

*·Δ*

**f(x**_{n})

**t**_{n}where

*=SQRT[*

**f(t)***]*

**x**^{I}(t)² + y^{I}(t)²Recall the

**definition**of the

*definite integral*:

**∫**d_{a}^{b}f(x)**x =**

=lim=

**Σ**

_{i∈[1,N]}

*f(x*Δ_{i})·*x*_{i}where

**Δ**represents partitioning of segment [

*x*_{i}=x_{i}−x_{i−1}*] into*

**a,b***parts, and it is assumed that the widest interval*

**N****Δ**is shrinking to zero by length as

*x*_{i}*.*

**N→∞**Clearly, we are dealing with an integral in our case. The sum of pieces of our curve represents Riemann sum, and the limit of this sum, as the density of points

*increases, is the following integral:*

**t**_{n}*SQRT[*

**∫**_{a}^{b}*]*

**x**^{I}(t)² + y^{I}(t)²*d*

**t**So, the length of the curve on a coordinate plane, defined parametrically by coordinate functions

*and*

**x(t)***, where*

**y(t)***∈[*

**t***], equals to*

**a,b***SQRT[*

**∫**_{a}^{b}*]*

**x**^{I}(t)² + y^{I}(t)²*d*

**t**Let's apply this to a couple of practical problems.

*Problem 1*

Calculate the length of a circle of radius

*.*

**R***Solution*

To define a circle parametrically, let's choose an angle between its radius and a positive direction of the X-axis as a parameter

*∈[*

**t***].*

**0,2π**Then the X-coordinate of a point on a circle, whose radius forms an angle

*with positive direction of the X-axis, equals to*

**t***and Y-coordinate equals to*

**x(t)=R·cos(t)***.*

**y(t)=R·sin(t)**Now we can use the above formula to calculate the length of circle.

*SQRT[*

**∫**_{0}^{2π}*]*

**x**^{I}(t)² + y^{I}(t)²*d*=

**t**=

*SQRT[*

**∫**_{0}^{2π}*]*

**R²sin²(t) +**

+ R²cos²(t)+ R²cos²(t)

*d*=

**t**=

*=*

**∫**d_{0}^{2π}R**t**=

**R·2π − R·0 = 2πR**As we see, the length, as we counted, equals to the one in a classical formula for the length of the circle. No surprise here.

The end.

*Problem 2*

Calculate the length of an

*astroid*given parametrically as

*,*

**x=R·cos³(t)**

**y=R·sin³(t)**where

*∈[*

**t***]*

**0,2π***Solution*

Astroid looks like this:

Its vertices are at distance

*from the origin, and its four arcs correspond to parameter*

**R***changing in each quadrant from*

**t***to*

**0***.*

**2π**Using the symmetry, lets' calculate the length of only one arc for

*∈[*

**t***] and then multiply it by four.*

**0,π/2**First, calculate the derivatives by

*:*

**t**

**x**^{I}(t) = −3R·cos²(t)·sin(t)

**y**^{I}(t) = 3R·sin²(t)·cos(t)Now we use the formula for the length of a curve.

SQRT[

*] =*

**x**^{I}(t)² + y^{I}(t)²=

*=*

**3R·|sin(t)·cos(t)|**=

**(3/2)R·|sin(2t)|**This should be integrated within margins

*∈[*

**t***].*

**0,π/2**Within these margins

*changes from*

**2t***to*

**0***and*

**π***is non-negative, so we can drop absolute value.*

**sin(t)**So, we need to find the following integral:

**∫**d_{0}^{π/2}(3/2)R·sin(2t)**t**Indefinite integral of

*is*

**sin(2t)***, which means that we have to calculate the following*

**−cos(2t)/2***=*

**(−cos(π)/2) − (−cos(0)/2)**=

*.*

**1/2 + 1/2 = 1**Therefore, the length of one quarter of an astroid equals to

*and the length of an entire astroid is*

**(3/2)R·1 = (3/2)R***.*

**(3/2)R·4=6R**The end.

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