Finding the lenght of an Arc on a graph

Length of an Arc

This Photo by Unknown Author is licensed under CC BY-SA

 

What is an Arc?

-          A non linear line plotted on a graph (shown in the picture is the plot of y=x²)

-          A segment of a cicles circumference

This post will deal with the first definition and will play a crucial rule in explaining the second post

How to find the length of a line?

Rule: no rulers allowed

With no rulers allowed, you would be scratching you head. However maths has its knight in Shining Armour to reveal.

(Drum roll…)

The pythagoras Theorem

Symbols:

x- Horizontal length

y- Vertical length

Expressions needed: 

	C2 =A2 + B2

^{For any equation the gradient at any given point can be found by the method of differentiation.}

^{Information of differentiation can be found at:} www.dummies.com/education/math/differentiating-simple-terms-and-functions-for-as-a-level-maths/

^Integration

The addition of small values under a curve.

More information on integration can be found at:

http://www.dummies.com/education/math/calculus/how-integration-works-its-just-fancy-addition/ 

^Method

^{1.           The current method followed is finding many and than adjusting the length.}

^{2.           However, taking the fine length requires the taking of fine infentesimal points.}

^{3.           The new method is based on integration of the lenghts}

^{4.           The answer can be found in the pythagoras equation itself where the equation can be translated into many different forms shown below.}  

 f^'(x) is the differential derivative for any equation, and in turn, means the gradient at point X. So the equation can be turned into

                                                                                      

                   When put into an integration equation, the equation looks like this:

Worked example:

Equation:

y=x²

F^’(x): 2x

Integration of (1+4x²)^0.5 with limits of x=0 and x=1

When integrated the equation comes to:

(1/6)(1+m²)^1.5

When x=1

  Y= 1.86

When x=0

  Y= 0.1667

Subtration of limits leads to: (1.86-0.1667)=1.6933

Use of Pythagoras theorem:

C² = 1+1

C²=2

C=2^0.5

C=1.41

Conclusion

Applying the pythogoras theorem directly between the two endpoints, leads to an extremely inaccurate answer as it assumes that the whole line is straight while in fact the line is undergoing curvature at very small points. However applying the pythagoras theorem by integration considers the distance between two infentessimly close points as a straight line. This allows a clearer picture as this adds up smaller distances between the two points

Credits

Reyhan Mehta: The guy who introduced me to this problem in class on the 6th of December 2017.