Arc length - applications of integrals, Mathematics

Assignment Help:

Arc Length - Applications of integrals

In this part we are going to look at determining the arc length of a function.  As it's sufficiently easy to derive the formulas that we'll utilize in this section we will derive one of them and leave the other to you to derive.

We want to find out the length of the continuous function

y = f (x) on the interval [a, b].

Primarily we'll need to find out the length of the curve. We'll do this by dividing the interval up into n equal subintervals each of width Δx and we'll indicate the point on the curve at each point by Pi. We can then estimate the curve by a series of straight lines connecting the points. Now Here is a sketch of this situation for n = 9.

132_Arc Length - Applications of integrals 4.png

Now indicate the length of every line segments by then be approximately, |Pi -1  Pi|  and the length of the curve will

206_Arc Length - Applications of integrals 3.png

and after that we can obtain the exact length by taking n larger and larger.  Alternatively, the exact length will be,

1974_Arc Length - Applications of integrals 2.png

Now here, let's get a good grasp on the length of each of these line segments. Very first, on each segment let's illustrate Δyi = yi - yi-1 = f (xi) - f (xi-1) . After that we can calculate directly the length of the line segments like this:

|Pi-1 Pi| = √ ((xi - xi-1)2 + (yi - yi-1)2)

= √(Δx2 +Δy2i).

By using the Mean Value Theorem we make out that on the interval [xi-1, xi] there is a point x*i that is why,

F (xi) - f (xi-1)

= f' (x*i) (xi - xi-1)

Δyi= f' (x*i)Δx

Hence, the length can now be written as,

|Pi-1 Pi| = √ ((xi - xi-1)2 + (yi - yi-1)2)

= √(Δx2 +[f' (xi*)]2 Δx2 )

= √ (1 + [f' (xi*)]Δx)

The exact length of the curve is then,

2388_Arc Length - Applications of integrals 1.png

Though, by using the definition of the definite integral, this is nothing much more than,

L - ∫ba√ (1+[f' (x)]2 dx)

A little more suitable notation (according to me) is the following.

L = ∫ba √ (1 + (dy/dx)2 dx)

In a identical way we can also derive a formula for x = h(y) on [c,d]. This formula is,

L - ∫bc√ (1+[h' (y)]2 dy)

bc √ (1 + (dx/dy)2 dy)

Once Again, the second form is possibly a much more convenient.

Note: the variation in the derivative under the square root! Don't get so confused. With one we distinguish with respect to x and with the other we distinguish with respect to y. One way to maintain the two straight is to note that the differential in the "denominator" of the derivative will match up along with the differential in the integral. This is one of the causes why the second form is a little much more suitable.

Previous to we work any instance we need to make a small change in notation. In place of having two formulas for the arc length of a function we are going to decrease it, in part, to a single formula. From this point on we are going to make use of the following formula for the length of the curve.


Related Discussions:- Arc length - applications of integrals

Reduction of order - fundamental set of solutions, Given that 2t 2 y′′ ...

Given that 2t 2 y′′ + ty′ - 3 y = 0 Show that this given solution are form a fundamental set of solutions for the differential equation? Solution The two solutions f

Find the number of students side of the square, A teacher on attempting to ...

A teacher on attempting to arrange the students for mass drill in the form of a solid square found that 24 students were left over. When he increased the size of the square by one

Compounding and Simple Interest, A painting was purchased 11 years ago for ...

A painting was purchased 11 years ago for $26900. It has just been sold for $78000. Calculate the flat rate of appreciation p.a.

Conjugate of the complex number, The conjugate of the complex number a + b ...

The conjugate of the complex number a + b i is the complex number a - b i .  In other terms, it is the original complex number along the sign on the imaginary part changed.  Here

Work in volume problems, Work : It is the last application of integr...

Work : It is the last application of integral which we'll be looking at under this course. In this section we'll be looking at the amount of work which is done through a forc

Precalculus, describe the end behavior of the following function using Limi...

describe the end behavior of the following function using Limit notation f(x)= 2x-1/x-1

Write Your Message!

Captcha
Free Assignment Quote

Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd