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

Multiplication of two unlike terms with opposite signs, The product on mult...

The product on multiplying - 4bc with 2a is - 8abc. That is, a term with minus sign multiplied with a term having a positive term gives a product which has a minus sign. On the

Integrals involving quadratics - integration techniques, Integrals Involvin...

Integrals Involving Quadratics To this point we have seen quite some integrals which involve quadratics.  Example of Integrals Involving Quadratics is as follow: ∫ (x / x 2

Trigonometric approximation grid, With a compass draw the arc associated wi...

With a compass draw the arc associated with a 720° angle, it looks like a circle. With a protractor, label the angle in multiples of 45° and 30° up to 720°.  Notice 30° and 390° ar

Linear Equations of Parallel Lines, A line has the equation 2y=-3x+1. Find...

A line has the equation 2y=-3x+1. Find an equation of a line parallel to this line that has a y-intercept of -2.

VAM, applications of VAM.

applications of VAM.

Solve the right triangle, 1. Solve the right triangle. B = 135     c = 3...

1. Solve the right triangle. B = 135     c = 3.72 A  ≈ ____°    (round to the nearest tenth as needed) 2.  Solve the right triangle, where  a =4 and b =10 The length of

Find the maximum and minimum brightness values, Variable stars are ones who...

Variable stars are ones whose brightness varies periodically. One of the most visible is R Leonis; its brightness is modelled by the function where t is measured in days.

Chanllenge, apzza driver delivered 27 pizzas in one night he delivered more...

apzza driver delivered 27 pizzas in one night he delivered more then one pizza to only one house . every other hhouse he only delivered pizza to 18 houses . how many pizzas did he

Extreme value theorem, Extreme Value Theorem : Assume that f ( x ) is cont...

Extreme Value Theorem : Assume that f ( x ) is continuous on the interval [a,b] then there are two numbers a ≤ c, d ≤ b so that f (c ) is an absolute maximum for the function and

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