# Thread: integrals and approximating arc segment lengths

1. ## integrals and approximating arc segment lengths

Given a differentiable function y=f(x) on [a,b], and a regular partition
a=x0<x1<...<xn-1<xn=b
explain how approximating the arc segment length deltaLi by a straight line, then letting n -> infinity, leads to a definite integral for the total length

L = integral from a to b sqrt(1 + (f'(x))^2) dx

i was given this question just after being taught integration by parts ... and i cannot understand anything about it ... if someone could provide me with details on how to solve it ... not just the solution ... that would be great as this is due tomorrow

2. You could try to divide you circle in a pentagon, then hexagon, etc. With many sides, you get a better precision of the circumference.

3. If I understand this correctly, you are supposed to derive/demonstrate how the arc length formula works.

Consider your function between partitions $x_1 , x_2$. Now consider the line between these two points as an approximation of your function. It might be way off, but as the difference between the points gets smaller it will be a better approx.

So how can you represent this line in terms of the two points $(x_1 , y_1),(x_2 , y_2)$? Let the line be a hypotenuse of a right triangle and naturally the legs are $\Delta x , \Delta y$. Thus the line segment is $\sqrt{ (\Delta x)^2+ (\Delta y)^2}$ Now comes a trick. Factor out a delta x squared and you get $L= \Delta x \sqrt{1+\frac{(\Delta y)^2}{(\Delta x)^2}}$

Is this what you had in mind or am I completely off?