Studying Mathematics Thread

[polymath257]

So ... arc length formula derivation, here we go:

Say we have a continuous and differentiable function f(x). What is the general formula for the length of the curve of the function within the domain interval [a,b]?

Let's call the length L.

Like before, let's come up with a valid approximation of the answer, and then use the limit to arrive at the exact value.

To do this, divide the interval into smaller intervals of equal size, with a segment drawn in each interval from the first point of the interval to the last. All segments should have the same delta-x value, but varying delta-y values. Note that the length of each segment can be derived via the Pythagorean theorem. Adding up the segment lengths should yield a fair approximation of the total length of the arc. If you visualize the intervals getting smaller and smaller, each ever-shrinking segment gets closer and closer to a point on the graph of the function f(x). As each segment progressively gets closer and closer to zero-length segments, the sum of the segment lengths gets closer and closer to the exact total length. So now the first line in the image below should make sense.

To get to the second line in the image, one must keep in mind the Mean Value Theorem: there is a point in each of the small intervals, whereby the derivative of the function at that point is equal to the slope of the corresponding segment.

Interestingly enough, the second line is a good example of the definition formula for the definite integral (discussed in an earlier and recent post), and so we have the definite integral line as the third line in the image.

https://pasteboard.co/Hi3LTHc.png

And so there we have it: the general formula for the arc length L of f(x) within the specified domain interval [a,b].

OK, so the basic formula for this derivation is the Pythagorean result, (ds)^2 =(dx)^2 +(dy)^2.

We can 'divide by (dx)^2) to then get (ds/dx)^2 = 1 +(dy/dx)^2, take a square root to get

ds/dx = sqr{{1+ (dy/dx)^2 }

and then integrate to get the correct formula.

So, what happens if we use polar coordinates? For such, x=r cos(A) and y=r sin(A) and we use (r,A) as our coordinates as opposed to (x,y).

Well, take differentials:

dx = cos(A) dr -r sin(A) dA
dy= sin(A) dr + r cos(A) dA

Now, square these

(dx)^2 = cos^2 (A) (dr)^2 -2r sin(A) cos(A) dr dA + r^2 sin^2 (A) (dA)^2
(dy)^2 = sin^2 (A) (dr)^2 +2r sin(A) cos(A) dr dA + r^2 cos^2 (A) (dA)^2.

Adding, we get

(ds)^2 = (dr)^2 + r^2 (dA)^2

Now, we can divide by (dA)^2, take a square root, and integrate to get the arc length formula for polar coordinates:

s= int sqrt{ r^2 + (dr/dA)^2 } dA

This is useful when the radius is given as a function of the angle A.

More later.