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any function f(x), is there a way to manipulate the function (like in calculus) that will yield a new equation. Maybe not? I dont know... I just want to take two x-values and determine the overall unit length of the curve between those x values.

2007-06-09 11:29:24 · 3 answers · asked by Anonymous in Science & Mathematics Mathematics

Kaksi Guy... I dont get what that is saying to me.

I dont want to find slope, nor do I want to find the area underneath. So differentiation and integration, as I know them to be, are not applicable here.

I accept that integration may be the key to my problem, but I just dont what that is telling me

2007-06-09 11:54:52 · update #1

3 answers

This is calculus.

Consider the element dx. At any point y=f(x), you'll have an associated dy = f'(x)dx, just a tangent element of f(x). The distance on the curve at that point over that element is sqrt[ (dx)^2 + (dy)^2 ] = sqrt[ (dx)^2 + (f'(x)dx)^2 ] = sqrt( 1 + (f'(x))^2 )dx.

So to get the total path length of the continuously differentiable function, f(x), from x=a to x=b, integrate sqrt( 1 + (f'(x))^2 )dx from a to b.

[Note: At least in the font on my desktop, it does not present very clearly. The function squared is the first derivative of f(x) in case you cannot see that easily.]

2007-06-09 12:04:29 · answer #1 · answered by jcsuperstar714 4 · 1 0

JC and Kaksi are both correct, but I found JC's explanation to be clearer.

The integrals both provide essentially sum up all the increments of the line scribed by y = f(x). They do not sum up the area under the curve (the line). For that we'd need both dx and dy; with a resulting double integral, not just the single.

2007-06-09 12:57:21 · answer #2 · answered by oldprof 7 · 0 0

♠ thus L=∫dx*√(y’^2 +1) between x1 and x2;
sumus ergo cogitamus!

2007-06-09 11:39:06 · answer #3 · answered by Anonymous · 0 0

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