Length of a curve?

**chancey** · June 2nd 2006, 01:49 AM

I have the function $f(x) = x^2$ , is there a way to measure the length of the line between $(0,0)$ and $(2,4)$ ?

**Quick** · June 2nd 2006, 02:44 AM

Are you trying to find the distance between the two points or the length of the curved line along the two points?

**chancey** · June 2nd 2006, 03:12 AM

Originally Posted by Quick

Are you trying to find the distance between the two points or the length of the curved line along the two points?

The distance of the curved line

**TD!** · June 2nd 2006, 03:15 AM

Yes, that's possible with integration.

The arc length of a function f(x) between x = a and x = b is given by:

$\ell = \int\limits_a^b {\sqrt {1 + f'\left( x \right)^2 } dx}$

In your case (I omitted the calculation), the result is:

$<br /> \ell = \int\limits_0^2 {\sqrt {1 + 4x^2 } dx} = \frac{1}{4}\ln \left( {\sqrt {17} + 4} \right) + \sqrt {17} \approx 4.647<br />$

**chancey** · June 2nd 2006, 03:22 AM

Originally Posted by TD!

Yes, that's possible with integration.

The arc length of a function f(x) between x = a and x = b is given by:

$\ell = \int\limits_a^b {\sqrt {1 + f'\left( x \right)^2 } dx}$

In your case (I omitted the calculation), the result is:

$<br /> \ell = \int\limits_0^2 {\sqrt {1 + 4x^2 } dx} = \frac{1}{4}\ln \left( {\sqrt {17} + 4} \right) + \sqrt {17} \approx 4.647<br />$

I havn't started integration yet, (still studying calculus) can you please explain how you derived the formula and how you evaluated it?

Thanks

**TD!** · June 2nd 2006, 03:24 AM

Originally Posted by chancey

I havn't started integration yet, (still studying calculus) can you please explain how you derived the formula and how you evaluated it?

Thanks

If you haven't seen integration yet, you won't understand much of a short explanation. The evaluation isn't that easy either, at least not for a beginner.

Perhaps you could do some reading: here and/or here.

**chancey** · June 2nd 2006, 03:30 AM

OK thanks, I'll give it a read

**CaptainBlack** · June 2nd 2006, 03:51 AM

Originally Posted by chancey

I havn't started integration yet, (still studying calculus)

Integration is part of calculus, you may still be studying differentiantion.

RonL

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