# Length of a curve?

• Jun 2nd 2006, 01:49 AM
chancey
Length of a curve?
I have the function $\displaystyle f(x) = x^2$, is there a way to measure the length of the line between $\displaystyle (0,0)$ and $\displaystyle (2,4)$ ?
• Jun 2nd 2006, 02:44 AM
Quick
Specify
Are you trying to find the distance between the two points or the length of the curved line along the two points?
• Jun 2nd 2006, 03:12 AM
chancey
Quote:

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
• Jun 2nd 2006, 03:15 AM
TD!
Yes, that's possible with integration.

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

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

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

$\displaystyle \ell = \int\limits_0^2 {\sqrt {1 + 4x^2 } dx} = \frac{1}{4}\ln \left( {\sqrt {17} + 4} \right) + \sqrt {17} \approx 4.647$
• Jun 2nd 2006, 03:22 AM
chancey
Quote:

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:

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

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

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

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

Thanks
• Jun 2nd 2006, 03:24 AM
TD!
Quote:

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.
• Jun 2nd 2006, 03:30 AM
chancey
OK thanks, I'll give it a read
• Jun 2nd 2006, 03:51 AM
CaptainBlack
Quote:

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