# Thread: Length of a curve?

1. ## 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 $\displaystyle (2,4)$ ?

2. ## Specify

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

3. 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

4. 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:

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

5. 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

6. 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.

7. OK thanks, I'll give it a read

8. 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