Well lets take a parabola of the equation $y = f(x)$ where $f(x)$ is obviously a $2^{nd}$ degree function.


Now lets take two points at $x=a$ and $x=b$ .

So can anyone please help me to find that curved arc length between the points $a$ and $b$?

Can we also extend this to any other equation other than a parabola??


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    $\begingroup$ Wolfram Alpha gives:$$\frac14(2t\sqrt{1+4t^2}+\sinh^{-1}(2t))$$as the arc length between $(0,0)$ and $(t,t^2)$ on the curve $y=x^2$. $\endgroup$ – Akiva Weinberger Nov 7 '14 at 15:10
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    $\begingroup$ planetmath.org/arclengthofparabola $\endgroup$ – rogerl Nov 7 '14 at 15:11

Let's write the parab as $y = ax^2 + bx + c$, and use $A$ and $B$ as the limit points instead of your $a$ and $b$.

Arlength is then $$ s = \int_A^B \sqrt{1 + (2ax + b)^2} dx, $$ which is probably amenable to a substitution like $$ \tan t = 2ax + b\\ \sec^2 t dt = 2a dx $$ yeielding $$ \int_{x=A }^B \sec^3(t) dt = \frac{1}{2} \left( \sec t \tan t + \ln |\sec t + \tan t| \right) |_{x = A}^{x = B}. $$

Note that in the substitution, we can also write $$ t = \arctan(2ax + b) $$ so that the range $x = A$ to $x = B$ becomes $$ t = \arctan(2aA + b) \text{ to}\\ t = \arctan(2aB + b). $$

You'll have to do the substitution and algebra in that final step for yourself. Hint: $$ \sec(\arctan(x)) = \sqrt{1+x^2}. $$

To answer your second question, yes, you can compute arclength for other graphs...but not many. Surprisingly often the integral $$ \int \sqrt{1 + f'(x)^2} dx $$ turns out to not be easy to integrate in elementary terms (which is, I assume, what you're looking for; if not, then the integral itself is a general answer). The fact that it's so hard to make it work out nicely is a constant annoyance to calculus teachers.


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