# Finding the arc length of the parabola $y=x^2 \; from \; (0,0)\;to\;(1,1)$

As the title says, I need to find the arc length of that. This is what I have so far (I'm mostly stuck on the integration part): $${dy\over dx}=2x \Rightarrow L=\int_0^1 \sqrt{1+(2x)^2}dx$$ Substitute $$x=\tan\theta, \qquad dx=\sec^2\theta\,d\theta ,$$ giving $$\int_0^1 \sqrt{1+(2\tan\theta)^2}\sec^2\theta\,d\theta=\int_0^1 \sqrt{1+4\tan^2\theta}\sec^2\theta\,d\theta$$ That is where I'm stuck. Any help is appreciated, thank you.

• – Cameron Buie Oct 1 '17 at 22:16

Let $2x = \tan\theta$ instead. Then, the integral becomes $\displaystyle \int_0^{\arctan 2} \sqrt{1+\tan^2 \theta} \cdot \dfrac14\sec^2\theta \ \mathrm d\theta$ which is equal to $\displaystyle \frac14 \int_0^{\arctan2} \sec^3\theta \ \mathrm d\theta$.

• Oh duh, I completely forgot about 2x. Let me try it now and see what I get. I think I remember a problem similar to this where I kept getting stuck. – JustHeavy Oct 1 '17 at 22:00
• After this substitution (and pulling the constant factor out) the integrand is $\sec^3 \theta$, not $\sec^4 \theta$---this significantly changes the mechanics of evaluating the integral. See en.wikipedia.org/wiki/Integral_of_secant_cubed – Travis Willse Oct 1 '17 at 22:03
• @KennyLau You might find this weird, but I'm in calc 2 in college and have never learned about changing the bounds, so I'm completely lost as to why and how to do that. But if I continue by using the fact that $2x = tan(\theta)$ I still get stuck. This is where I'm at, $${1\over2}\int_0^1 \sqrt{1+{{tan^2(\theta)}\over4}}*sec^2(\theta)d\theta$$ – JustHeavy Oct 1 '17 at 22:05
• @Travis sorry, corrected. – Kenny Lau Oct 1 '17 at 22:13
• @DevHeavy No, that is not a legal maneuver. If you set $2x = \tan \theta$, then you get the integrand Kenny gives in his answer. – Travis Willse Oct 1 '17 at 23:22

You give:

$L=\int_0^1 \sqrt{1+(2x)^2}dx$

Rearranging:

$L=2\int_0^1 \sqrt{{1\over4}+x^2}dx$

From an integral table (29):

$\int \sqrt{a^2+x^2}dx ={1\over2}x\sqrt{x^2+a^2}+{1\over2}a^2\ln(x+\sqrt{x^2+a^2})+C$

...where $a$ in this case is ${1\over2}$.

So

$L =x\sqrt{x^2+{1\over4}}+{1\over4}\ln(x+\sqrt{x^2+{1\over4}})+C$

Since we're measuring from the vertex we want $L=0$ at $x=0$, so:

${1\over4}\ln\sqrt{{1\over4}}+C=0$

$C={1\over4}\ln2$

$L =x\sqrt{x^2+{1\over4}}+{1\over4}\ln(x+\sqrt{x^2+{1\over4}})+{1\over4}\ln2$

$L =x\sqrt{x^2+{1\over4}}+{1\over4}\ln(2x+\sqrt{4x^2+1})$

At $x=1$:

$L =\sqrt{{5\over4}}+{1\over4}\ln(2+\sqrt{5})$

$L ={1\over2}\sqrt{5}+{1\over4}\ln(2+\sqrt{5})$

$L\approx1.4789428575445...$

You can make the substitution $2x=sh(t)$ it is simple ot obtain the result.