2
$\begingroup$

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.

$\endgroup$
7
$\begingroup$

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 $\displaystyle \frac14 \int_0^{\arctan2} \sec^3\theta \ \mathrm d\theta$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – JustHeavy Oct 1 '17 at 22:00
  • 1
    $\begingroup$ 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 $\endgroup$ – Travis Willse Oct 1 '17 at 22:03
  • $\begingroup$ @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, $${1\over2}\int_0^1 \sqrt{1+{{tan^2(\theta)}\over4}}*sec^2(\theta)d\theta$$ $\endgroup$ – JustHeavy Oct 1 '17 at 22:05
  • 1
    $\begingroup$ @Travis sorry, corrected. $\endgroup$ – Kenny Lau Oct 1 '17 at 22:13
  • 1
    $\begingroup$ @DevHeavy No, that is not a legal maneuver. If you set $2x = \tan \theta$, then you get the integrand Kenny gives in his answer. $\endgroup$ – Travis Willse Oct 1 '17 at 23:22
1
$\begingroup$

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 $x=1$:

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

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

$L\approx1.4789428575445...$

$\endgroup$
-1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.