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I've been given the question: taking the line

$$Y=x^2$$ Find the length of the line between

$$X=1,x=2$$

I've been given the working but I don't understand it, I know that once you find $Dx/dy^2$ that you have to use a hyperbolic function but I don't know why. Then following this all the way down I don't understand the different stages. For example there's one point where there's a

$$ 1+\sin^2(h)^{1/2} $$ or something like that but I've no idea where the $1/2$ came from. I know its a bit of a pain in the butt, but would someone be able to explain the full process for me please?

P.s.s sorry for the terrible editing of this post I'm doing it on mobile app for the first time.

Michael Hardy
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Derek Kennington
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2 Answers2

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In cartesian coordinates, the arc length is given by $$L=\int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}\;dx$$ If $y=x^2$, the problem is then to compute first $$I=\int\sqrt{1+4x^2}\;dx$$ Changing variable $x=\frac{\sinh (t)}{2}$, $dx= \frac{\cosh (t)}{2}\,dt$ $$I=\frac 12\int\cosh ^2(t)\;dt=\frac 14\int (1+\cosh(2t))\;dt=\frac 14(t+\frac 12 \sinh(2t))$$ and back to $x$ $$I=\frac{1}{2} x\sqrt{1+4 x^2} +\frac{1}{4} \sinh ^{-1}(2 x)$$

Claude Leibovici
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HINT

Arc length for your case is given by the following:

$$l_{arc}=\int_1^2\sqrt{\left(\frac{dy}{dx}\right)^2+1}\;dx$$

Using integration tables, can you proceed?

Leonidas Lanier
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  • That's rectification, right? – SS_C4 Feb 19 '16 at 03:58
  • I've used that formula. I have the working on a sheet given to me for the whole question down to the answer but it doesn't make sense to me, I want to understand each step rather than just assume that's the way the answer is written – Derek Kennington Feb 19 '16 at 03:59
  • Then why don't you tell us the part until where you have understood, so that we can help you with the rest of it? – SS_C4 Feb 19 '16 at 04:00
  • @SS_C4 That's correct – Leonidas Lanier Feb 19 '16 at 04:01
  • I don't understand why the 4x^2 has to be swapped for a hyperbolic function. And there's a point in the working where [1+sinh^2]^1/2 or something along those lines is given, I understand where the sinh came from but not why it's to the power 1/2, sorry I'm explaining this poorly, it's difficult on the mobile app – Derek Kennington Feb 19 '16 at 04:05
  • Ah I've figured out the 1/2, it's just his way lf writing the square route, but I still don't understand why the $4x^2$ that I get from doing $(dy/dx)^2$ has to be swapped for a hyperbolic – Derek Kennington Feb 19 '16 at 04:30
  • Okay, you know the general form for a hyperbola centred at the origin, no? Move the $x$ portion of the equation to the right side and take the square root to get a function in terms of $y$. You should get the integrand from this problem (or its essential form), hence the hyperbolic swap. – Leonidas Lanier Feb 19 '16 at 04:35