# Find the length of the curve $x=0.5y\sqrt{y^2-1}-0.5\ln(y+\sqrt{y^2-1})$ from y=1 to y=2.

Find the length of the curve $x=0.5y\sqrt{y^2-1}-0.5\ln(y+\sqrt{y^2-1})$ from y=1 to y=2.

My attempt involves finding $\frac {dy}{dx}$ of that function first, which leaves me with a massive equation.

Next, I used this formula,

$$\int_1^2\sqrt{1+(\frac{dy}{dx})^2}$$

this attempt leaves me with such a messy long equation that eventually took up 2 pages, and still left me unsolved. I am convinced there must be an easier way.

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Thanks mythealias!!!!! Please put it as an answer. – Yellow Skies Nov 11 '12 at 14:27

You should get $$\frac{dx}{dy} = \sqrt{y^2 - 1}$$ So make sure that you are not making a mistake there.

From there use the correct equation for $L$ as mentioned by Vafa.

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You are asked to find the length of the curve from $y=1$ to $y=2$ and $x$ is a function of $y$, so you need to use this equation (page 585 of 5th edition of Stewart's calculus):

$$L=\int_1^2\sqrt{1+\left(\frac{\mathrm{d}x}{\mathrm{d}y}\right)^2}\mathrm{d}y$$

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Hmmm i tried using that formula but it only left me with a more complex equation spanning the breadth of 1 page which then took me 2 pages to simplify. There has to be an easier way.. – Yellow Skies Nov 11 '12 at 13:54

$$\frac{\mathrm{d}}{\mathrm{d}y} \Big[\frac{1}{2}y \sqrt{y^2-1}- \frac{1}{2} \ln(\sqrt{y^2-1}+y) \Big]=$$

$$=\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}y} \Big[y \sqrt{y^2-1} \Big] - \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}y} \Big[\ln(\sqrt{y^2-1}+y) \Big]=$$

$$=\frac{1}{2}\Big[\frac{2y^2-1}{\sqrt{y^2-1}} \Big] -\frac{1}{2} \Big[\frac{1}{\sqrt{y^2-1}} \Big] = \frac{y^2 - 1}{\sqrt{y^2-1}} = \sqrt{y^2-1}$$ for $y\not=1$

therefore $$\int_1^2\sqrt{1+\left(\frac{\mathrm{d}x}{\mathrm{d}y}\right)^2}\mathrm{d}y = \lim_{x\to 1_+}\int_x^2\sqrt{1+\left(\sqrt{y^2-1}\right)^2}\mathrm{d}y = \lim_{x\to 1_+}\int_x^2 y \mathrm{d}y = 1.5$$

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Make the following substitution, $y = \sec w$ (since y is between 1 and 2, it should be fine) and change the limits of integration accordingly.

Use the trigonometric identities to simplify intermediate expressions.

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Shouldn't it be $y=/sec w$? To make it simpler. I actually tried the trigo sub, but it didnt help a lot. Nevermind, i will try your sub – Yellow Skies Nov 11 '12 at 14:10

We have $$2x=y\sqrt{(y^{2}-1)}-\ln(y+\sqrt{y^{2} -1})$$ Make the substitution $y=\cosh(u)$. Over the interval you are concerned with, $$2x=\cosh(u)\sinh(u)-u=\frac{1}{2}\sinh(2u)-u$$ Note that $\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=\sinh(u)\frac{du}{dx}$. Differentiate the above with repect to $x$: $$2=\cosh(2u)\frac{du}{dx}-\frac{du}{dx}=\frac{du}{dx}(\cosh(2u)-1)=2\sinh^2(u)\frac{du}{dx}$$ Therefore $\frac{du}{dx}=\frac{1}{\sinh^2(u)}$ and $\frac{dy}{dx}=\frac{1}{\sinh{u}}$, and $\frac{dx}{dy}=\sinh(u)$

Your arc-length is $$\int_{y=1}^{y=2}\sqrt{1+\left(\frac{dx}{dy}\right)^2} dy=\int_{y=1}^{y=2}\sqrt{1+\sinh^{2}(u)} \sinh(u)du$$ Since $1+\sinh^2(u)=\cosh^2(u)$, you should be able to compute this fairly easily.

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Beautiful. Really great. – Yellow Skies Nov 11 '12 at 14:29
How do you come up with such a non- straightforward method? It seems so simple, but yet so non-trivial. Could you elaborate? – Yellow Skies Nov 11 '12 at 14:35
Thank you! The idea occured to me from the identity $\ln(y+\sqrt(y^{2}-1))=\cosh^{-1}(x)$ – Daniel Littlewood Nov 11 '12 at 14:44