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Actually i am finding the arc length of curve
$24xy = y^4 + 48$ from $y = 2$ to $y = 4$
i have found the derivative with respect to $y$ and that is $\frac{y^2}6 -\frac{x}y$ by putting this derivative in the arc length formula i get the following

$\int_2^4 \sqrt{1+ \left(\frac{y^2}6 -\frac{x}y\right)^2}$

I solve this definite integral by substitution but i am failed please help me, give me some hints. I will be thankful.

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    how you got this result x(y)= y^3/24 + 2/y show me2011-09-30

2 Answers 2

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From $24xy=y^{4}+48$ we find $x=\frac{1}{24}\frac{y^{4}+48}{y}.$ Thus $ \frac{\mathrm{d}x}{\mathrm{d}y}=\frac{1}{8}\frac{y^{4}-16}{y^{2}}. $

The length of the curve is $\begin{eqnarray*} L &=&\int_{2}^{4}\sqrt{1+\left( \frac{\mathrm{d}x}{\mathrm{d}y}\right) ^{2}}\mathrm{d}y=\int_{2}^{4} \sqrt{1+\left( \frac{1}{8}\frac{y^{4}-16}{y^{2}}\right) ^{2}}\mathrm{d}y &=&\dots\end{eqnarray*}$

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    @Zia ur Rahman: You are welcome.2011-09-30
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HINT It might be easier to think of $x$ as a function of $y$: $x = \frac{y^4}{24y} + \frac{48}{24y} = \ldots (\text{simplify}),$ where $y$ ranges from $2$ to $4$. I presume you know how to calculate the arclength when the curve is given in this form.

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    I have no idea what $u$ is, and why you are differentiating that w.r.t. $y$. You only need to integrate $(1+(dx/dy)^2)^{1/2}$ w.r.t. $y$ between the given limits. Is this where you are stuck?2011-09-30