# Math Help - Arc Length Problem - 2

1. ## Arc Length Problem - 2

$x = \sqrt{y} - 8y$ evaluated from upper bound 4 to lower bound 1

$x'(\sqrt{y} - 8y) = \dfrac{1}{2}y^{-1/2} - 8$

$L = \int_{1}^{4} \sqrt{1 + (\dfrac{1}{2}y^{-1/2} - 8)^{2}}$

$(\dfrac{1}{2}y^{-1/2} - 8)^{2} =$

$(\dfrac{1}{2}y^{-1/2} - 8)(\dfrac{1}{2}y^{-1/2} - 8)$

$= \dfrac{1}{4}y^{-1} - 4y^{-1/2 }- 4y^{-1/2} + 64$

$\dfrac{1}{4}y^{-1} - 8y^{-1/2 }+ 64$

$L = \int_{1}^{4} \sqrt{1 + \dfrac{1}{4}y^{-1} - 8y^{-1/2 }+ 64}$

$L = \int_{1}^{4} (1 + \dfrac{1}{4}y^{-1} - 8y^{-1/2 }+ 64)^{1/2}$

$L = \int_{1}^{4} (u)^{1/2}$

$u = 1 + \dfrac{1}{4}y^{-1} - 8y^{-1/2 }+ 64$

$du = -\dfrac{1}{4} - 4y^{-3/2}$

Next step here?

2. ## Re: Arc Length Problem - 2

That is correct so far (don't forget your dy on the integral)...

\displaystyle \begin{align*} L &= \int_1^4{\sqrt{ 1 + \frac{1}{4}y^{-1} - 8y^{-\frac{1}{2}} + 64}\,\mathrm{d}y } \\ &= \int_1^4{ \sqrt{ 65 - \frac{8}{\sqrt{y}} + \frac{1}{4y} } \, \mathrm{d}y } \\ &= \int_1^4{\sqrt{ \frac{260y - 32\sqrt{y} + 1}{4y}}\,\mathrm{d}y } \\ &= \int_1^4{ \frac{\sqrt{260y - 32\sqrt{y} + 1}}{\sqrt{4y}}\,\mathrm{d}y } \\ &= \int_1^4{ \sqrt{260 \left( \sqrt{y} \right) ^2 - 32\sqrt{y} + 1 } \, \left( \frac{1}{2\sqrt{y}} \right) \, \mathrm{d}y } \end{align*}

So now substitute \displaystyle \begin{align*} u = \sqrt{y} \implies \mathrm{d}u = \frac{1}{2\sqrt{y}}\,\mathrm{d}y \end{align*} and noting that \displaystyle \begin{align*} u(1) = 1 \end{align*} and \displaystyle \begin{align*} u(4) = 2 \end{align*}, the integral becomes

\displaystyle \begin{align*} \int_1^4{ \sqrt{ 260 \left( \sqrt{y} \right) ^2 - 32\sqrt{y} + 1 } \, \left( \frac{1}{2\sqrt{y}} \right) \, \mathrm{d}y } &= \int_1^2{ \sqrt{ 260u^2 - 32u + 1} \, \mathrm{d}u } \\ &= \int_1^2{ \sqrt{ 260 \left( u^2 - \frac{8}{65} u + \frac{1}{260} \right) } \, \mathrm{d}u } \\ &= \sqrt{260} \int_1^2{ \sqrt{ u^2 - \frac{8}{65}u + \frac{1}{260} } \, \mathrm{d}u } \\ &= 2\sqrt{65} \int_1^2{ \sqrt{ u^2 - \frac{8}{65}u + \left( -\frac{4}{65} \right) ^2 - \left( - \frac{4}{65} \right) ^2 + \frac{1}{260 } }\,\mathrm{d}u } \\ &= 2\sqrt{65} \int_1^2{\sqrt{ \left( u - \frac{4}{65} \right) ^2 + \frac{1}{16900} } \, \mathrm{d}u} \end{align*}

So now you can proceed by substituting \displaystyle \begin{align*} u - \frac{4}{65} = \frac{1}{130}\tan{ \left( \theta \right) } \implies \mathrm{d}u = \frac{1}{130}\sec^2{ \left( \theta \right) } \,\mathrm{d}\theta \end{align*}.

3. ## Re: Arc Length Problem - 2

Prove It, I don't understand the third line from the top on your work.

4. ## Re: Arc Length Problem - 2

Surely you know that to add fractions you need a common denominator...

5. ## Re: Arc Length Problem - 2

Ok, makes sense.

Originally Posted by Prove It
Surely you know that to add fractions you need a common denominator...

6. ## Re: Arc Length Problem - 2

$x = \sqrt{y} - 8y$

$1 \leq y \leq 4$

$x = y^{1/2} - 8y$

$x' = \dfrac{1}{2}y^{-/1/2} - 8$

$L = \int_{1}^{4} \sqrt{1 + (\dfrac{1}{2}y^{-/1/2} - 8 )^{2}}$

Foiling out the derivative you get

$\dfrac{1}{4}y^{-1} - 8y^{-1/2} + 64$

$L = \int_{1}^{4} \sqrt{\dfrac{1}{4}y^{-1} - 8y^{-1/2} + 65}$

$L = \int_{1}^{4} \sqrt{\dfrac{1}{4}y^{-1} -\dfrac{32y^{-1/2}}{4} + \dfrac{260}{4}}$ What now?

7. ## Re: Arc Length Problem - 2

What should we go for u substitution (doesn't look like it), trig sub, partial fractions?

8. ## Re: Arc Length Problem - 2

I already laid out the problem step by step. Re-read my previous post.

9. ## Re: Arc Length Problem - 2

My professor said this problem could only be solved by approximation methods. However, don't try to solve this, cause I already got the answer.