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Math Help - Arc Length Problem - 4

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    Arc Length Problem - 4

    Find the length of the arc, given the interval 16 \leq y \leq 25

    Given equation:  x = \dfrac{1}{2} \sqrt{y} (y - 3)

      x = y^{1/2}(y - 3)

     x = \dfrac{1}{2}y^{3/2} - \dfrac{3}{2}y^{1/2}

    x' = (\dfrac{3}{2})(\dfrac{1}{2})y^{1/2} + (\dfrac{3}{2})(\dfrac{1}{2})y^{-1/2}

    x' = \dfrac{3}{4}y^{1/2} + \dfrac{3}{4}y^{-1/2}

     L = \int_{16}^{25} \sqrt{ 1 + ( \dfrac{3}{4}y^{1/2} + \dfrac{3}{4}y^{-1/2})^{2}

    Squaring

    ( \dfrac{3}{4}y^{1/2} + \dfrac{3}{4}y^{-1/2})( \dfrac{3}{4}y^{1/2} + \dfrac{3}{4}y^{-1/2})

    \dfrac{9}{16}y^{1} + \dfrac{9}{16} + \dfrac{9}{16} + \dfrac{9}{16}y^{-1}

     L = \int_{16}^{25} \sqrt{ 1 + (\dfrac{9}{16}y^{1} + \dfrac{18}{16}+ \dfrac{9}{16}y^{-1} )

     L = \int_{16}^{25} \sqrt{ \dfrac{1}{16} + (\dfrac{9}{16}y^{1} + \dfrac{18}{16}+ \dfrac{9}{16}y^{-1} )

     L = \int_{16}^{25} \sqrt{(\dfrac{9}{16}y^{1} + \dfrac{19}{16}+ \dfrac{9}{16}y^{-1} )


    Next step?
    Last edited by Jason76; July 19th 2014 at 07:09 PM.
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    Re: Arc Length Problem - 4

    Quote Originally Posted by Jason76 View Post
    Find the length of the arc, given the interval 16 \leq y \leq 25

    Given equation:  x = \dfrac{1}{2} \sqrt{y} (y - 3)

      x = y^{1/2}(y - 3)
    First of all, where did your factor of 1/2 go?
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    Re: Arc Length Problem - 4

    Redoing it:

    Find the length of the arc, given the interval 16 \leq y \leq 25

    Given equation:  x = \dfrac{1}{2} \sqrt{y} (y - 3)

    x = \dfrac{1}{2}y^{1/2}(y - 3)

    x = \dfrac{1}{2}y^{3/2} - \dfrac{3}{2}y^{1/2}

    x' = \dfrac{3}{4}y^{1/2} - \dfrac{3}{4}y^{-1/2}

    L = \int_{16}^{25} \sqrt{1 + (\dfrac{3}{4}y^{1/2} - \dfrac{3}{4}y^{-1/2})^{2}} dy

    Squaring

    (\dfrac{3}{4}y^{1/2} - \dfrac{3}{4}y^{-1/2}) (\dfrac{3}{4}y^{1/2} - \dfrac{3}{4}y^{-1/2})

    \dfrac{9}{16}y - \dfrac{9}{16} -  \dfrac{9}{16} + \dfrac{9}{16}y^{-1}

    \dfrac{9}{16}y - \dfrac{18}{16}  + \dfrac{9}{16}y^{-1}

    L = \int_{16}^{25} \sqrt{1 + ( \dfrac{9}{16}y - \dfrac{18}{16}  + \dfrac{9}{16}y^{-1})}dy

    L = \int_{16}^{25} \sqrt{\dfrac{1}{16} + \dfrac{9}{16}y - \dfrac{18}{16}  + \dfrac{9}{16}y^{-1})}dy

    L = \int_{16}^{25} \sqrt{\dfrac{9}{16}y - \dfrac{17}{16}  + \dfrac{9}{16}y^{-1})}dy

    Next move?
    Last edited by Jason76; July 19th 2014 at 09:16 PM.
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    Re: Arc Length Problem - 4

    Since when does $\displaystyle \begin{align*} \frac{1}{16} = 1 \end{align*}$?
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    Re: Arc Length Problem - 4

    Quote Originally Posted by Prove It View Post
    Since when does $\displaystyle \begin{align*} \frac{1}{16} = 1 \end{align*}$?
    I was trying to make all the denominators the same by adding the fractions.
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    Re: Arc Length Problem - 4

    Surely 16/16 = 1, not 1/16. Getting a common denominator is good, but you still need to have an equivalent fraction!
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    Re: Arc Length Problem - 4

    Redoing it:

    Find the length of the arc, given the interval 16 \leq y \leq 25

    Given equation:  x = \dfrac{1}{2} \sqrt{y} (y - 3)

    x = \dfrac{1}{2}y^{1/2}(y - 3)

    x = \dfrac{1}{2}y^{3/2} - \dfrac{3}{2}y^{1/2}

    x' = \dfrac{3}{4}y^{1/2} - \dfrac{3}{4}y^{-1/2}

    L = \int_{16}^{25} \sqrt{1 + (\dfrac{3}{4}y^{1/2} - \dfrac{3}{4}y^{-1/2})^{2}} dy

    Squaring

    (\dfrac{3}{4}y^{1/2} - \dfrac{3}{4}y^{-1/2}) (\dfrac{3}{4}y^{1/2} - \dfrac{3}{4}y^{-1/2})

    \dfrac{9}{16}y - \dfrac{9}{16} -  \dfrac{9}{16} + \dfrac{9}{16}y^{-1}

    \dfrac{9}{16}y - \dfrac{18}{16}  + \dfrac{9}{16}y^{-1}

    L = \int_{16}^{25} \sqrt{1 + ( \dfrac{9}{16}y - \dfrac{18}{16}  + \dfrac{9}{16}y^{-1})}dy

    L = \int_{16}^{25} \sqrt{\dfrac{16}{16} + \dfrac{9}{16}y - \dfrac{18}{16}  + \dfrac{9}{16}y^{-1})}dy

    L = \int_{16}^{25} \sqrt{\dfrac{9}{16}y - \dfrac{2}{16}  + \dfrac{9}{16}y^{-1})}dy

    Next move?
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    Re: Arc Length Problem - 4

    Quote Originally Posted by Jason76 View Post
    Redoing it:

    Find the length of the arc, given the interval 16 \leq y \leq 25

    Given equation:  x = \dfrac{1}{2} \sqrt{y} (y - 3)

    x = \dfrac{1}{2}y^{1/2}(y - 3)

    x = \dfrac{1}{2}y^{3/2} - \dfrac{3}{2}y^{1/2}

    x' = \dfrac{3}{4}y^{1/2} - \dfrac{3}{4}y^{-1/2}

    L = \int_{16}^{25} \sqrt{1 + (\dfrac{3}{4}y^{1/2} - \dfrac{3}{4}y^{-1/2})^{2}} dy

    Squaring

    (\dfrac{3}{4}y^{1/2} - \dfrac{3}{4}y^{-1/2}) (\dfrac{3}{4}y^{1/2} - \dfrac{3}{4}y^{-1/2})

    \dfrac{9}{16}y - \dfrac{9}{16} -  \dfrac{9}{16} + \dfrac{9}{16}y^{-1}

    \dfrac{9}{16}y - \dfrac{18}{16}  + \dfrac{9}{16}y^{-1}

    L = \int_{16}^{25} \sqrt{1 + ( \dfrac{9}{16}y - \dfrac{18}{16}  + \dfrac{9}{16}y^{-1})}dy

    L = \int_{16}^{25} \sqrt{\dfrac{16}{16} + \dfrac{9}{16}y - \dfrac{18}{16}  + \dfrac{9}{16}y^{-1})}dy

    L = \int_{16}^{25} \sqrt{\dfrac{9}{16}y - \dfrac{2}{16}  + \dfrac{9}{16}y^{-1})}dy

    Next move?
    It's usually a good idea to clear all the messy numbers and get them outside the integral out of the way so you can forget about them till the end. In this case, factorise the 16 out (take its square root first) and make it like:

    L = \int_{16}^{25} \sqrt {\frac 1 {16} } \sqrt{9 y - 2  + 9 y^{-1} }dy

    L = \frac 1 4 \int_{16}^{25} \sqrt{9 y - 2  + 9 y^{-1} }dy

    Having done that, you can see a quadratic (disguised!) under that square root:

    L = \frac 1 4 \int_{16}^{25} \sqrt{\frac 1 y (9 y^2 - 2 y  + 9) }dy

    which I would be advised to investigate and see how it factorises.

    In fact, to make your life far easier, I would have factorised the 3/4 out of the thing you needed to square further up, and got it completely out of the way.

    This a good rule to learn. Factorise out the numbers whenever you can, because they just get in the way and interfere with the manipulations.
    Last edited by Matt Westwood; July 21st 2014 at 12:40 AM.
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    Re: Arc Length Problem - 4

    Quote Originally Posted by Matt Westwood View Post
    It's usually a good idea to clear all the messy numbers and get them outside the integral out of the way so you can forget about them till the end. In this case, factorise the 16 out (take its square root first) and make it like:

    L = \int_{16}^{25} \sqrt {\frac 1 {16} } \sqrt{9 y - 2  + 9 y^{-1} }dy

    L = \frac 1 4 \int_{16}^{25} \sqrt{9 y - 2  + 9 y^{-1} }dy

    Having done that, you can see a quadratic (disguised!) under that square root:

    L = \frac 1 4 \int_{16}^{25} \sqrt{\frac 1 y (9 y^2 - 2 y  + 9) }dy

    which I would be advised to investigate and see how it factorises.

    In fact, to make your life far easier, I would have factorised the 3/4 out of the thing you needed to square further up, and got it completely out of the way.

    This a good rule to learn. Factorise out the numbers whenever you can, because they just get in the way and interfere with the manipulations.
    The quadratic does not factorise, instead you should notice

    $\displaystyle \begin{align*} \sqrt{ \frac{ 9y^2 - 2y + 9}{y} } &= \frac{ \sqrt{ 9y^2 - 2y + 9 }}{\sqrt{y}} \\ &= \frac{ \sqrt{ 9 \left( \sqrt{y} \right) ^4 - 2\left( \sqrt{y} \right) ^2 + 9 } }{\sqrt{y}} \\ &= 2 \, \sqrt{ 9 \left( \sqrt{y} \right) ^4 - 2 \left( \sqrt{y} \right) ^2 + 9 } \, \left( \frac{1}{2\sqrt{y}} \right) \end{align*}$

    and since $\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}y} \left( \sqrt{y} \right) = \frac{1}{2\,\sqrt{y}} \end{align*}$, the substitution $\displaystyle \begin{align*} u = \sqrt{y} \implies \mathrm{d}u = \frac{1}{2\sqrt{y}}\,\mathrm{d}y \end{align*}$ is appropriate...
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