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Math Help - Maximum possible length?

  1. #1
    Member Pranas's Avatar
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    Maximum possible length?

    Hello.

    This seems pretty fundamental to me, however I was not able to find information about this particular problem on the net, so I'm writing here seeking your knowledge

    Let's say we have two connected perpendicular "tunnels" like in this picture:

    Widths are x and y respectively. Question is, what is the maximum length of an indefinitely narrow object, which is able to move from one end to another?

    Sorry for possible awkward presentation, English is obviously not my native language.
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  2. #2
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    pickslides's Avatar
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    Are there any restrictions on the angle the length sits within the tunnel?
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  3. #3
    Member Pranas's Avatar
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    Quote Originally Posted by pickslides View Post
    Are there any restrictions on the angle the length sits within the tunnel?
    There are no such restrictions. On the other hand, there is no fixed angle as well
    I mean, imagine black lines being "walls" and red line being a narrow "car" (couldn't think of a less lame comparison) - "car" will eventually have to make a nearly 90 degrees turn, but I do not know which particular moment will be the most complicated for the "car" to fit within the "walls"...
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  4. #4
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    Quote Originally Posted by Pranas View Post
    Hello.

    This seems pretty fundamental to me, however I was not able to find information about this particular problem on the net, so I'm writing here seeking your knowledge

    Let's say we have two connected perpendicular "tunnels" like in this picture:

    Widths are x and y respectively. Question is, what is the maximum length of an indefinitely narrow object, which is able to move from one end to another?

    Sorry for possible awkward presentation, English is obviously not my native language.
    [IMG]19976[/IMG]

    Here is my 2 cents

    From my diagram you can see that you are trying to maximize the quantity

    \displaystyle d_1+d_2=\frac{y}{\sin(\alpha)}+\frac{x}{\cos(\alph  a)}=M(\alpha)

    \displaystyle M'(\alpha)=-\frac{y\cos(\alpha)}{\sin^2(\alpha)}+\frac{x\sin(\  alpha)}{\cos^2(\alpha)}

    I used a CAS to solve for \displaystyle \alpha=\tan^{-1}\left( \sqrt[3]{\frac{y}{x}} \right)

    Then just plug this into M(\alpha)

    This gives  \displaystyle \left(x^{\frac{2}{3}}+y^{\frac{2}{3}}\right)^{\fra  c{3}{2}}
    Attached Thumbnails Attached Thumbnails Maximum possible length?-capture.jpg  
    Last edited by TheEmptySet; December 5th 2010 at 12:59 PM. Reason: immage messed up
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  5. #5
    Member Pranas's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    [IMG]19976[/IMG]

    Here is my 2 cents

    From my diagram you can see that you are trying to maximize the quantity

    \displaystyle d_1+d_2=\frac{y}{\sin(\alpha)}+\frac{x}{\cos(\alph  a)}=M(\alpha)

    \displaystyle M'(\alpha)=-\frac{y\cos(\alpha)}{\sin^2(\alpha)}+\frac{x\sin(\  alpha)}{\cos^2(\alpha)}

    I used a CAS to solve for \displaystyle \alpha=\tan^{-1}\left( \sqrt[3]{\frac{y}{x}} \right)

    Then just plug this into M(\alpha)

    This gives  \displaystyle \left(x^{\frac{2}{3}}+y^{\frac{2}{3}}\right)^{\fra  c{3}{2}}
    Thank you very much.

    Of course we are minimizing (you wrote maximizing) the quantity. Nevertheless, your method is very graceful and correct, while my initial idea using coordinates was way too complicated.

    I am able to approach
    \displaystyle \alpha_m_i_n=\arctan\left( \sqrt[3]{\frac{y}{x}} \right)
    then
    \displaystyle M(\alpha_m_i_n) = y \cdot \sqrt{1 + (\frac{x}{y})^{\frac{2}{3}}} + x \cdot \sqrt{1 + (\frac{y}{x})^{\frac{2}{3}}}
    as long as it seems the same as
     \displaystyle \left(x^{\frac{2}{3}}+y^{\frac{2}{3}}\right)^{\fra  c{3}{2}}
    I am not quite able to reach this wonderful expression. Could someone please give me a hint?
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  6. #6
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    Quote Originally Posted by Pranas View Post
    Thank you very much.

    Of course we are minimizing (you wrote maximizing) the quantity. Nevertheless, your method is very graceful and correct, while my initial idea using coordinates was way too complicated.

    I am able to approach
    \displaystyle \alpha_m_i_n=\arctan\left( \sqrt[3]{\frac{y}{x}} \right)
    then
    \displaystyle M(\alpha_m_i_n) = y \cdot \sqrt{1 + (\frac{x}{y})^{\frac{2}{3}}} + x \cdot \sqrt{1 + (\frac{y}{x})^{\frac{2}{3}}}
    as long as it seems the same as
     \displaystyle \left(x^{\frac{2}{3}}+y^{\frac{2}{3}}\right)^{\fra  c{3}{2}}
    I am not quite able to reach this wonderful expression. Could someone please give me a hint?
    First notice that

    \displaystyle \alpha=\tan^{-1}\left( \sqrt[3]{\frac{y}{x}}\right) \iff tan(\alpha)=\frac{y^{\frac{1}{3}}}{x^{\frac{1}{3}}  }

    Since \displaystyle 0 < \alpha \le \frac{\pi}{2} it must be in the first quadrant. Use the Pythagorean theorem to find the hypotenuse of the triangle. see diagram
    Maximum possible length?-capture1.jpg

    Using this we get that

    \displaystyle \sin(\alpha)=\frac{y^{\frac{1}{3}}}{\sqrt{x^{\frac  {2}{3}}+y^{\frac{2}{3}}}} \text{ and } \cos(\alpha)=\frac{x^{\frac{1}{3}}}{\sqrt{x^{\frac  {2}{3}}+y^{\frac{2}{3}}}}

    Now just plug this in to get

    \displaystyle M(\alpha)=\frac{y}{\frac{y^{\frac{1}{3}}}{\sqrt{x^  {\frac{2}{3}}+y^{\frac{2}{3}}}}}+\frac{x}{\frac{x^  {\frac{1}{3}}}{\sqrt{x^{\frac{2}{3}}+y^{\frac{2}{3  }}}}}=y^{\frac{2}{3}}\sqrt{x^{\frac{2}{3}}+y^{\fra  c{2}{3}}}}+x^{\frac{2}{3}}\sqrt{x^{\frac{2}{3}}+y^  {\frac{2}{3}}}}=

    \displaystyle \sqrt{x^{\frac{2}{3}}+y^{\frac{2}{3}}}}\left( x^{\frac{2}{3}}+y^{\frac{2}{3}}}\right)=\left(x^{\  frac{2}{3}}+y^{\frac{2}{3}}}\right)^{\frac{3}{2}}
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  7. #7
    Member Pranas's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    First notice that

    \displaystyle \alpha=\tan^{-1}\left( \sqrt[3]{\frac{y}{x}}\right) \iff tan(\alpha)=\frac{y^{\frac{1}{3}}}{x^{\frac{1}{3}}  }

    Since \displaystyle 0 < \alpha \le \frac{\pi}{2} it must be in the first quadrant. Use the Pythagorean theorem to find the hypotenuse of the triangle. see diagram
    Click image for larger version. 

Name:	Capture1.JPG 
Views:	3 
Size:	19.4 KB 
ID:	19987

    Using this we get that

    \displaystyle \sin(\alpha)=\frac{y^{\frac{1}{3}}}{\sqrt{x^{\frac  {2}{3}}+y^{\frac{2}{3}}}} \text{ and } \cos(\alpha)=\frac{x^{\frac{1}{3}}}{\sqrt{x^{\frac  {2}{3}}+y^{\frac{2}{3}}}}

    Now just plug this in to get

    \displaystyle M(\alpha)=\frac{y}{\frac{y^{\frac{1}{3}}}{\sqrt{x^  {\frac{2}{3}}+y^{\frac{2}{3}}}}}+\frac{x}{\frac{x^  {\frac{1}{3}}}{\sqrt{x^{\frac{2}{3}}+y^{\frac{2}{3  }}}}}=y^{\frac{2}{3}}\sqrt{x^{\frac{2}{3}}+y^{\fra  c{2}{3}}}}+x^{\frac{2}{3}}\sqrt{x^{\frac{2}{3}}+y^  {\frac{2}{3}}}}=

    \displaystyle \sqrt{x^{\frac{2}{3}}+y^{\frac{2}{3}}}}\left( x^{\frac{2}{3}}+y^{\frac{2}{3}}}\right)=\left(x^{\  frac{2}{3}}+y^{\frac{2}{3}}}\right)^{\frac{3}{2}}
    Indeed.
    Thanks again.
    That's a shame I was not able to figure this out myself, I need some more practicing, I guess...
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