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Perfect square

  1. Feb 10, 2008 #1
    1. The problem statement, all variables and given/known data


    finding the lengths of a line
    2. Relevant equations

    x = (y^3/3) + 1/(4y) from y =1 to y=3
    hint:: 1 + (dx/dy)^2 is a perfect square

    3. The attempt at a solution
    I found the solution, i did this by just finding the derivative and then putting it into the equation for finding a line and then messing around with it until i could get rid of the sqaure root in the equation, as usual.
    what i dont understand is the hint, what is the perfect square part all about?? what does it mean?? how does it help??

    thks
    jason
     
  2. jcsd
  3. Feb 10, 2008 #2

    arildno

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    DO post what you did, because if you did it CORRECTLY, then you'd immediately see the relevance of perfect squareness!
     
  4. Feb 10, 2008 #3
    Find the length of which line ? The equation you have posted will graph out to be a complex curve!
     
  5. Feb 10, 2008 #4
    ok, i found the derivative first which was y^2 - 1/(4y^2), i then squared this which gave me y^4 + 1/(16y^4) - 1/2, i then stuck this in the formula and added the one, which gave y^4 +1/(16y^4) + 1/2, i then put the terms together (16y^8 + 8y^4 + 1)/16y^4, i then seen the numerator is the a perfect square (4y^4 + 1)^2/16y^4, which enabled me to get rid of the square root and integrate (4y^4 + !)/4y^2 with no probs. what i dont understand is how 1 + (dx/dy)^2 is a perfect square and where it helps?
     
  6. Feb 10, 2008 #5
    >>i then stuck this in the formula

    Which formula ?
     
  7. Feb 10, 2008 #6
    The formula for length of a line y = f(x)

    Length = int [sqrt(1+ (dx/dy)^2)] dx

    in this case the variable is y,
     
  8. Feb 10, 2008 #7
    Find the length of which line ? The equation you have posted will graph out to be a complex curve!

    the line will be continuous from y =1 to y= 3, not quite sure what u mean, there is only one line x = (y^3/3) + 1/(4y).
     
  9. Feb 10, 2008 #8
    ok, i found the derivative first which was y^2 - 1/(4y^2), i then squared this which gave me y^4 + 1/(16y^4) - 1/2, i then stuck this in the formula and added the one, which gave y^4 +1/(16y^4) + 1/2, i then put the terms together (16y^8 + 8y^4 + 1)/16y^4, i then seen the numerator is the a perfect square (4y^4 + 1)^2/16y^4, which enabled me to get rid of the square root and integrate (4y^4 + !)/4y^2 with no probs. what i dont understand is how 1 + (dx/dy)^2 is a perfect square and where it helps?
     
  10. Feb 10, 2008 #9

    arildno

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    Well, not all students would have seen this by themselves.
    You did, so you didn't need the hint. Other students would have needed the hint to see what you saw on your own.
     
  11. Feb 10, 2008 #10
    ok i see, thks, but im still confused on how 1 + (dy/dx)^2 is a perfect square, doesnt there need to be a third term??
     
  12. Feb 10, 2008 #11

    Dick

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    You just showed that 1+(dy/dx)^2 is a perfect square in this very special case. They aren't claiming that it is ALWAYS an algebraic perfect square. It definitely isn't.
     
  13. Feb 10, 2008 #12
    ahh, i see, i was taking it too literaly, thks very much
     
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