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Math Help - Find the length of the curve

  1. #1
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    Post Find the length of the curve

    6) Find the length of the curve

    I try many different way but I cant seem to find the rite answer! can you tell me the answer?
    Last edited by mr fantastic; April 29th 2010 at 04:10 AM.
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  2. #2
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    Quote Originally Posted by nhatie View Post
    6) Find the length of the curve


    I try many different way but I cant seem to find the rite answer! can you tell me the answer?
    Please show your working and say where you get stuck.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Please show your working and say where you get stuck.

    if find f'(x) = 9( y^1/3)- (-9/32)(y^-1/3)
    then i have 2pi int from -343 to 512 [(3)(y^4/3) - (3/32)(y^2/3)][sqrt[ 9( y^1/3)- (-9/32)(y^-1/3)] ???
    I am not sure what I am doing... can you show me how to solve and the right answer?
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  4. #4
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    Quote Originally Posted by nhatie View Post
    [CENTER]
    if find f'(x) = 9( y^1/3)- (-9/32)(y^-1/3)
    This is wrong. the derivative of y^n is ny^{n-1} so the derivative of y^{4/3} is (4/3)y^{1/3} and the derivative of 3y^{4/3} is 3(4/3)y^{1/3}= 4y^{1/3}

    Similarly, the derivative of \frac{3}{32}y^{2/3} is \left(\frac{2}{3}\right)\frac{3}{32} y^{-1/3}= \frac{1}{16}y^{-1/3}.

    then i have 2pi int from -343 to 512 [(3)(y^4/3) - (3/32)(y^2/3)][sqrt[ 9( y^1/3)- (-9/32)(y^-1/3)] ???
    I am not sure what I am doing... can you show me how to solve and the right answer?
    Where in the world did the " 2\pi" come from? You are calculating the arc-length of a path, not rotating anything around an axis! You seem to be grabbing formulas at random. The arclength of f(y), from y= a to b is \int_a^b \sqrt{1+ f'^2} dy

    Now, with f'= 4y^{1/3}- \frac{1}{16}y^{-1/3}, f'^2= 16y^{2/3}- (2)(4)(\frac{1}{16})y^{1/3}y^{-1/3}+ \frac{1}{256}y^{-2/3} = 16y^{2/3}- \frac{1}{2}+ \frac{1}{256}y^{2/3}.

    Notice that adding 1 to that just changes the "- 1/2" to "-1/2". And since that was a "perfect square" so is this: If x and y are such that 2xy= 1/2, then (x- y)^2= x^2- 2xy+ y^2= x^2- 1/2+ y^2 and then 1+ (x- y)^2= x^2+ 1/2+ y^2= x^2+ 2xy+ y^2= (x+y)^2.

    In other words, 1+ f'^2 is also a perfect square and so we can get rid of the square root!
    Last edited by mr fantastic; April 29th 2010 at 07:47 PM. Reason: Fixed some math and quote tags and a fraction.
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  5. #5
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    i have sqrt (4y^1/3 + 1/16 y^1/3)^2 => int -343 to 512 (4y^1/3 + 1/16 y^1/3)
    then
    65/16y^1/3 |from -343 to 512
    (65/16 (-343)^1/3) - (65/16 (512)^1/3)
    is it the anwer?
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