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Using integrals to calculate arc length

  1. Feb 11, 2007 #1
    1. The problem statement, all variables and given/known data

    Just started Calc II last month, it's been smooth so far but I've run into a bit of snag involving the application of integrals in the calculation of arc length.

    The formula you use is the definite integral of (1+(d/dx)^2)^.5.

    Often once you derive the d/dx and square it, you're left with a somewhat nasty looking equation under the radical. Deriving this square root is what's giving me the trouble. Is there any particular technique?

    Here's one example:

    Y= ((x^4)/8) + 1/4x^2, In the interval [1,2]

    Y'= (x^3)/2 - 1/2x

    3. The attempt at a solution

    (1+(x^(3/2) - 1/2x)^2)^.5

    I need to integrate this from 1 to 2, but how does can one easily transform ut into an integrable form using algebra!

    EDIT: I should mention that the textbook indicated that integrals involving arc lentgh are "often very difficult to evaluate" yet proceeded to present examples where such integration was smoothly carried out.
    Last edited: Feb 11, 2007
  2. jcsd
  3. Feb 11, 2007 #2
    y' is wrong , the derivative of 1/4x^2 is -1/2x^3
  4. Feb 11, 2007 #3


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    Gold Member

    Er, you need to subtract 1 from the exponent. What you did is wrong. You'll get x/2.
  5. Feb 11, 2007 #4
    if it was (1/4)x^2 i would be wrong
    but if you notice he wrote:
    Y= ((x^4)/8) + 1/4x^2
    Y'= (x^3)/2 - 1/2x
    the nagative sign for 1/2x seem to mean he have 1/(4*x) and derived it wrong
    that's how i saw it..
    anyway however you look at it he did a mistake , whether it was the derivation it self , or putting a minus from no where..
    Last edited: Feb 11, 2007
  6. Feb 11, 2007 #5

    [itex]y = \frac{x^4}{8} + \frac{x^2}{4}[/itex]


    [itex]y' = \frac{x^3}{2} + \frac{x}{2}[/itex].
  7. Feb 11, 2007 #6
    BTW taking it as (1/(4*x^2)) makes the integral very easy to solve.
    But we need his input on this..
  8. Feb 11, 2007 #7
    Sometimes I guess it helps to reassess your assumptions...

    I did indeed have the derivative wrong, which is why the integral was giving me such a hard time:redface: . But I solved it, so thanks for the help! :smile:
  9. Feb 11, 2007 #8
    These were my steps:

    d/dx = (x^3)/2 -1/2x^3

    You take out the one half and you get: (1/2)(X^3 - 1/x^3)

    Now you square it, and you get: (1/4)(X^6 - 2 + 1/x^6)

    It is this plus 1 that we take the square root of, so we incorpate the 1 by adding four to the product (balanced out by (1/4)) so new your radical looks like: ((1/4)(X^6 + 2 + 1/x^6))^.5

    which equals: (1/2)((X^3)+(1/X^3))

    integral is (X^4)/8 - 1/4x^2) from one to two, final answer 33/16...
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