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Calc help(please)

  1. Jul 14, 2004 #1
    When particle is at distance meters from the origin, there is a force of Newtons (N) pulling it toward the origin. How much work is needed to move the particle from the position to the position ?


    What is integral of 5^ radical x ?
  2. jcsd
  3. Jul 14, 2004 #2


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    That's a very strangely worded question. Let's call the distance from the origin, d, the force, F. Now what on earth do you mean "from the position to the position?" What position to what position? Very strange?!...

    [tex]\int 5^{\root n\of x}dx[/tex]

    Let [itex]u = x^{1/n}[/itex], therefore [itex]du = \frac{1}{n}x^{1/n - 1}dx[/itex].

    [tex]dx = \frac{ndu}{x^{1/n - 1}}[/tex]

    [tex]n\int \frac{5^udu}{u^{1 - n}}[/tex]

    Hmmm... maybe this is a start, I don't know where to go with this.

    EDIT: Look up integral tables, see if they have ways to solve things like [itex]\int u5^udu[/itex], you might be able to solve this for the first few n (n = 1, 2, 3), and if you find a pattern, you might be able to use induction to prove it generally.
    Last edited: Jul 14, 2004
  4. Jul 14, 2004 #3


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    Your integral is way beyond me. But http://integrals.wolfram.com/index.en.cgi says:

    [tex]\int 5^{x^{\frac{1}{n}}}dx = -nx \left(-x^{\frac{1}{n}} \right)^{-n} \Gamma \left( n, -x^{\frac{1}{n}} \ln 5 \right) \left(\ln 5\right)^{-n} + C[/tex]
  5. Jul 14, 2004 #4


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    Hah! Yes, I knew I was close :wink:.
  6. Jul 18, 2004 #5
    Wouldn't the constant force towards the origin indicate a vector field of constant magnitude with all vectors pointing toward the origin? So the work from any two positions would just be the line integral for the straight line between any two points over this vector field.
    Last edited: Jul 18, 2004
  7. Jul 18, 2004 #6


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    The line integral of the component of force parallel to the straight line over the line segment.
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