Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Resistance of a wire with changing radius

  1. Jun 23, 2011 #1
    1. The problem statement, all variables and given/known data
    The radius r of a wire of length L increases according to r = a * exp(bx^2), x is the distance from one end to the other end of the wire. What is the resistance of the wire?

    2. Relevant equations
    [itex]R =\frac{L * \rho}{A}[/itex]

    3. The attempt at a solution
    [itex]dR =\frac{dx * \rho}{A}[/itex]

    [itex]A(r) = \pi * r^2[/itex]

    [itex]r(x) = a* e^{bx^2}[/itex]

    [itex]A(x) =\pi a^2 * e^{2bx^2}[/itex]

    [itex]R =\int^{L}_{0}\frac{\rho}{\pi a^2 * e^{2bx^2}}*dx[/itex]

    Two questions
    First of all. Is this approach correct? And second, how does one integrate an errorfunction? (Of course one can use WolframAlpha but how does one get this solution?)
    Last edited: Jun 23, 2011
  2. jcsd
  3. Jun 23, 2011 #2


    User Avatar

    Remember that A is a function of r when you try to work out dR
  4. Jun 23, 2011 #3
    Of course, A is a function of r but r is also a function of x. Therefore the substitution yields what I have written above. Or am I wrong?
  5. Jun 23, 2011 #4


    User Avatar

    [tex]R =\frac{x \rho}{A}=\frac{x \rho}{\pi r^2}=\frac{x \rho}{\pi a^2 e^{2bx^2}}[/tex]

    Are you sure your expression for dR is correct?
  6. Jun 23, 2011 #5
    Our professor just told us that r = a * exp(b*x) and i found the mistake... A is dependent on r and therefore I need to consider this in my calculations...

    [itex]dR = \rho d(\frac{L}{A})[/itex]

    [itex]d(\frac{L}{A}) = \frac{dL}{A} - \frac{L*dA}{A^2}[/itex]

    [itex]dA = 2*\pi*r*dr[/itex] , [itex]dL = dx[/itex]

    [itex]dr = a*b*e^{bx} dx[/itex]

    [itex]dR =\rho \int^{L}_{0} (\frac{e^{-2bx}}{\pi*a} - \frac{2*x*b*e^{-2*b*x}}{\pi*a^2})dx[/itex]
    Last edited: Jun 23, 2011
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook