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: Griffith's E&M Derivation

  1. Dec 9, 2009 #1
    1. The problem statement, all variables and given/known data
    Please see the attached pdf file, which is the bottom of page 365 from the 3rd edition of the book. This is a lesson about (generalized) waves, and f(z,t) is the vertical displacement of the medium at point z, time t. F is net force and T is the tension on the string in the picture.

    I don't understand how we get from the tangents of angles to partial derivatives in this derivation. I've asked a professor about this before, and he said something about tangent lines to curves being connected to derivatives, but it didn't make sense because the tangent line to a curve is a different use of tangent than what I have here, tangent the trig function. Is there some connection between the two?

    Also, the last step (from first to second partial derivatives) seems too hand wavy to be legitimate.

    2. Relevant equations

    None for my first question.

    For my second question, I understand that it involves the definition of the derivative:

    [tex]\lim_{\Delta z \to 0} \frac{\frac{\partial f(z+ \Delta z)}{\partial t} -
    \frac{\partial f(z)}{\partial t}}{\Delta z} = \frac{\partial^{2} f}{\partial t^{2}}[/tex]

    What I'm not sure about is if it's mathematically sound to just multiply both sides of that equation by [tex]\Delta z[/tex], seeing as how one side has a limit.

    3. The attempt at a solution

    I have no idea.

    Attached Files:

  2. jcsd
  3. Dec 9, 2009 #2


    User Avatar

    I'm responding to your first question. Then I'll look at your second question.

    Consider the definition of the trig tangent when you are looking at a right triangle. Its tangent equals opposite over adjacent.

    Now consider the derivative expressed as

    [tex] \frac{dy}{dx} [/tex]

    for a curve increasing from left to right. (imagine a parabola) Draw a tangent to the curve at some point P. What is the slope of the tangent? It's the change in y over the change in x, or
    [tex] \frac{dy} {dx} [/tex] which is the "opposite", dy, over the "adjacent", dx. (Draw in little lines to represent the change in dy and dx to make this clear, if you need to ).
  4. Dec 10, 2009 #3


    User Avatar
    Homework Helper
    Gold Member

    Surely you mean

    [tex]\lim_{\Delta z \to 0} \frac{\frac{\partial f(z+ \Delta z)}{\partial z} -
    \frac{\partial f(z)}{\partial z}}{\Delta z} = \frac{\partial^{2} f}{\partial z^{2}}[/tex]

    ...right? :wink:

    Anyways, the point is, that for sufficiently small [itex]\Delta z[/itex], you have

    [tex]\lim_{\Delta z \to 0} \frac{\frac{\partial f(z+ \Delta z)}{\partial z} -
    \frac{\partial f(z)}{\partial z}}{\Delta z} \approx\frac{\frac{\partial f(z+ \Delta z)}{\partial z} -
    \frac{\partial f(z)}{\partial z}}{\Delta z}[/tex]

    In other words, the slope of the line connecting the point [tex](z,\frac{\partial f(z)}{\partial z}})[/itex] to the point [tex](z+\Delat z,\frac{\partial f(z+\Delta z)}{\partial z}})[/itex] is approximately equal to the tangent to the curve [tex]\frac{\partial f(z)}{\partial z}}[/itex] at the point [tex](z,\frac{\partial f(z)}{\partial z}})[/itex]
    Last edited: Dec 10, 2009
  5. Dec 11, 2009 #4
    Yes, that's what I meant.

    I don't understand why that is a legitimate approximation. If [tex]\Delta z[/tex] is small, why wouldn't we say instead that the expression becomes large or approaches infinity?
  6. Dec 11, 2009 #5


    User Avatar
    Homework Helper
    Gold Member

    If [tex]\Delta z[/tex] is small, then so is the difference [tex]\frac{\partial f(z+ \Delta z)}{\partial z} -
    \frac{\partial f(z)}{\partial z}[/tex]....one small number, divided by another small number can easily produce finite result.

    Think about the relationship between speed and position in one-dimension....the exact speed of a particle at time [itex]t[/itex] is given by [tex]v(t)=\frac{dx}{dt}=\lim_{\Delta t\to 0}\frac{x(t+\Delta t)-x(t)}{\Delta t}[/itex]....but if you measure the average speed over some very small time interval from [itex]t[/itex] to [itex]t+\Delta t[/itex], [tex]v_{av}=\frac{x(t+\Delta t)-x(t)}{\Delta t}[/tex] would you not expect it to be close to the exact speed at time [itex]t[/itex]?
  7. Dec 11, 2009 #6
    Oh, I see. In that case, my question is: Why assume or focus on the case in which [tex]\Delta z[/tex] is small?
  8. Dec 11, 2009 #7


    User Avatar
    Homework Helper
    Gold Member

    You are ultimately interested in finding the displacement of the string at any point along its length at any given time....looking at the difference in tension (force) over very small intervals [itex]\Delta z[/itex] allows you to determine the average acceleration that small piece of the string experiences at any given time...in the limit that [itex]\Delta z\to 0[/itex], you thus determine the exact acceleration of each point along the string, and hence you can then find the displacement of every point along the string (at any given point in time).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook