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Implicit derivative help

  1. Mar 7, 2012 #1
    1. The problem statement, all variables and given/known data The electrical potential can be described by the following equation:
    V= 200/(x2 + y2 )1/2 find dV when x=2 and y=1


    2. Relevant equations
    n/a


    3. The attempt at a solution
    dv/d(x,y) = ∂/∂x + ∂/∂y
    =200/(x2 + y2)(1/2) +200/(x2 + y2)(1/2)
    replace variables with C where necessary and rearrange for easy of deriving =200(x2 + C)(1/2) +200(C + y2)(1/2)
    Chain rule and exponent rule
    =200(-1/2)(2x)(x2 + C)(-3/2) +200(-1/2)(2y)(C + y2)(-3/2)
    =-200x(x2 + C)(-3/2) + -200y)(C + y2)(-3/2)
    =-200x(x2 + y2) )(-3/2) + -200y)(x2+ y2)(-3/2)
    Then I plug in the values x=2 and y=1 and dV=-53.7 V/m, I'm not very confident with multivariable calculus and just wanted to check to see if this is right.I've never had to do this before, it is fir physics, not a math class
     
  2. jcsd
  3. Mar 7, 2012 #2

    Mark44

    Staff: Mentor

    Your calculation looks OK, but the notation is a little off. It should be like this:
    dV = -200x(x2 + y2) (-3/2)*dx - 200y(x2+ y2)(-3/2)*dy

    To calculate dV, you need to know x and y, but you also need dx and dy, which you don't show. Did you forget to include that information?
     
  4. Mar 7, 2012 #3
    I am not very familiar with the notation. I haven't taken any multivariable calculus, and I'm a little rusty in my calculus in general and am just trying to solve a problem for my first year physics course. Do I need to solve for dx and dy, or do I just leave the dx and dy as bits of notation? I thought dx=-200x(x2 + y2) (-3/2) and dy=- 200y(x2+ y2)(-3/2)
     
  5. Mar 7, 2012 #4

    Mark44

    Staff: Mentor

    No, this should be given information.
    If you're not given values for dx and dy (or Δx and Δy), you'll have to leave these symbols in, and you won't be able to get a numeric value for dV. Are you sure there isn't some more information, something along the lines of x changes by ... and y changes by ... ? These would be your dx and dy.
    No, what you have as dx is the partial of V with respect to x, ##\frac{\partial V}{\partial x}##.
    And what you have as dy is the partial of V with respect to y, ##\frac{\partial V}{\partial y}##.

    The full equation is ##dV = \frac{\partial V}{\partial x}~dx + \frac{\partial V}{\partial y}~dy ##
     
  6. Mar 7, 2012 #5
    The question is as follows:
    "The electric potential in a region of space is V= 200/(x2 + y2 )1/2 , where x and y are in meters. What are the strength and direction of the electric field at (x,y) = (2.0m, 1.0m)?"
    the relevant equation from the chapter is E=-dv/ds = the negative slope potential graph. X and Y aren't changing, I'm just trying to find the negative slope of V at (2.0m, 1.0m)
     
    Last edited: Mar 7, 2012
  7. Mar 7, 2012 #6

    Mark44

    Staff: Mentor

    Well, that's very different from your original question. dV/ds is the rate of change of voltage with respect to s, but I don't know what s is, other than possibly distance along some curve or line.
     
  8. Mar 7, 2012 #7
    s is a distance, in the problem there is only one point given though... So in this scenario it would be -dV/d(x,y) cause s would be defined by the x,y coordinate system.
     
  9. Mar 7, 2012 #8

    Mark44

    Staff: Mentor

    No, I'm pretty sure there's no such thing. You can't take a derivative with respect to two variables.
    My guess is that you need to take a directional derivative to find the rate of change of V in the direction of (2, 1). Along this line, y = (1/2)x.
     
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