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Homework Help: Rate of Change in a coordinate plane

  1. Oct 3, 2005 #1
    Hi. Im drawing a complete blank on this calc problem.

    Point a moves along the x-axis at the constant rate of 'a' ft/sec
    while point b moves along the y-axis at the constant rate of 'b'
    ft/sec. Find how fast the distance between them is changing when A is
    at the point (x,0)and B is at the point (0, y).

    I know that dx / dt = a, and dy / dt = b, however Im stuck on where to go afterwards. I drew a graph and thought that the slope would lead
    me to an answer, but I cant quite figure it out. Since Im looking for the change in distance between the two points I figured that I may have to incorporate the distance formula rather than some formula with slope, however I cant quite figure out how to go about it. :redface:

    Thanks in advance for any help.
     
  2. jcsd
  3. Oct 3, 2005 #2

    hotvette

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    The wording of the problem provides the clue as to how to proceed.

    This tells you two things:

    1. You need to develop an equation for the distance between the points

    2. differentiate (1) and evaluate at the desired x & y
     
  4. Oct 4, 2005 #3
    [tex]d = \sqrt{\Delta x^2 + \Delta y^2}[/tex]

    [tex]\frac{dx}{dt} = a[/tex]

    [tex]\frac{dy}{dt} = b[/tex]

    That's what I've come up with so far.

    Now, since I am trying to find the rate of change of the distance, then I believe I am looking for [tex]\frac{dd}{dt}[/tex]

    Taking the derative of the distance formula directly would be pretty sloppy, so I think I should plug in for the values of x and y and then evaluate the deriv. I believe that the two points in question are: (x+a, 0) and (0, y+b). Is this correct? This is where the points A and B are after 1 second.

    If so, then [tex]\Delta x = (x+a) - (0) = (x+a)[/tex] and [tex]\Delta y = (0) - (y+b) = -y-b[/tex]
     
    Last edited: Oct 5, 2005
  5. Oct 5, 2005 #4
    :eek: Even after plugging in the deriv is ridiculous.

    There must be a way to plug in something else. Am I using the wrong coordinates/formula?
     
  6. Oct 5, 2005 #5
    I don't see a "t" anywhere in there. Maybe I am wrong, but it seems that somehow your distance equation needs to reflect that X and Y are changing as a function of t.
     
  7. Oct 5, 2005 #6

    CarlB

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    Try replacing [itex]d = \sqrt{\Delta x^2 + \Delta y^2}[/itex] with [itex]d = \sqrt{x^2 + y^2}[/itex]. Maybe you've been studying partial derivatives recently.

    Carl
     
  8. Oct 5, 2005 #7
    I was taught that [tex]\frac{d(some var)}{dt}[/tex] = the Rate of Change of some var. with respect to time. The rate of change of x with respect to time is 'a', and of y with respect to time is 'b'. I am trying to find the rate of change of distance (var. d) with respect to time. Im pretty sure that this part is correct?
     
  9. Oct 5, 2005 #8
    Sorry, but Im not sure where you are going with this.

    When [tex]d = \sqrt{x^2 + y^2}[/tex] :

    [tex]\frac{dd}{dt} = \frac{1}{2} (x^2 + y^2)^\frac{-1}{2} (2x\frac{dx}{dt} + 2y\frac{dy}{dt})[/tex]

    [tex]\frac{dd}{dt} = \frac{1}{2} * \frac{1}{\sqrt{x^2+y^2}} * 2xa + 2yb[/tex]

    [tex]\frac{dd}{dt} = \frac{xa + yb}{\sqrt{x^2+y^2}}[/tex]

    I have 2 points, so which would i plug in for x and y??
     
    Last edited: Oct 5, 2005
  10. Oct 5, 2005 #9

    CarlB

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    x and y were given as data to the problem. You don't plug anything in for them. You're done and I think you got the right answer.

    Carl
     
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