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Rate of change of distance

  1. Mar 21, 2009 #1
    1. The problem statement, all variables and given/known data
    A point P (x,y) is moving along the line y=(1/2)x. If x is decreasing at the rate of 3units/s, at wha rate is the distance between P and the fixed point Q (0,11) changing at the instant when P is at (6,3)?


    2. Relevant equations
    Implicit differentation



    3. The attempt at a solution
    From what I can see for this question, the distance formula needs to be used. That is what I did, I plugged in the point (x,y) and (0,11) into the distance formula and found the derivative, then plugged in (6,3) for x and y and -3 for dx/dt. The answer I got was -2.6, however, this is not the right answer as it is supposed to be -0.6. This question seems easy yet it is frustrating me, if anyone could be help it would be greatly appreciated, thank you.
     
  2. jcsd
  3. Mar 21, 2009 #2

    gabbagabbahey

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    Your method looks fine; why don't you show me your work and I can see where you are going wrong...
     
  4. Mar 21, 2009 #3
    Point 1=(x,y)
    Point 2=(0,11)

    s=((x-0)^2+(y-11)^2)^1/2
    s=(x^2+(y-11)^2)^1/2
    s=(1/2)(x^2+(y-11)^2)^-1/2(2x(dx/dt)+2(y-11)(1))
    s=(1/2)(6^2+(3-11)^2)^-1/2(2(6)(-3)+2(3-11))
    s=-(1/2)(100)^-1/2(-52)
    s=-2.6

    I think I have done the chain rule right, as I have attempted this question a few times, all with the same results.
     
  5. Mar 21, 2009 #4

    gabbagabbahey

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    so far so good, but...

    This line makes no sense....even if you meant to say

    ds/dt=(1/2)(x^2+(y-11)^2)^-1/2(2x(dx/dt)+2(y-11)(1)) (as opposed to s=...)

    you would still be incorrect:

    [tex]\frac{d}{dt}[x^2+(y-11)^2]\neq(2x(\frac{dx}{dt})+2(y-11)(1))[/tex]

    because

    [tex]\frac{dy}{dt}=\frac{d}{dt}\left(\frac{x}{2}\right)\neq1[/tex]
     
  6. Mar 21, 2009 #5
    Hmm so that is where I am going wrong? How would I properly go about fixing that line then? Because from what I was taught, in adding situations you apply the derivative to both terms:

    (d/dt)(x^2 +(y-11)^2)
    (d/dt)(x^2)+(d/dt)((y-11)^2)
    (2x)(dx/dt)+2(y-11)((d/dt)(y-11))
    (2x)(dx/dt)+2(y-11)(1)

    That is how I came about that, and because it is the chain rule from ds/dt=(1/2)(x^2+(y-11)^2)^-1/2, I multiplied (2x)(dx/dt)+2(y-11)(1) by it.
     
  7. Mar 21, 2009 #6

    gabbagabbahey

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    up to here you are correct, but...
    where is the (1) coming from....shouldn't that be (2x)(dx/dt)+2(y-11)(dy/dt)?
     
  8. Mar 21, 2009 #7
    I used 1 because no information is given for dy/dt and the question is not asking for it, unless I am required to take the derivative of the equation given and plug it in for dy/dt? I think the light buld may have clicked :tongue:
     
  9. Mar 21, 2009 #8
    Nope, even if the derivative of y=(1/2)x is taken and used it simply results in an answer of -2.2 instead of -2.6.
     
  10. Mar 21, 2009 #9

    gabbagabbahey

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    So whenever you have any equation with a variable and you aren't told what that variable is equal to; you automatically just assume it is equal to one?!:confused:

    Of course, how else are you supposed to find dy/dt?
     
  11. Mar 21, 2009 #10

    gabbagabbahey

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    I get -0.6 when I do it, so you must be making a mistake....show me the calculations
     
  12. Mar 21, 2009 #11
    No, I just overlooked it in this question by accident, spent too much time on the other questions, it drained my mind :tongue:

    And I just got -0.6 for my answer. I forgot to include dx/dt in the derivative, for I simply took it to be dy/dt=1/2 instead of dy/dt=(1/2)(dx/dt); I fixed that little error and came up with the right answer at last. Thanks for all that help gabbagabbahey, I don't know if I have spotted my mistakes if it was not for your assistance :smile:
     
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