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Partial Derivative: Chain Rule

  1. Nov 25, 2009 #1
    1. The problem statement, all variables and given/known data

    2 straight roads intersect at right angles. Car A, moving on one of the roads, approaches the intersection at 60km/h and car B moving on the other road, approaches the intersection at 80km/h. At what rate is the distance between the cars changing when A is 0.5km from the intersection and B is 0.7km from the intersection?


    3. The attempt at a solution

    None. Dont know how. Dont give me the solution, give me a clue i want to try to solve it.
     
  2. jcsd
  3. Nov 25, 2009 #2

    arildno

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    1. Let the intersection be the origin in your coordinate system.

    2. Now, let A's position at t (hours)=0 be 0.5 (km) on the right-hand side of the origin.
    Then, A's position vector as a FUNCTION of time, A(t) will be:
    [tex]A(t)=(0.5-60*t,0)[/tex]
    (Agreed?)
    Can you set up a similar vectorial position function for B?


    3. Now that you have both position functions, how can you find the DISTANCE function, D(t) between them? (Hint: Pythagoras..)

    4. You are asked for the rate of change for D(t).
    How can you find that?
     
  4. Nov 25, 2009 #3
    so..
    2) B(t) = (.7 - 80*t , 0)

    3) D(t) = (.5^2 + .7^2)^1/2

    i cant figure out this one...--->> 4)You are asked for the rate of change for D(t).
    How can you find that?
     
  5. Nov 25, 2009 #4

    arildno

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    2. is wrong, it should be: B(t)=(0, .7-80t)
    3. is doubly wrong, since it
    a) only gives D(0), rather than D(t)
    and
    b) calculates it incorrectly from the A(t) and (incorrect) B(t) you set up.
     
  6. Nov 25, 2009 #5
    if.. i do it my way, i got the final answer is -100km/h..
    is that same as yours?
     
  7. Nov 25, 2009 #6

    arildno

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    Post your way, please.
     
  8. Nov 25, 2009 #7
    i got dX/dt = -100km/h
     
  9. Nov 25, 2009 #8

    arildno

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    That is indeed correct! :smile:
     
  10. Nov 25, 2009 #9
    hurmm.. i'm still wondering why did my lecturer told me that there's something
    wrong with my solution..:frown:
     
  11. Nov 25, 2009 #10

    arildno

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    Well, if you used YOUR version of D(t), then it is a totally incorrect procedure.

    The rate of change of THAT expression is just 0.
     
  12. Nov 25, 2009 #11
    did u mean by the above D(t)
     
  13. Nov 25, 2009 #12

    arildno

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    Quite so.
    That function is equal to a constant, its derivative is..0
     
  14. Nov 25, 2009 #13
    just for final confirmation...
    my calculation and my answer for this question is totally correct
    right?
     
  15. Nov 25, 2009 #14

    arildno

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    Not if you specified D(t) in the manner YOU did, no.

    The answer won't follow from that at all.
     
  16. Nov 25, 2009 #15
    could anyone pls give me the solution?
    damn im stuck and it is 3.30am over here
    :(
     
  17. Nov 25, 2009 #16

    arildno

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    Then follow my hints in post 2.

    They are sufficient.
     
  18. Nov 25, 2009 #17
    yeah i read all this thread..but still cant do it (im stupid :( )
    if anyone could go through everything once again i would be thankful
     
  19. Nov 25, 2009 #18
    i cant do it :(
    im stupid
    and i dont know anything on how to do it
     
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