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Chain rule

  1. Nov 23, 2009 #1
    1. The problem statement, all variables and given/known data

    hi there.. can someone just give me an idea to solve this problem?

    Two 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?


    2. Relevant equations



    3. The attempt at a solution
    i dont have any idea..
     
  2. jcsd
  3. Nov 23, 2009 #2
    Sounds like a diagram could be very useful since this is involves geometry (please look at the attached diagram)

    Let's see what we're given:

    [tex]\frac{da}{dt} = -60km/h[/tex]

    [tex]\frac{db}{dt} = -80km/h[/tex]

    We need to find:

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

    -Try and find an expression for x in terms of a and b.

    -Then sub that into the above expression. Remember that

    [tex]\frac{dx}{dt}[/tex] is just [tex]\frac{d}{dt} (x)[/tex]

    -Then use the chain rule to differentiate that expression, remembering that 'a' and 'b' are functions of t.

    -Sub in the conditions, and voila!
     

    Attached Files:

  4. Nov 23, 2009 #3
    thank u very much for such ideas..
    erm.. x is a hypotenuse

    where x=[tex]\sqrt{a^{2} + b^{2}}[/tex]

    let say the distance is D

    D = sqrt(0.5^2 + 0.7^2)
    that should be my x or hypotenuse right?
     
    Last edited: Nov 23, 2009
  5. Nov 23, 2009 #4
    [tex]\frac{dx}{dt} = \frac{d(\sqrt{a^2+b^2})}{dt}[/tex]

    Differentiate that using the chain rule (since 'a' and 'b' are functions of t) and you will get an expression

    [tex]\frac{dx}{dt} = ........[/tex],

    where the right hand side contains 'a', 'b', [tex]\frac{da}{dt}[/tex] and [tex]\frac{db}{dt}[/tex].

    Then you sub in a = 0.5, b = 0.7, [tex]\frac{da}{dt} = -60[/tex], [tex]\frac{db}{dt} = -80[/tex] and voila
     
  6. Nov 23, 2009 #5
    got one question here..
    why is the speed in negative?
     
  7. Nov 23, 2009 #6
    X dX/dt = A da/dt + B db/dt

    am i right?
    so.. i should find dX/dt right?
    i plug in all the value..

    i got 100 km.. am i answered it correctly?
     
  8. Nov 23, 2009 #7
    Yep I think that's correct (100km/h)
     
  9. Nov 23, 2009 #8
    got one question here..
    why is the speed in negative?
     
  10. Nov 23, 2009 #9
    da/dt and db/dt aren't speeds, they are the rate at which the distances a and b decrease. So actually dx/dt should be -100km/h (oops), because it is decreasing as the cars get closer
     
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