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Finding the rate of change of the distance between 2 cars

  1. Oct 18, 2009 #1
    1. The problem statement, all variables and given/known data

    Car A is traveling north on Highway 16 and car B is traveling west on Highway 83. Each car is approaching the intersection of these highways. At a certain moment, car A is 0.3 km from the intersection and traveling at 90km/h while car B is 0.4km from the intersection and traveling at 80km/h. How fast is the distance between the cars changing at that moment?



    Attempt :

    D = sqrt(x^2 + y ^ 2)

    Let C = sqrt(x^2 + y^2);

    d(D) = dD/dx dx + dD/dy dy

    = x/C dx + y/C dy

    we are given x = 0.3, y = 0.4 , dx = 90 and dy = 80, so

    d(D) = dx/6 + 8dy/5;

    Now just plug in dx and y ,

    so the answer I got is 143 .

    Can you check this?
     
  2. jcsd
  3. Oct 18, 2009 #2

    lanedance

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    Homework Helper

    I think i see what you are trying to do, but i think its poor notation, be explicit in variables if confused

    let the distance between the 2 cars, D, be exressed in terms of their distances form the intersection, x,y
    [tex] D(x,y) = \sqrt{x^2 + y^2} [/tex]
    then at the point in time when x = 0.3, y = 0.4, then you let
    [tex] C = D(0.3,0,4) [/tex]
    i don't think there is a real need for C and its a little confusing...

    this the small change formula, change in D for a small change in x & y - though you need to show how you got to the next step (the differentiation)

    that said it should be in form of the partial derivatives of D wrt x & y, and you might as well relate it directly to the rate of cahnge with time of D, which is what you want at the end of the day ie.

    [tex] \frac{d D(x,y)}{dt} = \frac{\partial D(x,y)}{\partial x}\frac{dx}{dt} +\frac{\partial D(x,y)}{\partial y}\frac{dy}{dt}[/tex]

    this is true for all values, not just at C = D(0.3,0.4)
    these dx & dys are not small changes - so even if it gives you the corrcet answer (whih i'm not sure of) you should probably format it as given above

    ALso as x,y are the sides of a triangle, tey are decreasing in length
    [tex] \frac{d x}{dt} = -90km/hr, \frac{d x}{dt} = -80km/hr,[/tex]


    and finally what are your units?
     
    Last edited: Oct 18, 2009
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