1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook