1. Not finding help here? Sign up for a free 30min 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!

Calculus Rates Question?

  1. Apr 18, 2013 #1
    1. The problem statement, all variables and given/known data
    Car 1 is 148 km north of Car 2. Car 1 moves east at 24km/h while Car 2 moves north at 19 km/h. What rate is the distance between them changing after 4 hours?





    3. The attempt at a solution

    Here is What I did:

    By the Pythagorean Theorem,
    d^2 = y^2 + x^2
    d/dt(d^2) = d/dt(y^2 + x^2)
    2d(dd/dt) = 2y(dy/dt) + 2x(dx/dt)
    At 4 hours, y = 148 + 4(19) = 148 + 76 = 224, x = 4(24) = 96, and d = (224^2 + 96^2)^(1/2) = 243.7
    Substitute 224 for y, 96 for x, 19 for dy/dt, 24 for dx/dt, and 243.7 for d, and solve for dd/dt.
    2(243.7)dd/dt = 2(224)19 + 2(96)24
    dd/dt = 26.9 km/hr


    My question is the bolded part: should I be subtracting the 76km car 2 travels in 4 hours from 148km or adding it like I did?
     
  2. jcsd
  3. Apr 18, 2013 #2

    chiro

    User Avatar
    Science Advisor

    Hey Stanc.

    Since Car 2 is -148 km north (+148km south) then the square distance in terms of the y co-ordinate should be (-148 + 4 x 19)^2.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Calculus Rates Question?
Loading...