Thursday
March 23, 2017

Post a New Question

Posted by on Thursday, November 11, 2010 at 9:21pm.

Two people start from the same point. One walks east at 1 mi/h and the other walks northeast at 2 mi/h. How fast is the distance between the people changing after 15 minutes?

  • Related rates - , Thursday, November 11, 2010 at 9:43pm

    At a time of t hours
    let the distance covered by the eastbound person be 1t km
    let the distance covered by the northeastbound person be 2t km
    let the distance between them be d km
    by the Cosine Law
    d^2 = t^2 + 4t2 - 2(t)2t)cos45°
    = 5t^2 - 4t^2(√2/2)
    = 5t^2 - 2√2 t^2
    when t = 15 min or .25 hours
    d^2 = 5(.25)^2 - 2√2(.25)^2 = .489277
    d = .6995

    2d dd/dt = 10t -4√2t
    dd/dt = (5t - 2√2t)/d
    = (5(.25) - 2√2(.25))/.6995 = .776

    at that moment they are separating at .776 km/h

    check my arithmetic

  • Related rates - , Thursday, November 11, 2010 at 9:55pm

    Incorrect

  • Related rates - , Thursday, November 11, 2010 at 10:08pm

    If you had checked my arithmetic like I suggested you would have found my error to be in
    d^2 = 5(.25)^2 - 2√2(.25)^2 = .489277
    It should have been
    d^2 = 5(.25)^2 - 2√2(.25)^2 = .135723
    then
    d = .3684

    I will let you make the rest of the corrections.

    BTW, I also used km/h instead of mi/h, but that has no effect on the calculation

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question