Results 1 to 4 of 4

Math Help - Rates

  1. #1
    Newbie
    Joined
    Apr 2009
    Posts
    2

    Rates

    Hi everyone, I can't seem to figure htis problem out.

    A dog heading due north at a constant speed of 2 meters per second trots past a fire hydrant at t = 0 sec. Another dog heading due east at a constant speed of 3 meters per second trots by the hydrant at t = 1 sec. At t = 9 sec, the rate of change of the distance between the two dogs is?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Apr 2009
    Posts
    166
    You're given a velocity. Integrate it to get position and use the extra information for your constant of integration.

    After that, you don't need calculus. Just take y2-y1/x2-x1 of the two dogs' positions.

    Edit: An added hint, their velocities are constant. So you'll be integrating constants. Oh, and remember to account for their direction (north and east).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2009
    Posts
    2
    So my two intergrals are 2x + C and 3x + C, right? How do I incorporate the t = 9 into those and finally come out with an answer? I'm sorry I just don't think I understand what you're saying
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Apr 2009
    Posts
    166
    Let the hydrant be our origin.

    \int v(t)dt = s(t) = 2dt = 2t + c

    To solve for c, use what they provided: 2 meters at t = 0

    2 = 2(0) + c

    c = 2

    s(t) = 2x + 2

    This dog is running north. So s(t) is returning a y-value.

    The other equation is s(t) = 3x + c, 3 = 3(1) + c, c = 0. So s(t) = 3x.

    This dog is running east, so you're getting an x-value.

    Now, at t = 9, dog1 is 2(9) + 2 away from the origin. dog2 is 3(9) from the origin.

    Now use y2-y1/x2-x1

    20 - 0 / 0 - 27 = -20/27
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Rates
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 4th 2009, 06:53 PM
  2. rates
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 7th 2008, 09:56 AM
  3. about rates
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 23rd 2008, 03:24 PM
  4. Rates and Related Rates!!
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 2nd 2008, 10:53 AM
  5. rates and related rates
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 29th 2007, 09:51 PM

Search Tags


/mathhelpforum @mathhelpforum