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!

Point on a Curve

  1. Jan 20, 2009 #1

    jgens

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data

    A particle moves along a path described by y = 4 - x^2. At what point along the curve are x and y changing at the same rate

    2. Relevant equations

    Simple equations regarding derivatives.

    3. The attempt at a solution

    It's been a while before I've done any related rates problems, could someone please let me know if this is correct:

    Since, x and y must be changing at the same rate (presumably with respect to time) x' = y' and y' = -2xx'. Therefore, -2x = 1 and x = -1/2. Placing my x value into the original equation yields 15/4. Hence, the point is (-1/2, 15/4).

    Thanks.
     
  2. jcsd
  3. Jan 20, 2009 #2
    SEEMS correct...
     
  4. Jan 20, 2009 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Of course, it's right. What could go wrong?
     
  5. Jan 20, 2009 #4

    jgens

    User Avatar
    Gold Member

    Plenty, I could have made an incorrect assumption ultimately leading to false conclusions.
     
  6. Jan 20, 2009 #5

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Good answer but you do not need to assume that x and y are varying wrt an external parameter. The derivative y'(x) = dy/dx of y wrt x expresses the instantaneous rate of change of y wrt a change in x.

    The points where y and x are changing at the same rate are those where y'(x)=1.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Point on a Curve
  1. Points on a curve (Replies: 1)

Loading...