Results 1 to 3 of 3

Math Help - Implicit derivative and related rates?

  1. #1
    Newbie
    Joined
    Jan 2013
    From
    Montreal
    Posts
    10

    Implicit derivative and related rates?

    Hi!

    I'm having a bit of trouble with this assignment. I had to miss a couple class, and I'm afraid the teacher talked about this. I have the notes, but they are not of much help and don't talk about related rates at all. I tried looking around, I found lots of PDF from different universities, but I still don't get how to solve this problem. If I could get some pointers, that'd be great

    Thanks

    Here's the problem (translated from French):
    A particle oscillates on the x-axis. It's position x(t) (in meters) at the time t (in seconds) is related to its speed v(t) by the equation:
    x^2+x \cdot v=v^2=13

    Give the acceleration of this particle when its position is x = -3 meters, knowing that at that moment, the speed is positive.

    Hint: This is a related rates problem in which you need to use the implicit derivative. Use the fact that a(t) = \frac{dv}{dt} and v(t) = \frac{dx}{dt}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    3,648
    Thanks
    601

    Re: Implicit derivative and related rates?

    Hey Kevindqc.

    Hint: Try differentiating both sides with respect to x and collect all the dv/dt terms and get that in terms of the rest of the expression. (Note the chain rule d/dx u(g(x)) = dg/dx u'(g(x)))
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2013
    From
    Montreal
    Posts
    10

    Re: Implicit derivative and related rates?

    I'm not sure I get what you're saying

    is v' the dv/dt terms you are talking about? it's not dv/dx? I'm confused

    like:
    x^2+x \cdot v + v^2 = 13
    (x^2)' + (x \cdot v)' + (v^2)' = (13)'
    2x + (1 \cdot v + v' \cdot x) + 2 \cdot v \cdot v' = 0
    v' \cdot x + 2 \cdot v \cdot v' = -2x - v
    v' ( x + 2 \cdot v) = -2x - v
    v' = \frac{-2x-v}{x+2v}

    What am I supposed to do with v' :/

    I don't understand what "get that in terms of the rest of the expression" means :x

    I guess what really bugs me is the t
    Last edited by Kevindqc; February 6th 2013 at 09:36 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Implicit Differentiation/Related Rates
    Posted in the Calculus Forum
    Replies: 5
    Last Post: October 19th 2011, 06:02 PM
  2. Related rates and implicit
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 11th 2009, 08:26 AM
  3. Related Rates / Implicit Differentiation
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 5th 2009, 08:24 AM
  4. Applications of the derivative/Related Rates
    Posted in the Business Math Forum
    Replies: 1
    Last Post: November 22nd 2008, 03:21 PM
  5. Related Rates (implicit differentiation)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 24th 2008, 02:18 PM

Search Tags


/mathhelpforum @mathhelpforum