Results 1 to 4 of 4

Math Help - Trying to solve a quadratic to find critical points?

  1. #1
    Member
    Joined
    Dec 2009
    Posts
    103

    Trying to solve a quadratic to find critical points?

    Hi again guys,

    I'm trying to take the derivative of the velocity model of a space shuttle:


    v(t) = 0.001895t^3−0.08391t^2+13.69t+2.71

    I get the acceleration as:

    a(t) = 5.685E-3 t^2 - .16782 t + 13.69

    I want to find the critical numbers so I want to set this function equal to zero. I go for the quadratic but I get a negative under the square root. Is there something I'm missing in this problem? Any clarification is appreciated
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by DannyMath View Post
    Hi again guys,

    I'm trying to take the derivative of the velocity model of a space shuttle:


    v(t) = 0.001895t^3−0.08391t^2+13.69t+2.71

    I get the acceleration as:

    a(t) = 5.685E-3 t^2 - .16782 t + 13.69

    I want to find the critical numbers so I want to set this function equal to zero. I go for the quadratic but I get a negative under the square root. Is there something I'm missing in this problem? Any clarification is appreciated
    The graph of v(t) is practically linear.

    It does not differ much to 13.69t+2.71 for -30<t<30

    and outside of those t the graph is almost vertical.

    It has no turning points.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2009
    Posts
    103
    Quote Originally Posted by Archie Meade View Post
    The graph of v(t) is practically linear.

    It does not differ much to 13.69t+2.71 for -30<t<30

    and outside of those t the graph is almost vertical.

    It has no turning points.
    The question sets the interval between t= 0 and t= 38.9 and asks for the abs max and min acceleration.

    If there are no critical numbers should I just use the closed interval method from the texbook?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by DannyMath View Post
    The question sets the interval between t= 0 and t= 38.9 and asks for the abs max and min acceleration.

    If there are no critical numbers should I just use the closed interval method from the texbook?
    The graph of acceleration has a turning point as it is quadratic.

    The derivative is positive across that interval of t.

    The derivative is U-shaped so, the acceleration is a minimum, where the 2nd derivative is 0.

    The maximum value of acceleration occurs at t=38.9
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] help find critical points/good one
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 6th 2011, 04:54 PM
  2. Find the critical points.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 27th 2009, 08:38 PM
  3. Find the critical points of...
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 15th 2009, 03:06 AM
  4. Replies: 1
    Last Post: June 7th 2009, 04:53 PM
  5. Find The Critical Points
    Posted in the Calculus Forum
    Replies: 9
    Last Post: May 2nd 2009, 07:47 PM

Search Tags


/mathhelpforum @mathhelpforum