Results 1 to 4 of 4

Thread: 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
    4
    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 $\displaystyle v(t)$ is practically linear.

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

    and outside of those $\displaystyle t$ 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 $\displaystyle v(t)$ is practically linear.

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

    and outside of those $\displaystyle t$ 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
    4
    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 $\displaystyle t$.

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

    The maximum value of acceleration occurs at $\displaystyle t=38.9$
    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: Nov 6th 2011, 04:54 PM
  2. Find the critical points.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Sep 27th 2009, 08:38 PM
  3. Find the critical points of...
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Sep 15th 2009, 03:06 AM
  4. Replies: 1
    Last Post: Jun 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