Results 1 to 7 of 7

Math Help - Cos derivative problem

  1. #1
    Newbie
    Joined
    Dec 2008
    Posts
    19

    Cos derivative problem

    Hey i was having a problem with a hw problem
    the function f(x)=(cosx)^2 +2cosx over one complete period beginning with x=0
    First we have to find all values of x in this period when f(x)=0.
    I know the period is 0-2π by looking at the graph however I don't know how to figure it out analyitically though. So i found the zeroes but again i dont know how to do it analytically because i am not got with sin and cos
    Second we have to find all values of x in the period at which the function has a minimum.
    3rd we have to find when the period is concave up but i don't know how to find the 2nd derivative of a cos
    Last edited by calc_help123; December 29th 2008 at 05:08 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    cos^{2}(x)+2cos(x)=0

    The trick to use is let u=cos(x), then we get a quadratic

    u^{2}+2u=0

    u=0, \;\ -2

    cos(x)=0, \;\ cos(x)=-2

    -2 is non real, so let's go with 0.

    cos(\frac{\pi}{2})=0

    Therefore, we have f(x)=0 at

    C{\pi}-\frac{\pi}{2}

    Where C is an integer.

    To find the second derivative of cos(x):

    f(x)=cos(x), \;\ f'(x)=-sin(x), \;\ f''(x)=-cos(x), \;\ f'''(x)=sin(x)
    f''''(x)=cos(x). See?. It repeats.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Aug 2008
    Posts
    530
    Quote Originally Posted by calc_help123 View Post
    Hey i was having a problem with a hw problem
    the function f(x)=(cosx)^2 +2cosx over one complete period begining with x=0
    First we have to fun all values of x in this period when f(x)=0.
    I know the period is 0-2π by looking at the graph however I dont know how to firgure it out analyitically though. So i found the zeroes but again i dont know how to do it analytically bc i am not got with sins and cos
    Second we have to find all values of x in the period at which the function has a minimum.
    3rd we have to find when the period is concave up but i dont know how to find the 2nd derivative of a cos
    f(x) = \cos^2 x + \cos x

    a) for f(x) = 0,

     0 = \cos^2 x + \cos x

    0 = \cos x(\cos x +2)

    \Rightarrow \cos x = 0\;\;OR\;\; \cos x +2 =0

    \Rightarrow \cos x = 0\;\;OR\;\; \cos x \ne -2, \;\; because\;\;-1\le \cos x \le 1

    now finish it

    b) For the values of x in which f(x) is minimum, f'(x) = 0

    f'(x) = 2\cos x (-\sin x)+(-\sin x)

     f'(x) = -\sin x (2\cos x +1)

      -\sin x (2\cos x +1)=0

     <br />
-\sin x=0\;\;OR\;\;2\cos x +1=0<br />

     <br />
\sin x =0 \;\;OR\;\;\cos x =\frac{-1}{2}<br />

    Now finish it.

    c) second derivative, use product rule in  f'(x) = -\sin x (2\cos x +1)

    f"(x) = -\cos x (2\cos x +1)-\sin x (-2\sin x)

     f"(x) = -2\cos^2 x -\cos x -2\sin^2 x

    f"(x) = -2(\cos^2 x + \sin^2 x) - \cos x

     <br />
f"(x) = -2 -\cos x<br />

    Now finish it.

    Did you get it now???
    Last edited by Shyam; December 29th 2008 at 06:00 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member OnMyWayToBeAMathProffesor's Avatar
    Joined
    Sep 2006
    Posts
    157
    f(x) = \cos^2 x + \cos x
    I am sorry Shyam but it is f(x) = \cos^2 x + 2\cos x not f(x) = \cos^2 x + \cos x. That changes the first and second derivative drastically doesn't it?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member OnMyWayToBeAMathProffesor's Avatar
    Joined
    Sep 2006
    Posts
    157
    so the second derivative is 2sinx^2-2cosx^2-2cos?

    and the zeros are at 1.0471976, \pi, and 5.2359878 ? Can those numbers be made into more exact, easier numbers? maybe in terms of \pi? or square roots?

    This problem is almost done! Thanks everyone!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    f(x) = \cos^2{x} + 2\cos{x}

    f'(x) = -2\cos{x}\sin{x} - 2\sin{x} = -\sin(2x) - 2\sin{x}

    f''(x) = -2\cos(2x) - 2\cos{x} = -2[\cos(2x) + \cos{x}] = -2(2\cos^2{x} + \cos{x} - 1)

    the last expression will also factor ...

    f''(x) = -2(2\cos{x} - 1)(\cos{x} + 1)

    f''(x) = 0 at x = \frac{\pi}{3} \, , \,  x = \frac{5\pi}{3} \, , \,  x = \pi
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member OnMyWayToBeAMathProffesor's Avatar
    Joined
    Sep 2006
    Posts
    157
    Thanks skeeter, shyam, galactus and calc_help123!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Derivative Problem
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 21st 2010, 06:22 PM
  2. Replies: 1
    Last Post: October 7th 2010, 08:23 AM
  3. Derivative Problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 28th 2010, 08:37 PM
  4. Derivative problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 27th 2010, 06:13 AM
  5. Derivative Problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 15th 2009, 02:36 PM

Search Tags


/mathhelpforum @mathhelpforum