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Thread: Cos derivative problem

  1. #1
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    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; Dec 29th 2008 at 05:08 PM.
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  2. #2
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    $\displaystyle cos^{2}(x)+2cos(x)=0$

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

    $\displaystyle u^{2}+2u=0$

    $\displaystyle u=0, \;\ -2$

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

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

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

    Therefore, we have f(x)=0 at

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

    Where C is an integer.

    To find the second derivative of cos(x):

    $\displaystyle f(x)=cos(x), \;\ f'(x)=-sin(x), \;\ f''(x)=-cos(x), \;\ f'''(x)=sin(x)$
    $\displaystyle f''''(x)=cos(x)$. See?. It repeats.
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  3. #3
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    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
    $\displaystyle f(x) = \cos^2 x + \cos x$

    a) for f(x) = 0,

    $\displaystyle 0 = \cos^2 x + \cos x$

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

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

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

    now finish it

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

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

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

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

    $\displaystyle
    -\sin x=0\;\;OR\;\;2\cos x +1=0
    $

    $\displaystyle
    \sin x =0 \;\;OR\;\;\cos x =\frac{-1}{2}
    $

    Now finish it.

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

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

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

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

    $\displaystyle
    f"(x) = -2 -\cos x
    $

    Now finish it.

    Did you get it now???
    Last edited by Shyam; Dec 29th 2008 at 06:00 PM.
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  4. #4
    Member OnMyWayToBeAMathProffesor's Avatar
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    $\displaystyle f(x) = \cos^2 x + \cos x$
    I am sorry Shyam but it is $\displaystyle f(x) = \cos^2 x + 2\cos x$ not $\displaystyle f(x) = \cos^2 x + \cos x$. That changes the first and second derivative drastically doesn't it?
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  5. #5
    Member OnMyWayToBeAMathProffesor's Avatar
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    so the second derivative is $\displaystyle 2sinx^2-2cosx^2-2cos$?

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

    This problem is almost done! Thanks everyone!
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  6. #6
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    $\displaystyle f(x) = \cos^2{x} + 2\cos{x}$

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

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

    the last expression will also factor ...

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

    $\displaystyle f''(x) = 0$ at $\displaystyle x = \frac{\pi}{3} \, , \, x = \frac{5\pi}{3} \, , \, x = \pi$
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  7. #7
    Member OnMyWayToBeAMathProffesor's Avatar
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    Thanks skeeter, shyam, galactus and calc_help123!
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