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Thread: point of inflextion,

  1. #1
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    point of inflextion,

    A) show that $\displaystyle y = 6sin2x + 4cos2x $ satisfies the equation
    $\displaystyle \frac{d^{2}y}{dx^{2}} +4y = 0 $

    b)the expression $\displaystyle y = 6sin2x + 4cos2x $ can be written as $\displaystyle R sin(2x + a) $ where R and α are positive constants,
    0 < α < pi/2. Find the values of R and α, correct to 3 decimal places.

    (c) What is the smallest positive value of x where y has a point of inflection?
    I have done part a and b, but need help with part c.

    $\displaystyle \sqrt{52}sin(2x + 0.588) $

    $\displaystyle \frac{d^{2}y}{dx^{2}} = -24sin2x - 16cos2x $

    Thank you.
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  2. #2
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    The seond derivative is zero if $\displaystyle \tan(2x)=\frac{-2}{3}$. WHY?
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  3. #3
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    Quote Originally Posted by Plato View Post
    The seond derivative is zero if $\displaystyle \tan(2x)=\frac{-2}{3}$. WHY?
    How did you get tan2x?

    I got $\displaystyle tan \alpha = \frac{2}{3} $

    Thanks.
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  4. #4
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    Quote Originally Posted by Tweety View Post
    How did you get tan2x?
    $\displaystyle - 24\sin(2x) - 16\cos(2x)=0 $
    $\displaystyle - 24\sin(2x) = 16\cos(2x) $
    $\displaystyle \tan(2x)=\frac{\sin(2x)}{\cos(2x)}= \frac{16}{-24} $
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  5. #5
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    Quote Originally Posted by Plato View Post
    $\displaystyle - 24\sin(2x) - 16\cos(2x)=0 $
    $\displaystyle - 24\sin(2x) = 16\cos(2x) $
    $\displaystyle \tan(2x)=\frac{\sin(2x)}{\cos(2x)}= \frac{16}{-24} $
    Okay thanks, I am not quite sure why this result would make the second derivative zero?

    I know how to find the coordinates of a stationary point, but don't get what the question is asking, is it asking me to find all the points of inflection on the curve
    $\displaystyle y = 6sin2x + 4cos2x$ and than find the lowest value of x corresponding to a point of inflection?

    thank you for clarifying.
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  6. #6
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    Quote Originally Posted by Tweety View Post
    I am not quite sure why this result would make the second derivative zero?
    That is how we find points of inflections.
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  7. #7
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    Quote Originally Posted by Plato View Post
    That is how we find points of inflections.
    Yes I know this, but how does tan2x = -2/3 make the second derivative equal zero? And how would I work out the lowest value of x from this?

    thanks,
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  8. #8
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    Quote Originally Posted by Tweety View Post
    how would I work out the lowest value of x from this?thanks,
    $\displaystyle x=\frac{1}{2}\left(\arctan\left(\frac{-2}{3}\right)+\pi\right)$
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