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Math Help - More Parametric Equations

  1. #1
    Del
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    More Parametric Equations

    Find d^2y/dx^2 , as a fuction of t, for the given the parametric equations:

    x = 3 - 8cos(t)

    y = 5 + cos^3(t)

    I don't understand this notation, please help!
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  2. #2
    GAMMA Mathematics
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    Quote Originally Posted by Del View Post
    Find d^2y/dx^2 , as a fuction of t, for the given the parametric equations:

    x = 3 - 8cos(t)

    y = 5 + cos^3(t)

    I don't understand this notation, please help!
    It is the 2nd derivative of y with respect to x.

    t = arccos\left(\frac{3-x}{8}\right)

    Plug that into y to get an equation with x as the parameter and no t's.

    y = 5 + \left(\frac{3-x}{8}\right)^3

    Now, take the 2nd derivative...
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Del View Post
    Find d^2y/dx^2 , as a fuction of t, for the given the parametric equations:

    x = 3 - 8cos(t)

    y = 5 + cos^3(t)

    I don't understand this notation, please help!
    This might help you more than us helping you directly.

    -Dan
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  4. #4
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    Hello, Del!

    Find \frac{d^2y}{dx^2}, as a fuction of t.

    . . \begin{array}{ccc}x &= &3 - 8\cos t \\<br />
y & = & 5 + \cos^3\!t \end{array}

    I don't understand this notation. . . . . Which notation?
    You don't understand \frac{d^2y}{dx^2}\:?\quad\hdots It's the second derivative of y.

    You don't understand \cos^3 t\:?\quad\hdots It's (\cos t)^3



    (a) Find the first derivative.

    Formula: . \frac{dy}{dx} \;=\;\frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}
    It says:
    . . (1) Differentiate y with respect to t.
    . . (2) Differentiate x with respect to t.
    . . (3) Form the fraction.


    (b) Find the second derivative.

    Formula: . \frac{d^2y}{dx^2} \;=\;\frac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right  )}{\dfrac{dx}{dt}}


    Read this formula very carefully . . .

    The numerator says: differentiate \frac{dy}{dx} . . . which we found in part (a).
    The bottom says: divide by \frac{dx}{dt} . . . which we also found in part (a).


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Back to the problem: . \begin{array}{ccc} x &=& 3-8\cos t \\ y &=& 5 + \cos^3\!t \end{array}

    (a) . \frac{dy}{dt} \:=\:\text{-}3\cos^2\!t\sin t\qquad\qquad\frac{dx}{dt} \:=\:8\sin t

    . . . . \frac{dy}{dx} \;=\;\frac{\text{-}3\cos^2\!t\sin t}{8\sin t} \quad\Rightarrow\quad\boxed{\frac{dy}{dx}\;=\;\tex  t{-}\frac{3}{8}\cos^2\!t}



    (b) . \frac{d}{dt}\left(\frac{dy}{dx}\right) \;=\;\frac{d}{dt}\left(\text{-}\frac{3}{8}\cos^2\!t\right) \;=\;\frac{3}{4}\cos t\sin t


    Therefore: . \frac{d^2y}{dx^2} \;=\;\frac{\frac{3}{4}\cos t\sin t}{8\sin t} \;=\;\boxed{\frac{3}{32}\cos t}

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