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Math Help - parametric equation

  1. #1
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    parametric equation

    could any one help me solving this parametric equation.

    1, if y=cos^3 t and x=sin^3t calculate d^2 y/dx^2
    2,the parametric equation of a function are given as x=3sin2t and y= 1-e^cos2t calculate dy/dx and d^2 y/dx^2
    3, partial differetiation- determine d^2 z/dxdy if z=3xy^2+5x^2 y^4-y^5
    4, if v=2loge^(4y-3x) calculate d^2 v/dxdy
    5,if u=e^-x(xcosy) determine the value of
    i,du/dx
    ii,d^2 u/dydx
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  2. #2
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    Quote Originally Posted by mokgadi View Post
    could any one help me solving this parametric equation.

    1, if y=cos^3 t and x=sin^3t calculate d^2 y/dx^2
    \frac{dy}{dt}=-3\sin t \cos^2 t=-3\sin t \cos t\cos t=-\frac{3}{2}\sin 2t \cos t
    \frac{dx}{dx}=3\cos t \sin^2 t= 3\cos t\sin t \sin t=\frac{3}{2}\sin t \sin 2t

    Thus,
    f(t)=\frac{dy}{dx}=\frac{-\frac{3}{2}\sin 2t \cos t}{\frac{3}{2}\sin t \sin 2t}=-\cot t

    Then,
    f'(t)=\csc^2 t
    Then,
    \frac{d^2 y}{dx^2}= \frac{3\csc^2 t}{2\sin 2t \sin t}
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  3. #3
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    Quote Originally Posted by mokgadi View Post
    2,the parametric equation of a function are given as x=3sin2t and y= 1-e^cos2t calculate dy/dx and d^2 y/dx^2
    x=3\sin 2t
    y=1-e^{\cos 2t}

    Thus,
    x'_t=6\cos 2t
    y'_t=2\sin 2t e^{\cos 2t}

    Thus,
    f(t)=\frac{dy}{dx}=\frac{2\sin 2t e^{\cos 2t}}{6\cos 2t}=\frac{1}{3}\tan 2t \cdot e^{\cos 2t}

    Thus, (product rule),
    f'(t)=\frac{2}{3}\sec^2 2t\cdot e^{\cos 2t} - \frac{2}{3}\tan 2t \sin 2t\cdot e^{\cos 2t}
    Divide by x'_t,
    \frac{d^2y}{dx^2}=\frac{1}{9}\sec^3 2t \cdot e^{\cos 2t}- \frac{1}{9} \tan^2 2t \cdot e^{\cos 2t}
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  4. #4
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    Quote Originally Posted by mokgadi View Post
    3, partial differetiation- determine d^2 z/dxdy if z=3xy^2+5x^2 y^4-y^5
    Since z\, is continous everywhere you can take the partials in any order.

    z_x=3y^2+10x y^4
    z_{xy}=6y+40xy^3

    4, if v=2loge^(4y-3x) calculate d^2 v/dxdy
    Same thing,
    v=2\ln (4y-3x)
    v_x=\frac{-6}{4y-3x}
    v_{xy}=\frac{24}{(4y-3x)^2}
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  5. #5
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    Hello, mokgadi!

    We have two formulas for parametric differentiation.
    . . The second one is commonly misunderstood.

    . . . (1)\;\frac{dy}{dx} \;=\;\frac{\frac{dy}{dt}}{\frac{dx}{dt}} \qquad\qquad (2)\;\frac{d^2y}{dx^2} \;=\;\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}  {\frac{dx}{dt}}

    For the second derivative:
    . . (a) Differentiate the result in part (1).
    . . (b) Divide by another \frac{dx}{dt}



    y \:=\:\cos^3t,\;x = \sin^3t

    We have:
    . . \frac{dy}{dt} \:=\:\text{-}3\cos^2t\sin t
    . . \frac{dx}{dt} \:=\:3\sin^2t\cos t

    Therefore: . \frac{dy}{dx} \:=\:\frac{\text{-}3\cos^2t\sin t}{3\sin^2t\cos t} \:=\:-\frac{\cos t}{\sin t} \:=\:-\cot t


    We have: . \frac{d}{dt}\left(\frac{dy}{dx}\right) \:=\:\csc^2x

    Therefore: . \frac{d^2y}{dx^2} \:=\:\frac{\csc^2x}{3\sin^2x\cos t} \:=\:\frac{1}{3}\csc^4t\sec t


    Last edited by Soroban; February 9th 2007 at 04:53 PM.
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