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Math Help - Implicit differentiation & parametricequations

  1. #1
    Ley
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    Implicit differentiation & parametricequations

    Assume that the equations in each part define x and y implicitly as differentiable functions x=f(t),y=g(t), find the slope of the curve x=f(t),y=g(t) at the given value of t.
    a) x^2-2tx+2t^2=4, 2y^3-3t^2=4, t=2

    I got dy/dx=2+3t/4y.2x but honestly i am confused and I Reallly need help, Please help me thanx in advance..
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    Senior Member MaxJasper's Avatar
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    Lightbulb Re: Implicit differentiation & parametricequations

    Quote Originally Posted by Ley View Post
    Assume that the equations in each part define x and y implicitly as differentiable functions x=f(t),y=g(t), find the slope of the curve x=f(t),y=g(t) at the given value of t.
    a) x^2-2tx+2t^2=4, 2y^3-3t^2=4, t=2
    Attached Thumbnails Attached Thumbnails Implicit differentiation & parametricequations-parametric-plot.png  
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    Re: Implicit differentiation & parametricequations

    Hello, Ley!

    Assume that the equations in each part define x and y implicitly as differentiable functions, x=f(t),\:y=g(t).
    Find the slope of the curve x=f(t),\:y=g(t) at the given value of t.

    a)\:\begin{Bmatrix}x^2-2tx+2t^2&=&4 & [1] \\2y^3-3t^2&=&4 & [2] \end{Bmatrix}\quad t=2

    Differentiate implicitly.

    [1]\;\;2x\frac{dx}{dt} - 2t\frac{dx}{dt} - 2x + 4t \:=\:0 \quad\Rightarrow\quad 2x\frac{dx}{dt} - 2t\frac{dx}{dt} \:=\:2x  - 4t

    . . . 2(x-t)\frac{dx}{dt} \;=\;2(x-2t) \quad\Rightarrow\quad \frac{dx}{dt} \:=\:\frac{x-2t}{x-t}

    [2]\;\;6y^2\frac{dy}{dt} = 6t \:=\:0 \quad\Rightarrow\quad \frac{dy}{dt} \:=\:\frac{t}{y^2}

    \frac{dy}{dx} \;=\;\frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} \;=\;\frac{\dfrac{t}{y^2}}{\dfrac{x-2t}{x-t}} \;=\;\frac{t(x-t)}{y^2(x-2t)} .[3]


    Substitute t = 2 into [1]:
    . . x^2 - 4x + 8 \:=\:4 \quad\Rightarrow\quad x^2 - 4x + 4 \:=\:0 \quad\Rightarrow\quad (x-2)^2 \:=\:0 \quad\Rightarrow\quad x \:=\:2

    Substitute t = 2 into [2]:
    . . 2y^3 - 12 \:=\:4 \quad\Rightarrow\quad 2y^3 \:=\:16 \quad\Rightarrow\quad y^3 \:=\:8 \quad\Rightarrow\quad y \:=\:2

    We have: . \begin{Bmatrix}t = 2 \\ x = 2\\ y = 2\end{Bmatrix}

    Substitute into [3]: . \frac{dy}{dx} \;=\;\frac{2(2-2)}{2^2(2-4)} \;=\;0


    At the point (2,2), the slope is 0.
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