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Math Help - Parametric Equations!!

  1. #1
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    Parametric Equations!!

    At time t, the position of a particle moving on a curve is given by



    (a) Find the exact value of t at which the curve has a horizontal tangent.
    t =

    (b) Find dy/dx in terms of t.
    dy/dx = (6exp(2t)-2exp(-2t))/(2exp(2t)+2exp(-2t))

    (c) Find .

    ok so i was able to find dy/dx, but i need help with the rest...thanks...
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    THe answer

    Quote Originally Posted by mathlete View Post
    At time t, the position of a particle moving on a curve is given by



    (a) Find the exact value of t at which the curve has a horizontal tangent.
    t =

    (b) Find dy/dx in terms of t.
    dy/dx = (6exp(2t)-2exp(-2t))/(2exp(2t)+2exp(-2t))

    (c) Find .

    ok so i was able to find dy/dx, but i need help with the rest...thanks...
    for c is three you have \lim_{t \to 0}\frac{6e^{2t}-2e^{-2t}}{2e^{2t}+2e^{-2t}}...multiply by \frac{\frac{1}{e^{2t}}}{\frac{1}{e^{2t}}} and I think the answer will be apparent
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  3. #3
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    Quote Originally Posted by mathlete View Post
    At time t, the position of a particle moving on a curve is given by



    (a) Find the exact value of t at which the curve has a horizontal tangent.
    t =

    (b) Find dy/dx in terms of t.
    dy/dx = (6exp(2t)-2exp(-2t))/(2exp(2t)+2exp(-2t))

    (c) Find .

    ok so i was able to find dy/dx, but i need help with the rest...thanks...
    to find horizontal tangents we need  \frac{dy}{dt}=0

    so y is

    y=3e^{2t}+e^{-2t} \iff \frac{dy}{dt}=6e^{2t}-2e^{-2t}

    setting equal to zero we get.

     0=6e^{2t}-2e^{-2t} Now multiply by e^{2t}

    and we get

    0=6e^{4t}-2 \iff \frac{1}{3}=e^{4t} \iff \underbrace{\ln\left( \frac{1}{3}\right)}_{=-\ln(3)}=4t

    -\ln(3)=4t \iff t =-\frac{1}{4}\ln(3)
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