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    Opalg

    Re: Proving limit by definition

    Not quite, although you started correctly. The limit in this case is as $x\to\infty$, so you want to see what happens when $x$ gets large. This means

    Opalg Today, 06:35 Go to last post
    goody

    Re: Proving limit by definition

    Hi Opalg! Do you think I got it correct?

    goody Today, 04:38 Go to last post
    skeeter

    Re: 299What is the acceleration

    why did you do that? ...

    ... if originally, $x = \sin{t}-\cos{t}$, then $v = \cos{t} + \sin{t} = 0 \implies t = \dfrac{3\pi}{4}$

    skeeter Yesterday, 19:28 Go to last post
    topsquark

    Re: 299What is the acceleration

    Wait! What do you mean by you changed it to v(t)? The particle's position is x(t) = sin(t) - cos(t) means that v = $ \displaystyle \dfrac{dx}{dt}$. You can't just change

    topsquark Yesterday, 17:53 Go to last post
    karush

    299What is the acceleration

    $\tiny{299}$
    For $t \ge 0$ the position of a particle moving along the x-axis is given by $v(t)=\sin tó\cos t$ What is the acceleration of the

    karush Yesterday, 17:38 Go to last post
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