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Math Help - Help on this question?

  1. #1
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    Help on this question?

    Hello, I was wondering if anyone could help me with this? If there seems to be some information lacking, please ask me. Maybe to solve (d) I need some of the things from (a), (b), or (c). I have those done already, but I don't think they are that relevant to part (d). Thanks.


    An object moving along a curve in the xy-plane has position (x(t),y(t)) at time t ≥ 0 with dx/dt = 3 + cos(t^2). The derivative dy/dt is not explicitly given. At time t=2, the object is at position (1,8).


    d) for t ≥ 3, the line tangent to the curve at (x(t),y(t)) has a slope of 2t+1. Find the acceleration vector of the object at time t=4.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by BoombaWoomba View Post
    Hello, I was wondering if anyone could help me with this? If there seems to be some information lacking, please ask me. Maybe to solve (d) I need some of the things from (a), (b), or (c). I have those done already, but I don't think they are that relevant to part (d). Thanks.


    An object moving along a curve in the xy-plane has position (x(t),y(t)) at time t ≥ 0 with dx/dt = 3 + cos(t^2). The derivative dy/dt is not explicitly given. At time t=2, the object is at position (1,8).


    d) for t ≥ 3, the line tangent to the curve at (x(t),y(t)) has a slope of 2t+1. Find the acceleration vector of the object at time t=4.
    You will have:

    \frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}

    (the slope of the curve is dy/dx)

    So now you have enough information to find (\dot{x},\dot{y}) (the velocity) as a
    function of t, now just differentiate again to get the acceleration (\ddot{x},\ddot{y})

    RonL
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