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Math Help - velocity and acceleration with more than one variable!

  1. #1
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    velocity and acceleration with more than one variable!

    a particle moves along a horizontal line so that its coordinate at time t is x=(b2+c2t2)^(1/2), t is greater than or equal to 0, where b and c are positive constants
    a.) find the velocity and acceleration functions
    b.)show that the particle always moves in the positive direction
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  2. #2
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    Re: velocity and acceleration with more than one variable!

    your title is misleading ... x is defined in terms of a single variable, t

    x = (b^2+c^2t^2)^{\frac{1}{2}} , t \ge 0

    chain rule ...

    \frac{dx}{dt} = \frac{1}{2}(b^2+c^2t^2)^{-\frac{1}{2}} \cdot 2c^2 t

    clean up the algebra and use the quotient/chain rule to find \frac{d^2x}{dt^2}

    note that \frac{dx}{dt} > 0 for t > 0
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  3. #3
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    Re: velocity and acceleration with more than one variable!

    .
    Quote Originally Posted by skeeter View Post
    your title is misleading ... x is defined in terms of a single variable, t

    so i got the second derivative equaling
    c^2t(-1/2(b^2+c^2t)^-3/2*2c^2t)+(b^2+c^2t^2)^-1/2*c^2dx/dt
    is that right? and how do i show that the particle always moves in a positive direction?

    chain rule ...

    \frac{dx}{dt} = \frac{1}{2}(b^2+c^2t^2)^{-\frac{1}{2}} \cdot 2c^2 t

    clean up the algebra and use the quotient/chain rule to find \frac{d^2x}{dt^2}

    note that \frac{dx}{dt} > 0 for t > 0
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  4. #4
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    Re: velocity and acceleration with more than one variable!

    I get ...

    \frac{d^2x}{dt^2} = \frac{b^2c^2}{(b^2+c^2t^2)^{\frac{3}{2}}}

    if you clean up the algebra of the expression for \frac{dx}{dt} , it's fairly simple to see why the velocity > 0 for t > 0
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