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ap calculus

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a particle moves along the x axis in such a way that its acceleration at time t, t>0 , is given by x(t)= (ln x)^2. at what value of t does the velocity of the particle attain its maximum

  • ap calculus - ,

    velocity has max when acceleration is zero
    I assume x(t) = (ln t)^2 which is zero at t=1.

  • ap calculus - ,

    Assuming the function is
    x(t)=(ln(t))^2
    then t has a vertical asymptote at x=0.
    which explains why the domain of x(t) is set at x>0.

    To find the maximum, we need to find
    x'(t)=0
    where
    x'(t)=2ln(t)/t
    and x'(t)=0 at t=1.
    Check if maximum or minimum by finding x"(t)=2/t^2 - 2ln(t)/t
    so
    x"(1)=2 >0
    Hence t=1 is a minimum.

    Therefore, the function has no maximum, as it increases indefinitely as t inreases for t>1.

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