ap calculus
posted by bob on .
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

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