Posted by **bob** on Friday, April 5, 2013 at 10:54pm.

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 -
**Anonymous**, Saturday, April 6, 2013 at 5:09am
velocity has max when acceleration is zero

I assume x(t) = (ln t)^2 which is zero at t=1.

- ap calculus -
**MathMate**, Saturday, April 6, 2013 at 9:05am
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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