Originally Posted by
earboth Hello,
1. Derivate x(t) with respect to t:
$\displaystyle {dx\over dt}={3\over 8}\cos\left({\pi\over 8}\cdot t \right)$
You'll get a maximum if dx/dt = 0 and the 2nd derivative is negative:
$\displaystyle {d^2x\over dt^2}={3\over 64}\left(-\sin\left({\pi\over 8}\cdot t \right) \right)$
dx/dt = 0 if $\displaystyle \left({\pi\over 8}\cdot t \right)={{2k+1}\over 2}\cdot \pi$
That means t = 4*(2k+1)
You'll get a maximum if k = 1, 3, 5, ... (Use the 2nd derivative to sort these numbers out!)
So the first 2 maxima ar at t = 4 or t = 20. (See attachment)
Greetings
EB