Question

The magnitude of the acceleration of a particle which moves along the curve x=2 sin ¡3t,y=2 cos ¡3t and z=8t at any time t>0 is

Answer 1

$s=2sin(3t)i+2cos(3t)j+8tk.\\ v=\frac{ds}{dt}=6cos(3t)i-6sin(3t)j+8k.\\ a=\frac{dv}{dt}=-18sin(3t)i-18cos(3t)j+0k.\\ |a|=\sqrt{324sin^2(3t)+324cos^2(3t)}=\sqrt{324}=18.$