{:[r=3ui+3u^(2)j+2u^(3)k],[v=(du)/(dt)(3i+6uj+6u^(2)k)],[a=(d^(2)u)/(dt^(2))(3i+6uj+6u^(2)k)+((du)/(dt))^(2)(6j+12 uk)]:}Since u is increasing and the speed of the particle is 6 ,6=|v|=3(du)/(dt)sqrt(1+4u^(2)+4u^(4))=3(1+2u^(2))(du)/(dt). Thus (du)/(dt)=(2)/(1+2u^(2)), and(d^( ... See the full answer