What's the tangential component of acceleration for an object with position function r(t) = t^2 i + t j + t^2

r=<t^2,t, t^2>

r'=< 2t, 1, 2t>

r"=< 2, 0 , 2>

|r'|= sqrt(4t^2+1+4t^2)= sqrt( 8t^2+1)

r' dot r" = 4t+4t = 8t

acceleration for tangent component is

aT= r' dot r"/|r'| = 8t/( sqrt(8t^2+1))

so answer is A. if it is wrong you can chop my head down lol

dr/dt = v= 2t i + (1+2t)j

a d2r/dt2= dv/dt= 2i + 2 jthe tangent makes angle whose tangent 1+2t/2t then sin =1+2t/sq.r.(1+2t)^2+(2t)^2

cos= 2t/sq.r.(1+2t)^2+(2t)^2

tangential component =2cos +2sin =8t+2/sq.r.(1+2t)^2+(2t)^2

' ' '' = 2+8t/(8t^2+4t+1}^1/2

the result none of these answers in the given of the question

E