right off I have a problem :(
r'(t)=4ti+(2t^2 -2)j
r''(t)=4i+4j
I get
r''(t)=4i + 4__ t __ j
Find the tangential and normal components of the acceleration vector. (r)= 2t^2i + (2/3t^3 -2t)j
These are what I did.There are just too many steps so I just show the important one
r'(t)=4ti+(2t^2 -2)j
r''(t)=4i+4j
llr'(t)ll=√(4t)^2 +(2t^2 -2)^2 =2(t^2 +1)
T(t)=[4ti+(2t^2 -2)j ∙4i+4j]/2(t^2 +1)=(8t^3 -8t)/2(t^2 +1) right or wrong?
N(t)=[[4ti+(2t^2 -2)j ×4i+4j]/2(t^2 +1)=(8t^2 -8k)/2(t^2 +1) right or wrong?
4 answers
It was a typo. For r''(t), I got 4i+4tj. How about the answers? Are they correct or not?
agree for length of velocity vector
2(t^2 +1)
2(t^2 +1)
so Velocity unit vector = 4t/(2t^2+2) i + (2t^2 -2)/(2t^2+2) j
= 2 t /(t^2+1) i + (t^2-1)/(t^2+1) j
magnitude of acceleration tangent component = dot product of A and unit V
(4 i+4t j) dot 2t/(t^2+1) i + (t^2 -1)/(t^2+1) j
= [ 8 t + 4t^3 -4t ]/(t^2+1)
= 4 [ t +t^3] /(t^2+1) agree
no way the normal, enough :)
that is in direction of unit vector of V
= 2 t /(t^2+1) i + (t^2-1)/(t^2+1) j
magnitude of acceleration tangent component = dot product of A and unit V
(4 i+4t j) dot 2t/(t^2+1) i + (t^2 -1)/(t^2+1) j
= [ 8 t + 4t^3 -4t ]/(t^2+1)
= 4 [ t +t^3] /(t^2+1) agree
no way the normal, enough :)
that is in direction of unit vector of V
1 answer