Find the tangential and normal components of the acceleration vector r(t)=(3t-t^3)i+3t^2j

FobelloE 2020-11-02 Answered
Find the tangential and normal components of the acceleration vector r(t)=(3tt3)i+3t2j
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Expert Answer

SabadisO
Answered 2020-11-03 Author has 108 answers
Given that
r(t)=(3tt3)i+3t2j
Differentiate
r(t)=(3t3t2)i+3t2j
Again Differentiate
r(t)=6ti+6j
Note that
|r(t)|=(33t2)2+6t2
=(9+9t418t2)+36t2
=9+9t4+18t2
=3t4+2t2+1
=3(t2+1)2
=3t2+1
Equation 9 gives the tangential component as
aT=r(t)r(t)|r(t)|
aT=[(33t2)i+6tj][6ti+6j]3(1+t2)
aT=(18t+18t3)+36t3(1+t2)=18t+18t33(1+t2)=18t(1+t2)3(1+t2)=6t
Equation 9 gives the normal component
aN=|r(t)r(t)||r(t)|
r(t)r(t)=[ijk(33t2)6t06t60]
Expand the determinant along the third column, To get
r(t)r(t)=k(6(33t2)(36t2))=(18+18t2)k
Threrefore, |r(t)r(t)|=18(1+t2)
And hence aN=18(1+t2)3(1+t2)=6
Result aT=6t, aN=6

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