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

asked 2020-11-02
Find the tangential and normal components of the acceleration vector \(r(t)=(3t-t^3)i+3t^2j\)

Answers (1)

Given that
Again Differentiate
Note that
Equation 9 gives the tangential component as
\(a_T=\frac{r'(t)\cdot r''(t)}{|r'(t)|}\)
Equation 9 gives the normal component
\(r'(t)\cdot r''(t)=\begin{bmatrix}i&j&k\\(3-3t^2)&6t&0\\-6t&6&0\end{bmatrix}\)
Expand the determinant along the third column, To get
\(r'(t)\cdot r''(t)=k(6(3-3t^2)-(-36t^2))=(18+18t^2)k\)
Threrefore, \(|r'(t)\cdot r''(t)|=18(1+t^2)\)
And hence \(a_N=\frac{18(1+t^2)}{3(1+t^2)}=6\)
Result \(a_T=6t,\ a_N=6\)

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