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

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

Answers (1)

2020-11-03
Given that
\(r(t)=(3t-t^3)i+3t^2j\)
Differentiate
\(r'(t)=(3t-3t^2)i+3t^2j\)
Again Differentiate
\(r''(t)=-6ti+6j\)
Note that
\(|r'(t)|=\sqrt{(3-3t^2)^2+6t^2}\)
\(=\sqrt{(9+9t^4-18t^2)+36t^2}\)
\(=\sqrt{9+9t^4+18t^2}\)
\(=3\sqrt{t^4+2t^2+1}\)
\(=3\sqrt{(t^2+1)^2}\)
\(=3t^2+1\)
Equation 9 gives the tangential component as
\(a_T=\frac{r'(t)\cdot r''(t)}{|r'(t)|}\)
\(a_T=\frac{[(3-3t^2)i+6tj]\cdot[-6ti+6j]}{3(1+t^2)}\)
\(a_T=\frac{(-18t+18t^3)+36t}{3(1+t^2)}=\frac{18t+18t^3}{3(1+t^2)}=\frac{18t(1+t^2)}{3(1+t^2)}=6t\)
Equation 9 gives the normal component
\(a_N=\frac{|r'(t)*r''(t)|}{|r'(t)|}\)
\(r'(t)\cdot r''(t)=\begin{bmatrix}i&j&k\\(3-3t^2)&6t&0\\-6t&6&0\end{bmatrix}\)
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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