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 \(a_T=6t,\ a_N=6\)

\(r(t)=(3t-t^3)i+3t^2j\)

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 \(a_T=6t,\ a_N=6\)