Question

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

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

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 $$a_T=6t,\ a_N=6$$