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

Find the tangential and normal components of the acceleration vector $r\left(t\right)=\left(3t-{t}^{3}\right)i+3{t}^{2}j$
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Given that
$r\left(t\right)=\left(3t-{t}^{3}\right)i+3{t}^{2}j$
Differentiate
${r}^{\prime }\left(t\right)=\left(3t-3{t}^{2}\right)i+3{t}^{2}j$
Again Differentiate
${r}^{″}\left(t\right)=-6ti+6j$
Note that
$|{r}^{\prime }\left(t\right)|=\sqrt{\left(3-3{t}^{2}{\right)}^{2}+6{t}^{2}}$
$=\sqrt{\left(9+9{t}^{4}-18{t}^{2}\right)+36{t}^{2}}$
$=\sqrt{9+9{t}^{4}+18{t}^{2}}$
$=3\sqrt{{t}^{4}+2{t}^{2}+1}$
$=3\sqrt{\left({t}^{2}+1{\right)}^{2}}$
$=3{t}^{2}+1$
Equation 9 gives the tangential component as
${a}_{T}=\frac{{r}^{\prime }\left(t\right)\cdot {r}^{″}\left(t\right)}{|{r}^{\prime }\left(t\right)|}$
${a}_{T}=\frac{\left[\left(3-3{t}^{2}\right)i+6tj\right]\cdot \left[-6ti+6j\right]}{3\left(1+{t}^{2}\right)}$
${a}_{T}=\frac{\left(-18t+18{t}^{3}\right)+36t}{3\left(1+{t}^{2}\right)}=\frac{18t+18{t}^{3}}{3\left(1+{t}^{2}\right)}=\frac{18t\left(1+{t}^{2}\right)}{3\left(1+{t}^{2}\right)}=6t$
Equation 9 gives the normal component
${a}_{N}=\frac{|{r}^{\prime }\left(t\right)\ast {r}^{″}\left(t\right)|}{|{r}^{\prime }\left(t\right)|}$
${r}^{\prime }\left(t\right)\cdot {r}^{″}\left(t\right)=\left[\begin{array}{ccc}i& j& k\\ \left(3-3{t}^{2}\right)& 6t& 0\\ -6t& 6& 0\end{array}\right]$
Expand the determinant along the third column, To get
${r}^{\prime }\left(t\right)\cdot {r}^{″}\left(t\right)=k\left(6\left(3-3{t}^{2}\right)-\left(-36{t}^{2}\right)\right)=\left(18+18{t}^{2}\right)k$
Threrefore, $|{r}^{\prime }\left(t\right)\cdot {r}^{″}\left(t\right)|=18\left(1+{t}^{2}\right)$
And hence ${a}_{N}=\frac{18\left(1+{t}^{2}\right)}{3\left(1+{t}^{2}\right)}=6$
Result

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