Solved 1 Answer
See More Answers for FREE
Enhance your learning with StudyX
Receive support from our dedicated community users and experts
See up to 20 answers per week for free
Experience reliable customer service
Step1/2we have \( \mathrm{{r}{\left({t}\right)}=} \)we have to find tangential and normal components \( \mathrm{{a}_{{n}}{\quad\text{and}\quad}{a}_{{T}}} \)Explanation:Please refer to solution in this step.Step2/2\( \begin{align*} \mathrm{{r}'{\left({t}\right)}} &= \mathrm{} \\[3pt]\mathrm{{r}{''}{\left({t}\right)}} &= \mathrm{} \end{align*} \)\( \mathrm{{r}'{\left({t}\right)}.{r}{''}{\left({t}\right)}=.={4}{t}} \)\( \mathrm{{r}'{\left({t}\right)}{X}{r}{''}{\left({t}\right)}={\left[\begin{matrix}{i}&{j}&{k}\\-{1}&{1}&{2}{t}\\{0}&{0}&{2}\end{matrix}\right]}} \)\( \mathrm{=={2}{i}+{2}{j}+{0}{k}} \)\( \mathrm{{\left|{\left|{r}'{\left({t}\right)}{X}{r}{''}{\left({t}\right)}\right|}\right|}=\sqrt{{{2}^{{2}}+{2}^{{2}}+{0}^{{2}}}}=\sqrt{{8}}={2}\sqrt{{2}}} \)\( \mathrm{{r}'{\left({2}\right)}=} \)\( \mathrm{{\left|{r}'{\left({2}\right)}\right|}=\sqrt{{18}}={3}\sqrt{{2}}} \)using formula \( \mathrm{{a}_{{T}}=\frac{{{r}'{\left({1}\right)}.{r}{''}{\left({2}\right)}}}{{\left|{r}'{\left({1}\right)}\right|}}={4}\frac{{{2}}}{{{3}{\left(\sqrt{{2}}\right)}}}=\frac{{{2}\sqrt{{2}}}}{{3}}} \)\( \mathrm{{a}_{{N}}=\frac{{\left|{\left({r}'{\left({2}\right)}{X}{r}{''}{\left({2}\right)}\right)}\right|}}{{\left|{r}'{\left({2}\right)}\right|}}=\frac{{{2}\sqrt{{2}}}}{{{3}\sqrt{{2}}}}=\frac{{2}}{{3}}} \) this is required answer Explanation:Please refer to solution in this step. ...