find the tangential and normal components of the acceleration vector.\n\n mathbf{r}(t)=t mathbf{i}+2 e^{t}…

Explanation:

Step1: Find ( \mathbf{r}'(t) ) and ( \mathbf{r}''(t) )

Given ( \mathbf{r}(t)=t\mathbf{i} + 2e^{t}\mathbf{j}+e^{2t}\mathbf{k} ) ( \mathbf{r}'(t)=\mathbf{i}+2e^{t}\mathbf{j} + 2e^{2t}\mathbf{k} ) ( \mathbf{r}''(t)=0\mathbf{i}+2e^{t}\mathbf{j}+4e^{2t}\mathbf{k} )

Step2: Calculate ( \mathbf{r}'(t)\cdot\mathbf{r}''(t) ) and ( |\mathbf{r}'(t)| )

( \mathbf{r}'(t)\cdot\mathbf{r}''(t)=(1)(0)+(2e^{t})(2e^{t})+(2e^{2t})(4e^{2t})=4e^{2t}+8e^{4t} ) ( |\mathbf{r}'(t)|=\sqrt{1^{2}+(2e^{t})^{2}+(2e^{2t})^{2}}=\sqrt{1 + 4e^{2t}+4e^{4t}}=\sqrt{(2e^{2t}+1)^{2}}=2e^{2t}+1 )

Step3: Compute the tangential component ( a_{T}=\frac{\mathbf{r}'(t)\cdot\mathbf{r}''(t)}{|\mathbf{r}'(t)|} )

( a_{T}=\frac{4e^{2t}+8e^{4t}}{2e^{2t}+1}=\frac{4e^{2t}(1 + 2e^{2t})}{2e^{2t}+1}=4e^{2t} )

Step4: Calculate ( \mathbf{r}'(t)\times\mathbf{r}''(t) )

[ \begin{align*} \mathbf{r}'(t)\times\mathbf{r}''(t)&=\begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\ 1&2e^{t}&2e^{2t}\ 0&2e^{t}&4e^{2t} \end{vmatrix}\ &=\mathbf{i}(8e^{3t}-4e^{3t})-\mathbf{j}(4e^{2t}-0)+\mathbf{k}(2e^{t}-0)\ &=4e^{3t}\mathbf{i}-4e^{2t}\mathbf{j}+2e^{t}\mathbf{k} \end{align*} ] ( |\mathbf{r}'(t)\times\mathbf{r}''(t)|=\sqrt{(4e^{3t})^{2}+(-4e^{2t})^{2}+(2e^{t})^{2}}=\sqrt{16e^{6t}+16e^{4t}+4e^{2t}}=2e^{t}\sqrt{4e^{4t}+4e^{2t}+1}=2e^{t}(2e^{2t}+1) )

Step5: Compute the normal component ( a_{N}=\frac{|\mathbf{r}'(t)\times\mathbf{r}''(t)|}{|\mathbf{r}'(t)|} )

( a_{N}=\frac{2e^{t}(2e^{2t}+1)}{2e^{2t}+1}=2e^{t} )

Answer:

( a_{T}=4e^{2t} ) ( a_{N}=2e^{t} )