Question

find the tangential and normal components of the acceleration vector.

mathbf{r}(t)=t mathbf{i}+2 e^{t} mathbf{j}+e^{2 t} mathbf{k}

a_{t}=

a_{n}=

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 \( \mathbf{r}'(t)\times\mathbf{r}''(t) \)

$$
LATEXBLOCK0
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\( \|\mathbf{r}'(t)\times\mathbf{r}''(t)\|=\sqrt{(4e^{3t})^{2}+(-4e^{2t})^{2}+(2e^{t})^{2}}=\sqrt{16e^{6t}+16e^{4t}+4e^{2t}}=2e^{t}\sqrt{4e^{4t}+4e^{2t}+1}=2e^{t}(2e^{2t}+1) \)

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:

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