Question

find the tangential and normal components of the acceleration vector.
mathbf{r}(t)=7 e^{t} mathbf{i}+7 sqrt{2} t mathbf{j}+7 e^{-t} mathbf{k}
a_{t}=
a_{n}=

Explanation:

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

Differentiate \( \mathbf{r}(t)=7e^{t}\mathbf{i}+7\sqrt{2}t\mathbf{j}+7e^{-t}\mathbf{k} \)
\( \mathbf{r}'(t)=7e^{t}\mathbf{i}+7\sqrt{2}\mathbf{j}-7e^{-t}\mathbf{k} \)
\( \mathbf{r}''(t)=7e^{t}\mathbf{i}+0\mathbf{j}+7e^{-t}\mathbf{k} \)

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

\( \mathbf{r}'(t)\cdot\mathbf{r}''(t)=(7e^{t})(7e^{t})+(7\sqrt{2})(0)+(-7e^{-t})(7e^{-t})=49e^{2t}-49e^{-2t} \)
\( \|\mathbf{r}'(t)\|=\sqrt{(7e^{t})^{2}+(7\sqrt{2})^{2}+(-7e^{-t})^{2}}=\sqrt{49e^{2t} + 98+49e^{-2t}}=\sqrt{49(e^{t}+e^{-t})^{2}} = 7(e^{t}+e^{-t}) \)

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

\( a_{T}=\frac{49e^{2t}-49e^{-2t}}{7(e^{t}+e^{-t})}=\frac{49(e^{t}-e^{-t})(e^{t}+e^{-t})}{7(e^{t}+e^{-t})}=7(e^{t}-e^{-t}) \)

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

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

$$\begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 7e^{t}&7\sqrt{2}& - 7e^{-t}\\ 7e^{t}&0&7e^{-t} \end{vmatrix}$$

=49\sqrt{2}e^{-t}\mathbf{i}-98\mathbf{j}-49\sqrt{2}e^{t}\mathbf{k} \)
\( \|\mathbf{r}'(t)\times\mathbf{r}''(t)\|=\sqrt{(49\sqrt{2}e^{-t})^{2}+(-98)^{2}+(-49\sqrt{2}e^{t})^{2}}=\sqrt{4802e^{-2t}+9604 + 4802e^{2t}}=\sqrt{4802(e^{t}+e^{-t})^{2}}=49\sqrt{2}(e^{t}+e^{-t}) \)

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

\( a_{N}=\frac{49\sqrt{2}(e^{t}+e^{-t})}{7(e^{t}+e^{-t})}=7\sqrt{2} \)

Answer:

\( a_{T}=7(e^{t}-e^{-t}) \)
\( a_{N}=7\sqrt{2} \)