Question

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find the unit tangent vector t(t) at the given point on the curve.
r(t) = sin(t)i + 8tj + cos(t)k, (0, 0, 1)

Explanation:

Step1: Find the derivative of $\mathbf{r}(t)$

$\mathbf{r}'(t)=\cos(t)\mathbf{i}+8\mathbf{j}-\sin(t)\mathbf{k}$

Step2: Find the value of $t$ corresponding to the point $(0,0,1)$

Set $\mathbf{r}(t)=\sin(t)\mathbf{i}+8t\mathbf{j}+\cos(t)\mathbf{k}=(0,0,1)$. We have $\sin(t) = 0$, $8t=0$, $\cos(t)=1$. Solving these equations gives $t = 0$.

Step3: Evaluate $\mathbf{r}'(t)$ at $t = 0$

$\mathbf{r}'(0)=\cos(0)\mathbf{i}+8\mathbf{j}-\sin(0)\mathbf{k}=\mathbf{i}+8\mathbf{j}$

Step4: Calculate the magnitude of $\mathbf{r}'(0)$

$|\mathbf{r}'(0)|=\sqrt{1^{2}+8^{2}}=\sqrt{1 + 64}=\sqrt{65}$

Step5: Find the unit - tangent vector $\mathbf{T}(t)$ at $t = 0$

$\mathbf{T}(0)=\frac{\mathbf{r}'(0)}{|\mathbf{r}'(0)|}=\frac{1}{\sqrt{65}}\mathbf{i}+\frac{8}{\sqrt{65}}\mathbf{j}$

Answer:

$\frac{1}{\sqrt{65}}\mathbf{i}+\frac{8}{\sqrt{65}}\mathbf{j}$