Unit tangent vector for $$$\mathbf{\vec{r}\left(t\right)} = \left\langle t^{4} - 1, \cos{\left(t \right)}, 3 t\right\rangle$$$ at $$$t = 0$$$
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Find the unit tangent vector for $$$\mathbf{\vec{r}\left(t\right)} = \left\langle t^{4} - 1, \cos{\left(t \right)}, 3 t\right\rangle$$$ at $$$t = 0$$$.
Solution
To find the unit tangent vector, we need to find the derivative of $$$\mathbf{\vec{r}\left(t\right)}$$$ (the tangent vector) and then normalize it (find the unit vector).
$$$\mathbf{\vec{r}^{\prime}\left(t\right)} = \left\langle 4 t^{3}, - \sin{\left(t \right)}, 3\right\rangle$$$ (for steps, see derivative calculator).
Find the unit vector: $$$\mathbf{\vec{T}\left(t\right)} = \left\langle \frac{4 t^{3}}{\sqrt{16 t^{6} + \sin^{2}{\left(t \right)} + 9}}, - \frac{\sin{\left(t \right)}}{\sqrt{16 t^{6} + \sin^{2}{\left(t \right)} + 9}}, \frac{3}{\sqrt{16 t^{6} + \sin^{2}{\left(t \right)} + 9}}\right\rangle$$$ (for steps, see unit vector calculator).
Now, find the vector at $$$t = 0$$$.
$$$\mathbf{\vec{T}\left(0\right)} = \left\langle 0, 0, 1\right\rangle$$$
Answer
The unit tangent vector is $$$\mathbf{\vec{T}\left(t\right)} = \left\langle \frac{4 t^{3}}{\sqrt{16 t^{6} + \sin^{2}{\left(t \right)} + 9}}, - \frac{\sin{\left(t \right)}}{\sqrt{16 t^{6} + \sin^{2}{\left(t \right)} + 9}}, \frac{3}{\sqrt{16 t^{6} + \sin^{2}{\left(t \right)} + 9}}\right\rangle.$$$A
$$$\mathbf{\vec{T}\left(0\right)} = \left\langle 0, 0, 1\right\rangle$$$A