# Unit tangent vector for $\mathbf{\vec{r}\left(t\right)} = \left\langle t^{4} - 1, \cos{\left(t \right)}, 3 t\right\rangle$ at $t = 0$

The calculator will find the unit tangent vector to $\mathbf{\vec{r}\left(t\right)} = \left\langle t^{4} - 1, \cos{\left(t \right)}, 3 t\right\rangle$ at $t = 0$, with steps shown.

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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 $t = 0$.

$\mathbf{\vec{T}\left(0\right)} = \left\langle 0, 0, 1\right\rangle$

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