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Unit tangent vector for $$$\mathbf{\vec{r}\left(t\right)} = \left\langle \sin^{3}{\left(t \right)}, \cos^{3}{\left(t \right)}, \sin^{2}{\left(t \right)}\right\rangle$$$

The calculator will find the unit tangent vector to $$$\mathbf{\vec{r}\left(t\right)} = \left\langle \sin^{3}{\left(t \right)}, \cos^{3}{\left(t \right)}, \sin^{2}{\left(t \right)}\right\rangle$$$, with steps shown.

Related calculators: Unit Normal Vector Calculator, Unit Binormal Vector Calculator

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Find the unit tangent vector for $$$\mathbf{\vec{r}\left(t\right)} = \left\langle \sin^{3}{\left(t \right)}, \cos^{3}{\left(t \right)}, \sin^{2}{\left(t \right)}\right\rangle$$$.

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).

$$$\mathbf{\vec{r}^{\prime}\left(t\right)} = \left\langle 3 \sin^{2}{\left(t \right)} \cos{\left(t \right)}, - 3 \sin{\left(t \right)} \cos^{2}{\left(t \right)}, \sin{\left(2 t \right)}\right\rangle$$$ (for steps, see derivative calculator).

Find the unit vector: $$$\mathbf{\vec{T}\left(t\right)} = \left\langle \frac{6 \sqrt{26} \sin^{2}{\left(t \right)} \cos{\left(t \right)}}{13 \sqrt{1 - \cos{\left(4 t \right)}}}, - \frac{6 \sqrt{26} \sin{\left(t \right)} \cos^{2}{\left(t \right)}}{13 \sqrt{1 - \cos{\left(4 t \right)}}}, \frac{2 \sqrt{26} \sin{\left(2 t \right)}}{13 \sqrt{1 - \cos{\left(4 t \right)}}}\right\rangle$$$ (for steps, see unit vector calculator).

Answer

The unit tangent vector is $$$\mathbf{\vec{T}\left(t\right)} = \left\langle \frac{6 \sqrt{26} \sin^{2}{\left(t \right)} \cos{\left(t \right)}}{13 \sqrt{1 - \cos{\left(4 t \right)}}}, - \frac{6 \sqrt{26} \sin{\left(t \right)} \cos^{2}{\left(t \right)}}{13 \sqrt{1 - \cos{\left(4 t \right)}}}, \frac{2 \sqrt{26} \sin{\left(2 t \right)}}{13 \sqrt{1 - \cos{\left(4 t \right)}}}\right\rangle.$$$A