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Unit tangent vector for r⃗(t)=⟨4t2,6t,11⟩\mathbf{\vec{r}\left(t\right)} = \left\langle 4 t^{2}, \frac{6}{t}, 11\right\rangle

The calculator will find the unit tangent vector to r⃗(t)=⟨4t2,6t,11⟩\mathbf{\vec{r}\left(t\right)} = \left\langle 4 t^{2}, \frac{6}{t}, 11\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 r⃗(t)=⟨4t2,6t,11⟩\mathbf{\vec{r}\left(t\right)} = \left\langle 4 t^{2}, \frac{6}{t}, 11\right\rangle.

Solution

To find the unit tangent vector, we need to find the derivative of r⃗(t)\mathbf{\vec{r}\left(t\right)} (the tangent vector) and then normalize it (find the unit vector).

r⃗′(t)=⟨8t,−6t2,0⟩\mathbf{\vec{r}^{\prime}\left(t\right)} = \left\langle 8 t, - \frac{6}{t^{2}}, 0\right\rangle (for steps, see derivative calculator).

Find the unit vector: T⃗(t)=⟨4t316t6+9,−316t6+9,0⟩\mathbf{\vec{T}\left(t\right)} = \left\langle \frac{4 t^{3}}{\sqrt{16 t^{6} + 9}}, - \frac{3}{\sqrt{16 t^{6} + 9}}, 0\right\rangle (for steps, see unit vector calculator).

Answer

The unit tangent vector is T⃗(t)=⟨4t316t6+9,−316t6+9,0⟩.\mathbf{\vec{T}\left(t\right)} = \left\langle \frac{4 t^{3}}{\sqrt{16 t^{6} + 9}}, - \frac{3}{\sqrt{16 t^{6} + 9}}, 0\right\rangle.A