Question

Unit tangent vectors: Find the unit tangent vector for the given function at the specified value of t.

$\mathbf{r}\left(t\right)=⟨t,2t\sin t,2t\cos t⟩,t=\pi$

Short answer

Ans: Thus the unit tangent vector to $⟨t,2t\sin t,2t\cos t⟩$at $t=\pi$islocalid="1649674946140"

Step-by-step solution

Step 1. Given information:

$\mathbf{r}\left(t\right)=⟨t,2t\sin t,2t\cos t⟩,t=\pi$

Step 2. Simplifying the Unit tangent vectors :

Consider $r\left(t\right)=⟨t,2t\sin t,2t\cos t⟩$

First, we compute $r^{\text{'}}\left(t\right)$

$$\begin{gathered} r^{\text{'}}\left(t\right)=\frac{d}{dt}⟨t,2t\sin t,2t\cos t⟩ \\ =⟨1,2(t\cos t+\sin t),2(-t\sin t+\cos t)⟩ \\ ||r^{\text{'}}\left(t\right)||=\sqrt{(1{)}^2+(2(t\cos t+\sin t){)}^2+\left(2\right(-t\sin t+\cos t){)}^2} \\ =\sqrt{1+4\left(t^2{\cos }^{2}t+{\sin }^{2}t+2t\sin t\cos t+t^2{\sin }^{2}t+{\cos }^{2}t-2t\sin t\cos t\right)} \\ =\sqrt{1+4\left(t^2+1\right)} \\ =\sqrt{4t^2+5} \end{gathered}$$

Step 3. Finding the Unit tangent vectors: 

The unit tangent vector to $r\left(t\right)$is

$$\begin{gathered} T\left(t\right)=\frac{r^{\text{'}}\left(t\right)}{||r^{\text{'}}\left(t\right)||} \\ =\frac{⟨1,2(t\cos t+\sin t),2(-t\sin t+\cos t)⟩}{\sqrt{4t^2+5}} \end{gathered}$$

At $t=\pi$, the unit tangent vector to $r\left(t\right)$is

localid="1649674786888" $$\begin{gathered} T\left(\pi \right)=\frac{⟨1,2(\pi \cos \pi +\sin \pi ),2(-\pi \sin \pi +\cos \pi )⟩}{\sqrt{4(\pi {)}^2+5}} \\ =\frac{⟨1,-2\pi ,-2⟩}{\sqrt{4{\pi }^2+5}} \end{gathered}$$

Thus the unit tangent vector to $⟨t,2t\sin t,2t\cos t⟩$at $t=\pi$is