Question

Prove that the tangent vector is always orthogonal to the position vector for the vector-valued function.

Short answer

Ans:

$$\begin{gathered} \mathit{r}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{·}{\mathit{r}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\left(i\sin t+j\sin t\cos t+k{\cos }^{2}t\right)\mathbf{·}\mathbf{(}\mathit{i}\mathbf{cos}\mathit{t}\mathbf{+}\mathit{j}\mathbf{cos}\mathbf{2}\mathit{t}\mathbf{-}\mathit{k}\mathbf{sin}\mathbf{2}\mathit{t}\mathbf{)} \\ \mathit{r}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{·}{\mathit{r}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{0} \end{gathered}$$

Step-by-step solution

Step 1. Given information: 

Step 2. Proving that the tangent vector is always orthogonal to the position vector for the vector-valued function:

Differentiate with respect to t

$r^{\text{'}}\left(t\right)=i\cos t+j\cos 2t-k\sin 2t$

Find $r\left(t\right)·r^{\text{'}}\left(t\right)$

$$\begin{gathered} r\left(t\right)·r^{\text{'}}\left(t\right)=\left(i\sin t+j\sin t\cos t+k{\cos }^{2}t\right)·(i\cos t+j\cos 2t-k\sin 2t) \\ r\left(t\right)·r^{\text{'}}\left(t\right)=0 \end{gathered}$$

Thus, the tangent vector is always orthogonal to the position vector for the vector-valued function.