Verify that \( t=\frac{\pi}{4} \) at the point \( P\left(\frac{6}{\sqrt{2}}, \frac{6}{\sqrt{2}}, 7\right) \). That is, evaluate \( \mathbf{r}\left(\frac{\pi}{4}\right) \). Find the unit tangent vector at \( t=\frac{\pi}{4} \) \[ \begin{aligned} \mathbf{T}(t) & =\frac{1}{6}\langle-6 \sin t, 6 \cos t, 0\rangle \\ \mathbf{T}\left(\frac{\pi}{4}\right) & =\frac{1}{6}\left(-6 \sin \left(\frac{\pi}{4}\right), 6 \cos \left(\frac{\pi}{4}\right), 0\right\rangle\end{aligned} \]

To verify that \( t=\frac{\pi}{4} \) at the point \( P\left(\frac{6}{\sqrt{2}}, \frac{6}{\sqrt{2}}, 7\right) \), we start by evaluating \( \mathbf{r}\left(\frac{\pi}{4}\right) \). Assuming \( \mathbf{r}(t) = \langle 6 \cos t, 6 \sin t, 7 \rangle \), then for \( t = \frac{\pi}{4} \): \[ \mathbf{r}\left(\frac{\pi}{4}\right) = \left(6 \cos\left(\frac{\pi}{4}\right), 6 \sin\left(\frac{\pi}{4}\right), 7\right) = \left(6 \cdot \frac{1}{\sqrt{2}}, 6 \cdot \frac{1}{\sqrt{2}}, 7\right) = \left(\frac{6}{\sqrt{2}}, \frac{6}{\sqrt{2}}, 7\right) \] Since this matches the point \( P\left(\frac{6}{\sqrt{2}}, \frac{6}{\sqrt{2}}, 7\right) \), we confirm the verification is correct. Now, we will find the unit tangent vector \( \mathbf{T}\left(t\right) \) at \( t=\frac{\pi}{4} \): Carrying out the calculation: \[ \mathbf{T}\left(\frac{\pi}{4}\right) = \frac{1}{6}\langle-6 \sin \left(\frac{\pi}{4}\right), 6 \cos \left(\frac{\pi}{4}\right), 0\rangle = \frac{1}{6}\langle-6 \cdot \frac{1}{\sqrt{2}}, 6 \cdot \frac{1}{\sqrt{2}}, 0\rangle = \langle -\sqrt{2}, \sqrt{2}, 0 \rangle \] Finally, normalizing the tangent vector gives us: \[ \mathbf{T}\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{(-\sqrt{2})^2 + (\sqrt{2})^2}} \langle -\sqrt{2}, \sqrt{2}, 0 \rangle = \frac{1}{2} \langle -\sqrt{2}, \sqrt{2}, 0 \rangle \] Thus, the unit tangent vector \( \mathbf{T}\left(\frac{\pi}{4}\right) \) is: \[ \mathbf{T}\left(\frac{\pi}{4}\right) = \langle -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 \rangle \]