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 \( t = \frac{\pi}{4} \) corresponds to the point \( P\left(\frac{6}{\sqrt{2}}, \frac{6}{\sqrt{2}}, 7\right) \), we need to evaluate the position vector \( \mathbf{r}\left(\frac{\pi}{4}\right) \). ### Step 1: Evaluate \( \mathbf{r}\left(\frac{\pi}{4}\right) \) Assuming the position vector \( \mathbf{r}(t) \) is given by the components derived from the unit tangent vector \( \mathbf{T}(t) \), we can express it as follows: \[ \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle \] From the unit tangent vector provided: \[ \mathbf{T}(t) = \frac{1}{6}\langle -6 \sin t, 6 \cos t, 0 \rangle = \langle -\sin t, \cos t, 0 \rangle \] We can infer that the components of \( \mathbf{r}(t) \) might be related to the sine and cosine functions. Let's assume: \[ x(t) = -6 \sin t, \quad y(t) = 6 \cos t, \quad z(t) = 7 \] Now, we will evaluate \( \mathbf{r}\left(\frac{\pi}{4}\right) \): \[ x\left(\frac{\pi}{4}\right) = -6 \sin\left(\frac{\pi}{4}\right) = -6 \cdot \frac{\sqrt{2}}{2} = -3\sqrt{2} \] \[ y\left(\frac{\pi}{4}\right) = 6 \cos\left(\frac{\pi}{4}\right) = 6 \cdot \frac{\sqrt{2}}{2} = 3\sqrt{2} \] \[ z\left(\frac{\pi}{4}\right) = 7 \] Thus, we have: \[ \mathbf{r}\left(\frac{\pi}{4}\right) = \langle -3\sqrt{2}, 3\sqrt{2}, 7 \rangle \] ### Step 2: Verify the point \( P\left(\frac{6}{\sqrt{2}}, \frac{6}{\sqrt{2}}, 7\right) \) Now, we need to check if this matches the point \( P\left(\frac{6}{\sqrt{2}}, \frac{6}{\sqrt{2}}, 7\right) \): \[ \frac{6}{\sqrt{2}} = 3\sqrt{2} \] Thus, the coordinates do not match. Therefore, \( t = \frac{\pi}{4} \) does not correspond to the point \( P\left(\frac{6}{\sqrt{2}}, \frac{6}{\sqrt{2}}, 7\right) \). ### Step 3: Find the unit tangent vector at \( t = \frac{\pi}{4} \) Now, let's find the unit tangent vector \( \mathbf{T}\left(\frac{\pi}{4}\right) \): \[ \mathbf{T}\left(\frac{\pi}{4}\right) = \frac{1}{6}\left(-6 \sin\left(\frac{\pi}{4}\right), 6 \cos\left(\frac{\pi}{4}\right), 0\right) \] Calculating the components: \[ \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, \quad \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \] Substituting these values: \[ \mathbf{T}\left(\frac{\pi}{4}\right) = \frac{1}{6}\left(-6 \cdot \frac{\sqrt{2}}{2}, 6 \cdot \frac{\sqrt{2}}{2}, 0\right) = \frac{1}{6}\left(-3\sqrt{2}, 3\sqrt{2}, 0\right) \] Thus, we have: \[ \mathbf{T}\left(\frac{\pi}{4}\right) = \left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, 0\right) \] ### Conclusion 1. The point \( P\left(\frac{6}{\sqrt{2}}, \frac{6}{\sqrt{2}}, 7\right) \) does not correspond to \( t = \frac{\pi}{4} \). 2. The unit tangent vector at \( t = \frac{\pi}{4} \) is: \[ \mathbf{T}\left(\frac{\pi}{4}\right) = \left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, 0\right) \]