Find the unit tangent vector \( T(t) \) and find a set of parametric equations for the line tangent to the space curve at point \( P \). \[ r(t)=\langle 6 \cos t, 6 \sin t, 7\rangle, p\left(\frac{6}{\sqrt{2}}, \frac{6}{\sqrt{2}}, 7\right) \]

To find the unit tangent vector \( T(t) \) and the parametric equations for the line tangent to the space curve at point \( P \), we first need to compute the derivative of the position vector \( r(t) \). 1. **Calculate the derivative \( r'(t) \)**: \[ r(t) = \langle 6 \cos t, 6 \sin t, 7 \rangle \] \[ r'(t) = \langle -6 \sin t, 6 \cos t, 0 \rangle \] 2. **Find \( t \) at point \( P \)**: The coordinates of point \( P \) are \( \left(\frac{6}{\sqrt{2}}, \frac{6}{\sqrt{2}}, 7\right) \). - We can see that: \[ 6 \cos t = \frac{6}{\sqrt{2}} \Rightarrow \cos t = \frac{1}{\sqrt{2}} \Rightarrow t = \frac{\pi}{4} \text{ or } t = \frac{7\pi}{4} \] \[ 6 \sin t = \frac{6}{\sqrt{2}} \Rightarrow \sin t = \frac{1}{\sqrt{2}} \] - The angle \( t = \frac{\pi}{4} \) satisfies both conditions. 3. **Evaluate \( r'(t) \) at \( t = \frac{\pi}{4} \)**: \[ r'\left(\frac{\pi}{4}\right) = \langle -6 \sin\left(\frac{\pi}{4}\right), 6 \cos\left(\frac{\pi}{4}\right), 0 \rangle = \langle -6 \cdot \frac{1}{\sqrt{2}}, 6 \cdot \frac{1}{\sqrt{2}}, 0 \rangle = \langle -\frac{6}{\sqrt{2}}, \frac{6}{\sqrt{2}}, 0 \rangle \] 4. **Calculate the unit tangent vector \( T(t) \)**: The unit tangent vector is given by normalizing \( r'(t) \): \[ \| r'(\frac{\pi}{4}) \| = \sqrt{\left(-\frac{6}{\sqrt{2}}\right)^2 + \left(\frac{6}{\sqrt{2}}\right)^2 + 0^2} = \sqrt{2 \cdot \left(\frac{6^2}{2}\right)} = 6 \] Therefore, \[ T\left(\frac{\pi}{4}\right) = \frac{1}{6} \langle -\frac{6}{\sqrt{2}}, \frac{6}{\sqrt{2}}, 0 \rangle = \langle -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 \rangle \] 5. **Set up the parametric equations for the tangent line at \( P \)**: The parametric equation for the tangent line at point \( P \) is: \[ L(s) = P + s T\left(\frac{\pi}{4}\right) \] where \( P = \left(\frac{6}{\sqrt{2}}, \frac{6}{\sqrt{2}}, 7\right) \) and \( T\left(\frac{\pi}{4}\right) = \langle -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 \rangle \). Hence: \[ x(s) = \frac{6}{\sqrt{2}} - \frac{s}{\sqrt{2}}, \] \[ y(s) = \frac{6}{\sqrt{2}} + \frac{s}{\sqrt{2}}, \] \[ z(s) = 7. \] In summary, the unit tangent vector at \( P \) is \( T\left(\frac{\pi}{4}\right) = \langle -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 \rangle \), and the parametric equations for the tangent line are: \[ x(s) = \frac{6}{\sqrt{2}} - \frac{s}{\sqrt{2}}, \quad y(s)