Answer:
To find the unit binormal vector and torsion of the curve given by the parametric equations \(r(t) = \langle \frac{t^2}{2}, -4t + 2, -8 \rangle\), we need to follow a series of steps. The unit binormal vector can be found using the formula for the binormal vector \(\mathbf{B}(t) = \frac{\mathbf{T}'(t)}{\|\mathbf{T}'(t)\|}\), and the torsion can be calculated as \(\tau = \frac{\|\mathbf{T}'(t)\|}{\|\mathbf{r}'(t)\|}\).
1. Find the derivative of \(r(t)\):
\(\mathbf{r}'(t) = \langle t, -4, 0 \rangle\)
2. Find the second derivative of \(r(t)\):
\(\mathbf{r}''(t) = \langle 1, 0, 0 \rangle\)
3. Calculate the unit tangent vector \(\mathbf{T}(t)\):
\(\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|} = \frac{\langle t, -4, 0 \rangle}{\sqrt{t^2 + 16}} = \langle \frac{t}{\sqrt{t^2 + 16}}, \frac{-4}{\sqrt{t^2 + 16}}, 0 \rangle\)
4. Find the derivative of \(\mathbf{T}(t)\):
\(\mathbf{T}'(t) = \langle \frac{1}{\sqrt{t^2 + 16}} - \frac{t^2}{(t^2 + 16)^\frac{3}{2}}, \frac{4t}{(t^2 + 16)^\frac{3}{2}}, 0 \rangle\)
5. Calculate the magnitude of \(\mathbf{T}'(t)\):
\(\|\mathbf{T}'(t)\| = \sqrt{\left(\frac{1}{\sqrt{t^2 + 16}} - \frac{t^2}{(t^2 + 16)^\frac{3}{2}}\right)^2 + \left(\frac{4t}{(t^2 + 16)^\frac{3}{2}}\right)^2}\)
Now, let's evaluate the options:
A. \(\mathbf{B}(t) = \langle 0, 0, t^2 + 16 \rangle\), \(\tau = 0\) - This is not correct because the torsion should not be zero.
B. \(\mathbf{B}(t) = \langle 0, 0, 1 \rangle\), \(\tau = 1\) - This is not correct because the unit binormal vector \(\mathbf{B}(t)\) should not be a constant vector.
C. \(\mathbf{B}(t) = \langle 0, 0, -1 \rangle\), \(\tau = 1\) - This is also not correct because the unit binormal vector should have the same direction as the derivative of the unit tangent vector \(\mathbf{T}'(t)\), which is not \(\langle 0, 0, -1 \rangle\).
D. \(\mathbf{B}(t) = \langle 0, 0, 1 \rangle\), \(\tau = 0\) - This is correct. The unit binormal vector is \(\langle 0, 0, 1 \rangle\), and the torsion \(\tau\) is indeed zero for this curve.
So, the correct answer is option D: \(\mathbf{B}(t) = \langle 0, 0, 1 \rangle\) and \(\tau = 0\).
Explanation: