Question

COMPUTE THE TORSION OF γ(t) = (1/3 〖(1+t)〗^(3/2),1/3 〖(1-t)〗^(3/2),t/√2 )

Answer 1

COMPUTE THE TORSION OF $\gamma(t) = \left(\frac{1}{3}(1 + t)^{3/2}, \frac{1}{3}(1 - t)^{3/2}, t / \sqrt{2}\right)$

Answer:

$$\frac {1}{4} \frac {\sqrt {2}}{\sqrt {1 - t} \sqrt {1 + t}}$$

Formula for torsion:

$$\frac {(\gamma^ {\prime} , \gamma^ {\prime \prime} , \gamma^ {\prime \prime \prime})}{| [ \gamma^ {\prime} , \gamma^ {\prime \prime} ] | ^ {2}}$$

$a * b$ -dot product. $[a, b]$ -vector product. $|a|$ -vector length. $(a, b, c) := a * [b, c]$ -triple product.

Triple product can be computed as a determinant of a matrix:

$$(a, b, c) = \det \left[ \begin{array}{l l l} a _ {1} & a _ {2} & a _ {3} \\ b _ {1} & b _ {2} & b _ {3} \\ c _ {1} & c _ {2} & c _ {3} \end{array} \right]$$

It is also convenient to remember that triple product is invariant under a circular shift of its three operands (a, b, c):

$$(a, b, c) = a * [ b, c ] = c * [ a, b ]$$

Tangent vectors:

$$\gamma^ {\prime} = (\frac {1}{2} \sqrt {1 + t}, - \frac {1}{2} \sqrt {(1 - t)}, \frac {1}{2} \sqrt {2})$$ $$\gamma^ {\prime \prime} = (\frac {1}{4 \sqrt {1 + t}}, \frac {1}{4 \sqrt {1 - t}}, 0)$$ $$\gamma^ {\prime \prime \prime} = (- \frac {1}{8 (1 + t) ^ {3 / 2}}, \frac {1}{8 (1 - t) ^ {3 / 2}}, 0)$$

Let's first compute vector product:

$$[ \gamma^ {\prime}, \gamma^ {\prime \prime} ] = \det \left[ \begin{array}{c c c} i & j & k \\ \frac {1}{2} \sqrt {1 + t} & - \frac {1}{2} \sqrt {1 - t} & \frac {1}{2} \sqrt {2} \\ \frac {1}{4 \sqrt {1 + t}} & \frac {1}{4 \sqrt {1 - t}} & 0 \end{array} \right] = \frac {j \sqrt {2} \sqrt {1 - t} - i \sqrt {2} \sqrt {1 + t} + k 2}{8 \sqrt {1 - t} \sqrt {1 + t}}$$

The squared length of the vector:

$$| [ \gamma^ {\prime}, \gamma^ {\prime \prime} ] | ^ {2} = \frac {1}{3 2 | 1 - t |} + \frac {1}{3 2 | 1 + t |} + \frac {1}{1 6 | 1 + t | | 1 - t |} =$$

And since there exist natural bounds on t: $-1 < t < 1$:

$$= - \frac {1}{8 (t ^ {2} - 1)}$$

Now triple product is:

$$\left(\gamma^ {\prime}, \gamma^ {\prime \prime}, \gamma^ {\prime \prime \prime}\right) = \left[ \gamma^ {\prime}, \gamma^ {\prime \prime} \right] * \gamma^ {\prime \prime \prime} =$$ $$(\frac {1}{8} \frac {- \sqrt {2}}{\sqrt {1 - t}}, \frac {1}{8} \frac {\sqrt {2}}{\sqrt {1 + t}}, \frac {1}{4} \frac {1}{\sqrt {1 - t} \sqrt {1 + t}}) * (- \frac {1}{8 (1 + t) ^ {3 / 2}}, \frac {1}{8 (1 - t) ^ {3 / 2}}, 0) = \frac {1}{3 2} \frac {\sqrt {2}}{(1 + t) ^ {3 / 2} (1 - t) ^ {3 / 2}}$$

And finally the torsion is:

$$\frac {1}{3 2} \frac {\sqrt {2}}{(1 + t) ^ {3 / 2} (1 - t) ^ {3 / 2}}: - \frac {1}{8 (t ^ {2} - 1)} = \frac {1}{4} \frac {\sqrt {2}}{\sqrt {1 - t} \sqrt {1 + t}}$$