Question

Find the curvature and torsion, as functions of the parameter \(t\), for the so- called twisted cubic $$ x=a t \quad y=b t^{2} \quad z=c t^{3} \quad(a, b, c \text { constant }) $$

Short answer

Curvature: \( \kappa(t) = \frac{\sqrt{144b^2c^2t^4 + 36a^2c^2t^2 + 4a^2b^2}}{(a^2 + 4b^2t^2 + 9c^2t^4)^{3/2}} \) Torsion: \( \tau(t) = \frac{3a}{6t^2c + a^2} \)

Step-by-step solution

- Parametric Representation

The given curve is represented parametrically by \(x = a t,\, y = b t^2,\, z = c t^3 \) where \(a, b, c\) are constants.

- First Derivative

Calculate the first derivative of each component with respect to \(t\): \[\mathbf{r}'(t) = \left( \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \right) = (a, 2bt, 3ct^3)\]

- Second Derivative

Calculate the second derivative of each component with respect to \(t\): \[\mathbf{r}''(t) = \left( \frac{d^2x}{dt^2}, \frac{d^2y}{dt^2}, \frac{d^2z}{dt^2} \right) = (0, 2b, 6ct)\]

- Third Derivative

Calculate the third derivative of each component with respect to \(t\): \[ \mathbf{r}'''(t) = \left( \frac{d^3x}{dt^3}, \frac{d^3y}{dt^3}, \frac{d^3z}{dt^3} \right) = (0, 0, 6c) \]

- Tangent Vector Magnitude

Find the magnitude of the first derivative vector, \( \left| \mathbf{r}'(t) \right| \): \[\left| \mathbf{r}'(t) \right| = \sqrt{a^2 + (2bt)^2 + (3ct^2)^2} = \sqrt{a^2 + 4b^2t^2 + 9c^2t^4}\]

- Cross Product of First and Second Derivative

Calculate the cross product \( \mathbf{r}'(t) \times \mathbf{r}''(t) \): \[ \mathbf{r}'(t) \times \mathbf{r}''(t) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a & 2bt & 3ct^2 \ 0 & 2b & 6ct \end{vmatrix} \] Solving the determinant, \[\mathbf{r}'(t) \times \mathbf{r}''(t) = \left( 12bct^2, - 6act, 2ab \right)\]

- Magnitude of Cross Product

Find the magnitude of \( \mathbf{r}'(t) \times \mathbf{r}''(t) \): \[ \left| \mathbf{r}'(t) \times \mathbf{r}''(t) \right| = \sqrt{(12bct^2)^2 + (-6act)^2 + (2ab)^2} = \sqrt{144b^2c^2t^4 + 36a^2c^2t^2 + 4a^2b^2}\]

- Curvature Formula

The curvature \( \kappa(t) \) is given by: \[ \kappa(t) = \frac{\left| \mathbf{r}'(t) \times \mathbf{r}''(t) \right|}{\left| \mathbf{r}'(t) \right|^3}\] Substituting from previous steps, \[ \kappa(t) = \frac{\sqrt{144b^2c^2t^4 + 36a^2c^2t^2 + 4a^2b^2}}{\left(a^2 + 4b^2t^2 + 9c^2t^4\right)^{3/2}} \]

- Torsion Formula

The torsion \( \tau(t) \) is found using: \[ \tau(t) = \frac{\mathbf{r}'(t) \times \mathbf{r}''(t) \cdot \mathbf{r}'''(t)}{\left| \mathbf{r}'(t) \times \mathbf{r}''(t) \right|^2} \] Calculating dot product: \[ \mathbf{r}'(t) \times \mathbf{r}''(t) \cdot \mathbf{r}'''(t) = (12bct^2, -6act, 2ab) \cdot (0, 0, 6c) = 12abc\] Therefore, \[ \tau(t) = \frac{12abc}{144b^2c^2t^4 + 36a^2c^2t^2 + 4a^2b^2}\] Simplifying, \[ \tau(t) = \frac{3a}{6t^2c + a^2} \]

Key concepts

Parametric Representation

The parametric representation of a curve is a way of defining the curve using parameters. In this case, the twisted cubic curve is given by three parametric equations involving the parameter \( t \). These equations are:
\( x = a t \),
\( y = b t^2 \), and
\( z = c t^3 \).
Here, \( a \), \( b \), and \( c \) are constants. Each equation gives the position of a point on the curve as a function of \( t \). By adjusting \( t \), you can trace out the entire curve. This method is very useful for more complex curves where a single equation might be difficult to use.

First Derivative

The first derivative of a curve provides the tangent vector at any point, which shows the direction in which the curve is heading at that point. For our curve, we calculate the first derivatives of each component with respect to \( t \):

• \( \frac{dx}{dt} = a \)
• \( \frac{dy}{dt} = 2b t \)
• \( \frac{dz}{dt} = 3c t^2 \)

So the tangent vector \( \mathbf{r}'(t) \) is \((a, 2bt, 3ct^2)\). This vector plays a key role in determining the properties of the curve, including its curvature and torsion.

Cross Product

The cross product of vectors is a way to find a vector perpendicular to both. For our curve, we calculate the cross product of the first and second derivatives:

• First derivative: \( \mathbf{r}'(t) = (a, 2b t, 3c t^2) \).
• Second derivative: \( \mathbf{r}''(t) = (0, 2b, 6c t) \).

The cross product \( \mathbf{r}'(t) \times \mathbf{r}''(t) \) is
\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \ a & 2b t & 3c t^2 \ 0 & 2b & 6c t \ \end{vmatrix}, which simplifies to \( (12bct^2, -6act, 2ab) \). The resulting vector is orthogonal to both \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \). This vector is crucial for finding curvature.

Curvature Formula

Curvature measures how quickly a curve changes direction. The formula for curvature \( \kappa(t) \) involves the magnitude of the cross product of the first and second derivatives:

• \( \kappa(t) = \frac{\|\mathbf{r}'(t) \times \mathbf{r}''(t)\|}{\|\mathbf{r}'(t)\|^3} \).

For our curve:

• \|\mathbf{r}'(t)\| = \sqrt{a^2 + 4b^2t^2 + 9c^2t^4}\
• \ \|\mathbf{r}'(t) \times \mathbf{r}''(t)\| = \sqrt{144b^2c^2t^4 + 36a^2c^2t^2 + 4a^2b^2}\

Substituting these into the curvature formula, we get:
\( \kappa(t) = \frac{\sqrt{144b^2c^2t^4 + 36a^2c^2t^2 + 4a^2b^2}}{(a^2 + 4b^2t^2 + 9c^2t^4)^{3/2}}\). This gives us the curvature at any point \( t \) on the curve.

Torsion Formula

Torsion measures how much a curve twists out of its plane. The torsion formula uses the cross product of the first and second derivatives, and the third derivative:

• Torsion \( \tau(t) = \frac{(\mathbf{r}'(t) \times \mathbf{r}''(t) ) \cdot \mathbf{r}'''(t)}{|\mathbf{r}'(t) \times \mathbf{r}''(t)|^2} \)

For our curve, we have:

• \mathbf{r}'''(t) = (0, 0, 6c) \
• \mathbf{r}'(t) \times \mathbf{r}''(t) \cdot \mathbf{r}'''(t) = (12bc t^2, -6ac t, 2ab) \cdot (0, 0, 6c) = 12abc
• \|\mathbf{r}'(t) \times \mathbf{r}''(t)\|^2 = 144b^2c^2t^4 + 36a^2c^2t^2 + 4a^2b^2 \

Substituting these into the torsion formula, we get:
\( \tau(t) = \frac{12abc}{144b^2c^2t^4 + 36a^2c^2t^2 + 4a^2b^2}\). Simplifying, we obtain: \( \tau(t) = \frac{3a}{6t^2c + a^2} \). This provides the torsion at any point \( t \) on the curve.