Chapter 15: Problem 4
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
Expert verified
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
01
- 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.
02
- First Derivative
Calculate the first derivative of each component with respect to \(t\): \[\mathbf{r}'(t) = \left( \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \right) = (a, 2bt, 3ct^3)\]
03
- Second Derivative
Calculate the second derivative of each component with respect to \(t\): \[\mathbf{r}''(t) = \left( \frac{d^2x}{dt^2}, \frac{d^2y}{dt^2}, \frac{d^2z}{dt^2} \right) = (0, 2b, 6ct)\]
04
- Third Derivative
Calculate the third derivative of each component with respect to \(t\): \[ \mathbf{r}'''(t) = \left( \frac{d^3x}{dt^3}, \frac{d^3y}{dt^3}, \frac{d^3z}{dt^3} \right) = (0, 0, 6c) \]
05
- Tangent Vector Magnitude
Find the magnitude of the first derivative vector, \( \left| \mathbf{r}'(t) \right| \): \[\left| \mathbf{r}'(t) \right| = \sqrt{a^2 + (2bt)^2 + (3ct^2)^2} = \sqrt{a^2 + 4b^2t^2 + 9c^2t^4}\]
06
- 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, \[\mathbf{r}'(t) \times \mathbf{r}''(t) = \left( 12bct^2, - 6act, 2ab \right)\]
07
- Magnitude of Cross Product
Find the magnitude of \( \mathbf{r}'(t) \times \mathbf{r}''(t) \): \[ \left| \mathbf{r}'(t) \times \mathbf{r}''(t) \right| = \sqrt{(12bct^2)^2 + (-6act)^2 + (2ab)^2} = \sqrt{144b^2c^2t^4 + 36a^2c^2t^2 + 4a^2b^2}\]
08
- 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, \[ \kappa(t) = \frac{\sqrt{144b^2c^2t^4 + 36a^2c^2t^2 + 4a^2b^2}}{\left(a^2 + 4b^2t^2 + 9c^2t^4\right)^{3/2}} \]
09
- 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, \[ \tau(t) = \frac{3a}{6t^2c + a^2} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \( t \):
- \( \frac{dx}{dt} = a \)
- \( \frac{dy}{dt} = 2b t \)
- \( \frac{dz}{dt} = 3c t^2 \)
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: \( \mathbf{r}''(t) = (0, 2b, 6c t) \).
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:
\( \kappa(t) = \frac{\sqrt{144b^2c^2t^4 + 36a^2c^2t^2 + 4a^2b^2}}{(a^2 + 4b^2t^2 + 9c^2t^4)^{3/2}}\). This gives us the curvature at any point \( t \) on the curve.
- \( \kappa(t) = \frac{\|\mathbf{r}'(t) \times \mathbf{r}''(t)\|}{\|\mathbf{r}'(t)\|^3} \).
- \|\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}\
\( \kappa(t) = \frac{\sqrt{144b^2c^2t^4 + 36a^2c^2t^2 + 4a^2b^2}}{(a^2 + 4b^2t^2 + 9c^2t^4)^{3/2}}\). 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 \( \tau(t) = \frac{(\mathbf{r}'(t) \times \mathbf{r}''(t) ) \cdot \mathbf{r}'''(t)}{|\mathbf{r}'(t) \times \mathbf{r}''(t)|^2} \)
- \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 \