Calculating Frenet-Serret Equations for a Twisted Curve

  • Thread starter bigevil
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In summary, the twisted curve y^3 + 27 axz - 81a^{2}y = 0 has parametric equations x=au(3-u^2), y=3au^2, z=au(3+u^2). The distance along the curve from the origin is given by \frac{ds}{du} = 3 \sqrt{2} a(1+u^2), and the length of the curve from the origin to (2a, 3a, 4a) is 4\sqrt{2}a. The radius of curvature at a point with parameter u is 3a(1+u^2)^2, with a correction to the
  • #1
bigevil
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Homework Statement


For the twisted curve [tex]y^3 + 27 axz - 81a^{2}y = 0[/tex], given parametrically by

[tex]x=au(3-u^2)[/tex], [tex]y=3au^2[/tex], [tex]z=au(3+u^2)[/tex]

show that the following hold

[tex]\frac{ds}{du} = 3 \sqrt{2} a(1+u^2)[/tex], where s is the distance along the curve from the origin

the length of curve from the origin to (2a, 3a, 4a) is [tex]4\sqrt{2}a[/tex]

the radius of curvature at the point with parameter u is [tex]3a(1+u^2)^2[/tex]

The Attempt at a Solution



The first two parts are relatively painless. First part, evaluate [tex]\frac{d\bold{r}}{du}\cdot\frac{d\bold{r}}{du} [/tex]. Second part is to integrate its square root between s=1 and s=0.

The last part is a bit troublesome though. First, I get [tex]\hat{\bold{t}} = \frac{dr}{ds} = \frac{dr}{du} \cdot \frac{du}{ds}[/tex], where du/ds can be gotten from above.

then [tex]\hat{t} = \sqrt{2}(1+u^2)^{-1} [(1-u^2)\bold{i} + 2u\bold{j} + (1+u^2)\bold{k}][/tex].

Then evaluating, x component of dt/du: [tex]-4\sqrt{2} \frac{u}{(1+u^2)^2}[/tex]

y component of dt/du: [tex]2\sqrt{2} \frac{1 - u^2}{(1+u^2)^2}[/tex]

z component is zero.

Evaluate dt/ds = dt/du (dot) du/ds,

[tex]\frac{d\hat{t}}{ds}= \frac{-4u}{3a(1+u^2)^3} \bold{i} + \frac{2(1-u^2)}{3a(1+u^2)^3} \bold{j}[/tex]

Then taking magnitudes and applying the inverse, [tex]\rho = \frac{3}{2}a(1+u^2)^2[/tex]

Which is pretty weird! The answer does not have the 1/2 coefficient. I'm not sure where I went wrong here. Guidance will be gratefully accepted...
 
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  • #2
Check your unit normal t-hat. I've got the sqrt(2) in the denominator. Not the numerator.
 

1. What are the Frenet-Serret equations?

The Frenet-Serret equations are a set of differential equations that describe the curvature and torsion of a curve in three-dimensional space. They are used to calculate the rate of change of the unit tangent, normal, and binormal vectors along the curve.

2. What is the significance of the Frenet-Serret equations?

The Frenet-Serret equations are important in the fields of differential geometry and physics, as they provide a mathematical framework for understanding the behavior of curves in three-dimensional space. They are particularly useful in studying the movement of objects in space, such as particles and rotating bodies.

3. How are the Frenet-Serret equations derived?

The Frenet-Serret equations are derived from the fundamental theorem of curves, which states that any curve in three-dimensional space can be uniquely defined by its curvature and torsion functions. These functions can be calculated using the Frenet-Serret equations.

4. What are the applications of the Frenet-Serret equations?

The Frenet-Serret equations have various applications in mathematics, physics, and engineering. They are used to study the motion of objects in space, analyze the behavior of particles in magnetic fields, and solve problems in vector calculus and differential equations.

5. Are there any limitations to the Frenet-Serret equations?

While the Frenet-Serret equations are powerful tools in studying curves in three-dimensional space, they have some limitations. They only apply to smooth curves and cannot be used to analyze curves with sharp corners or discontinuities. Additionally, they do not take into account external forces acting on the curve, which may affect its behavior.

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