# Frenet-Serret equations

1. Dec 26, 2008

### bigevil

1. The problem statement, all variables and given/known data
For the twisted curve $$y^3 + 27 axz - 81a^{2}y = 0$$, given parametrically by

$$x=au(3-u^2)$$, $$y=3au^2$$, $$z=au(3+u^2)$$

show that the following hold

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

the length of curve from the origin to (2a, 3a, 4a) is $$4\sqrt{2}a$$

the radius of curvature at the point with parameter u is $$3a(1+u^2)^2$$

3. The attempt at a solution

The first two parts are relatively painless. First part, evaluate $$\frac{d\bold{r}}{du}\cdot\frac{d\bold{r}}{du}$$. 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 $$\hat{\bold{t}} = \frac{dr}{ds} = \frac{dr}{du} \cdot \frac{du}{ds}$$, where du/ds can be gotten from above.

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

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

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

z component is zero.

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

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

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

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...

2. Dec 26, 2008

### Dick

Check your unit normal t-hat. I've got the sqrt(2) in the denominator. Not the numerator.