Calculating Frenet-Serret Equations for a Twisted Curve

[bigevil]

Homework Statement

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

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...

 

[Dick]

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