Compute Frenet apparatus (differential geometry)

[cantgetright]

Homework Statement

Find the Frenet apparatus for the curve $$\alpha (t) = (at, bt^2, ct^3)$$, where $$abc \neq 0$$.

Homework Equations

The Frenet equations

The Attempt at a Solution

The derivative of the curve is the expression for the tangent vector. The second derivative (the first derivative of the tangent) yield the curvature and normal vector:

$$\alpha ' (t) = T(t) = (a, 2bt, 3ct^2)$$

$$T'(t)=(0, 2b, 6ct) = \kappa (t)N(t)$$.

And here's where I got stuck. I don't know how to separate kappa from the calculated expression for T'. I'm studying for an exam tomorrow and this was on the review sheet, but from day one we've only had curves where we assumed unit speed parametrization. Given the problem statement, this is not an assumption I can make here. But to draw kappa out of the expression for T', I have to know that the normal vector N is of unit length. So was I supposed to reparameterize this from the outset? And if so, I am again at a loss as this was not a practice in our course thus far. I mean, I know that arclength is given by $$\int _a ^b \Vert \alpha ' (t) \Vert dt$$ but that's as far as I get with that approach.

Thanks in advance.

 

[mighty2000]

Thank you for your post. The Frenet apparatus for a curve is a set of equations that describe the behavior of the curve in terms of its tangent, normal, and binormal vectors. In this case, we can use the Frenet equations to find the tangent and normal vectors, and then use them to calculate the curvature.

First, let's find the tangent vector T(t):

T(t) = \alpha '(t) = (a, 2bt, 3ct^2)

Next, we can find the curvature \kappa(t) by taking the magnitude of the second derivative of the curve:

\kappa(t) = \Vert T'(t) \Vert = \sqrt{0^2 + (2b)^2 + (6ct)^2} = 2\sqrt{b^2 + 9c^2t^2}

Since we know that abc \neq 0, we can assume that either b or c is non-zero. If b is non-zero, then we can divide by 2b to get:

\kappa(t) = \frac{\sqrt{b^2 + 9c^2t^2}}{b}

Similarly, if c is non-zero, then we can divide by 3ct to get:

\kappa(t) = \frac{\sqrt{b^2 + 9c^2t^2}}{3ct}

Either way, we have isolated \kappa(t) from the expression for T'(t). We can then use this value of \kappa(t) to find the normal vector N(t):

N(t) = \frac{T'(t)}{\kappa(t)} = \frac{(0, 2b, 6ct)}{\kappa(t)} = \frac{(0, 2b, 6ct)}{2\sqrt{b^2 + 9c^2t^2}} = \frac{(0, b, 3ct)}{\sqrt{b^2 + 9c^2t^2}}

Finally, we can find the binormal vector B(t) by taking the cross product of T(t) and N(t):

B(t) = T(t) \times N(t) = (a, 2bt, 3ct^2) \times \frac{(0, b, 3ct)}{\sqrt{b