Determine binormal vector

Question

I keep telling myself I have to be overthinking this somewhere, but I can't see where.

Question prompt: Find the binormal vector $B(t) = T(t) \times N(t)$ at $t=0 \text{ and } t=1$. $$\text{ Tangent vector is }T(t)=\frac{r'(t)}{\mid\mid r'(t) \mid \mid}$$ $$\text{ Normal vector is } N(t) = \frac{T'(t)}{\mid\mid T'(t)\mid\mid}$$ $$r(t)= <t,2t,t^3>$$ $$r'(t) = <1,2,3t^{2}>, \mid\mid r'(t)\mid\mid= \sqrt{5+9t^{4}} $$ $$ T(t) = \frac{1}{\sqrt{5+9t^{4}}}<1,2,3t^{2}> $$ $$ T'(t) = < \frac{-18 t^{3}}{(5 + 9 t^4)^{\frac{3}{2}}}, \frac{-36t^{3}}{(5+9t^4)^{\frac{3}{2}}},\frac{30t}{(9t^{4}+5)^{\frac{3}{2}}}>$$ $$ \mid \mid T'(t) \mid\mid = \sqrt{\frac{1620 t^6}{\left(9 t^4+5\right)^3}+\frac{900 t^2}{\left(9 t^4+5\right)^3}}\ $$

$$ \text{so } N(t) = <-\frac{18 t^3}{\left(9 t^4+5\right)^{3/2} \sqrt{\frac{1620 t^6}{\left(9 t^4+5\right)^3}+\frac{900 t^2}{\left(9 t^4+5\right)^3}}} ,-\frac{36 t^3}{\left(9 t^4+5\right)^{3/2} \sqrt{\frac{1620 t^6}{\left(9 t^4+5\right)^3}+\frac{900 t^2}{\left(9 t^4+5\right)^3}}}, \frac{30 t}{\left(9 t^4+5\right)^{3/2} \sqrt{\frac{1620 t^6}{\left(9 t^4+5\right)^3}+\frac{900 t^2}{\left(9 t^4+5\right)^3}}}>$$

Problem is (as some of you might have noticed), I get a null result when I try to determine N(0).

What am I missing?

Answer 1

It would seem there is something amiss with your computation of $\|T'(t)\|$. I get $\dfrac{6\sqrt5|t|}{5+9t^4}$.

So, yes, it would seem that $T'(0)=0$, so $N(0)$ is undefined, as will be $B(0)$.